3.7.3 \(\int \frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [603]

Optimal. Leaf size=385 \[ \frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

-1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/2*(a+b)*(a^2-4*a*b+b^2)
*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1
/2)+tan(d*x+c))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2
+b^2)^3/d*2^(1/2)+1/4*(3*a^4-26*a^2*b^2+3*b^4)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/(a^2+b^2)^3/d/a^(1/2)/
b^(1/2)+1/2*a*tan(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+1/4*(3*a^2-5*b^2)*tan(d*x+c)^(1/2)/(a^2+b^2)^2/d
/(a+b*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.54, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3648, 3730, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} \frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {a \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {b} d \left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(3/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a + b)*(a^2
- 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + ((3*a^4 - 26*a^2*b^2 + 3*b
^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*Sqrt[a]*Sqrt[b]*(a^2 + b^2)^3*d) + ((a - b)*(a^2 + 4*a*b
+ b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2 + 4*a*b
+ b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) + (a*Sqrt[Tan[c + d*x]
])/(2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + ((3*a^2 - 5*b^2)*Sqrt[Tan[c + d*x]])/(4*(a^2 + b^2)^2*d*(a + b*T
an[c + d*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d
^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\int \frac {\frac {a}{2}-2 b \tan (c+d x)-\frac {3}{2} a \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 \left (a^2+b^2\right )}\\ &=\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {\frac {1}{4} a \left (5 a^2-3 b^2\right )-4 a^2 b \tan (c+d x)-\frac {1}{4} a \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a \left (a^2+b^2\right )^2}\\ &=\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {2 a^2 \left (a^2-3 b^2\right )-2 a b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a \left (a^2+b^2\right )^3}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 \left (a^2+b^2\right )^3}\\ &=\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {2 a^2 \left (a^2-3 b^2\right )-2 a b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}\\ &=\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 \left (a^2+b^2\right )^3 d}\\ &=\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}+\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}\\ &=\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}\\ &=\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4-26 a^2 b^2+3 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {b} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {a \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 3.78, size = 326, normalized size = 0.85 \begin {gather*} \frac {b^{9/2} \tan ^{\frac {5}{2}}(c+d x)-b^{7/2} \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))+\frac {(a+b \tan (c+d x)) \left (a^{5/2} b^{5/2} \left (a^2+b^2\right )^2 \sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^{5/2} b^{5/2} \left (3 a^2-5 b^2\right ) \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}+\left (4 \sqrt [4]{-1} a^{5/2} (a+i b)^3 b^{5/2} \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+a^2 b^2 \left (3 a^4-26 a^2 b^2+3 b^4\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+4 \sqrt [4]{-1} a^{5/2} (a-i b)^3 b^{5/2} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right ) (a+b \tan (c+d x))\right )\right )}{a^{3/2} \left (a^2+b^2\right )^2}}{2 a b^{5/2} \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(3/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

(b^(9/2)*Tan[c + d*x]^(5/2) - b^(7/2)*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]) + ((a + b*Tan[c + d*x])*(a^(5/2)
*b^(5/2)*(a^2 + b^2)^2*Sqrt[Tan[c + d*x]] + (a^(5/2)*b^(5/2)*(3*a^2 - 5*b^2)*(a^2 + b^2)*Sqrt[Tan[c + d*x]] +
(4*(-1)^(1/4)*a^(5/2)*(a + I*b)^3*b^(5/2)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + a^2*b^2*(3*a^4 - 26*a^2*b^2
+ 3*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + 4*(-1)^(1/4)*a^(5/2)*(a - I*b)^3*b^(5/2)*ArcTanh[(-1)^
(3/4)*Sqrt[Tan[c + d*x]]])*(a + b*Tan[c + d*x]))/2))/(a^(3/2)*(a^2 + b^2)^2))/(2*a*b^(5/2)*(a^2 + b^2)*d*(a +
b*Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 339, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (\frac {3}{8} a^{4} b -\frac {1}{4} a^{2} b^{3}-\frac {5}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\frac {a \left (5 a^{4}+2 a^{2} b^{2}-3 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (3 a^{4}-26 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (-a^{3}+3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(339\)
default	\(\frac {\frac {\frac {2 \left (\left (\frac {3}{8} a^{4} b -\frac {1}{4} a^{2} b^{3}-\frac {5}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\frac {a \left (5 a^{4}+2 a^{2} b^{2}-3 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (3 a^{4}-26 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (-a^{3}+3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/(a^2+b^2)^3*(((3/8*a^4*b-1/4*a^2*b^3-5/8*b^5)*tan(d*x+c)^(3/2)+1/8*a*(5*a^4+2*a^2*b^2-3*b^4)*tan(d*x+c)
^(1/2))/(a+b*tan(d*x+c))^2+1/8*(3*a^4-26*a^2*b^2+3*b^4)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))+2/
(a^2+b^2)^3*(1/8*(-a^3+3*a*b^2)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2
)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*(3*a^2*b-b^3)*2
^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)
*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 402, normalized size = 1.04 \begin {gather*} \frac {\frac {{\left (3 \, a^{4} - 26 \, a^{2} b^{2} + 3 \, b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (3 \, a^{2} b - 5 \, b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((3*a^4 - 26*a^2*b^2 + 3*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*
sqrt(a*b)) - (2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) +
2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*(a^
3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(a^3 + 3*a^2*b - 3*a
*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + ((3*a^2
*b - 5*b^3)*tan(d*x + c)^(3/2) + (5*a^3 - 3*a*b^2)*sqrt(tan(d*x + c)))/(a^6 + 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 +
2*a^2*b^4 + b^6)*tan(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*tan(d*x + c)))/d

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 11404 vs. \(2 (337) = 674\).
time = 12.89, size = 22816, normalized size = 59.26 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/16*(16*sqrt(2)*((a^23*b + 3*a^21*b^3 - 17*a^19*b^5 - 123*a^17*b^7 - 342*a^15*b^9 - 546*a^13*b^11 - 546*a^11
*b^13 - 342*a^9*b^15 - 123*a^7*b^17 - 17*a^5*b^19 + 3*a^3*b^21 + a*b^23)*d^5*cos(d*x + c)^4 + 2*(3*a^21*b^3 +
26*a^19*b^5 + 99*a^17*b^7 + 216*a^15*b^9 + 294*a^13*b^11 + 252*a^11*b^13 + 126*a^9*b^15 + 24*a^7*b^17 - 9*a^5*
b^19 - 6*a^3*b^21 - a*b^23)*d^5*cos(d*x + c)^2 + (a^19*b^5 + 9*a^17*b^7 + 36*a^15*b^9 + 84*a^13*b^11 + 126*a^1
1*b^13 + 126*a^9*b^15 + 84*a^7*b^17 + 36*a^5*b^19 + 9*a^3*b^21 + a*b^23)*d^5 + 4*((a^22*b^2 + 8*a^20*b^4 + 27*
a^18*b^6 + 48*a^16*b^8 + 42*a^14*b^10 - 42*a^10*b^14 - 48*a^8*b^16 - 27*a^6*b^18 - 8*a^4*b^20 - a^2*b^22)*d^5*
cos(d*x + c)^3 + (a^20*b^4 + 9*a^18*b^6 + 36*a^16*b^8 + 84*a^14*b^10 + 126*a^12*b^12 + 126*a^10*b^14 + 84*a^8*
b^16 + 36*a^6*b^18 + 9*a^4*b^20 + a^2*b^22)*d^5*cos(d*x + c))*sin(d*x + c))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b
^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^17*b + 8*a^15*b^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*
a^9*b^9 - 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*
a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8
- 30*a^2*b^10 + b^12))*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^1
2)/((a^24 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10
*b^14 + 495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*(1/((a^12 + 6*a^10*b^2 + 15*a^8*
b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(3/4)*arctan(-((a^24 - 6*a^22*b^2 - 84*a^20*b^4 - 322
*a^18*b^6 - 603*a^16*b^8 - 540*a^14*b^10 + 540*a^10*b^14 + 603*a^8*b^16 + 322*a^6*b^18 + 84*a^4*b^20 + 6*a^2*b
^22 - b^24)*d^4*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/((a^2
4 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10*b^14 +
495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4
+ 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + sqrt(2)*((a^27 + 9*a^25*b^2 + 30*a^23*b^4 + 22*a^21*b^6
- 165*a^19*b^8 - 693*a^17*b^10 - 1452*a^15*b^12 - 1980*a^13*b^14 - 1881*a^11*b^16 - 1265*a^9*b^18 - 594*a^7*b
^20 - 186*a^5*b^22 - 35*a^3*b^24 - 3*a*b^26)*d^7*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^
4*b^8 - 30*a^2*b^10 + b^12)/((a^24 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 +
924*a^12*b^12 + 792*a^10*b^14 + 495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4))*sqrt(1/
((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + (3*a^20*b + 26*a^18*b^
3 + 99*a^16*b^5 + 216*a^14*b^7 + 294*a^12*b^9 + 252*a^10*b^11 + 126*a^8*b^13 + 24*a^6*b^15 - 9*a^4*b^17 - 6*a^
2*b^19 - b^21)*d^5*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/((
a^24 + 12*a^22*b^2 + 66*a^20*b^4 + 220*a^18*b^6 + 495*a^16*b^8 + 792*a^14*b^10 + 924*a^12*b^12 + 792*a^10*b^14
+ 495*a^8*b^16 + 220*a^6*b^18 + 66*a^4*b^20 + 12*a^2*b^22 + b^24)*d^4)))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4
+ 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^17*b + 8*a^15*b^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*a^
9*b^9 - 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^
6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 -
30*a^2*b^10 + b^12))*sqrt(((a^18 - 27*a^16*b^2 + 168*a^14*b^4 + 224*a^12*b^6 - 366*a^10*b^8 - 366*a^8*b^10 +
224*a^6*b^12 + 168*a^4*b^14 - 27*a^2*b^16 + b^18)*d^2*sqrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 1
5*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))*cos(d*x + c) + sqrt(2)*((3*a^20*b - 82*a^18*b^3 + 531*a^16*b^5 + 504*a^14
*b^7 - 1322*a^12*b^9 - 732*a^10*b^11 + 1038*a^8*b^13 + 280*a^6*b^15 - 249*a^4*b^17 + 30*a^2*b^19 - b^21)*d^3*s
qrt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))*cos(d*x + c) + (a^
15 - 33*a^13*b^2 + 345*a^11*b^4 - 1217*a^9*b^6 + 1611*a^7*b^8 - 795*a^5*b^10 + 91*a^3*b^12 - 3*a*b^14)*d*cos(d
*x + c))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^17*b + 8*
a^15*b^3 - 12*a^13*b^5 - 72*a^11*b^7 - 110*a^9*b^9 - 72*a^7*b^11 - 12*a^5*b^13 + 8*a^3*b^15 + 3*a*b^17)*d^2*sq
rt(1/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/(a^12 - 30*a^10*b^
2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12))*sqrt(sin(d*x + c)/cos(d*x + c))*(1/((a^12 +
6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(1/4) + (a^12 - 30*a^10*b^2 + 25
5*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)*sin(d*x + c))/cos(d*x + c))*(1/((a^12 + 6*a^10*b^2
+ 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4))^(3/4) + sqrt(2)*((a^33 - 6*a^31*b^2 - 90*a^
29*b^4 - 294*a^27*b^6 - 54*a^25*b^8 + 2082*a^23...

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(3/2)/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)**(3/2)/(a + b*tan(c + d*x))**3, x)

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 9.16, size = 2500, normalized size = 6.49 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(3/2)/(a + b*tan(c + d*x))^3,x)

[Out]

atan((((((518*a*b^15*d^2 - 18*a^15*b*d^2 - 4494*a^3*b^13*d^2 + 3022*a^5*b^11*d^2 + 17194*a^7*b^9*d^2 + 5298*a^
9*b^7*d^2 - 3338*a^11*b^5*d^2 + 506*a^13*b^3*d^2)/(2*(a^16*d^5 + b^16*d^5 + 8*a^2*b^14*d^5 + 28*a^4*b^12*d^5 +
56*a^6*b^10*d^5 + 70*a^8*b^8*d^5 + 56*a^10*b^6*d^5 + 28*a^12*b^4*d^5 + 8*a^14*b^2*d^5)) + (((((4224*a^4*b^18*
d^4 - 320*a^2*b^20*d^4 - 192*b^22*d^4 + 22272*a^6*b^16*d^4 + 51072*a^8*b^14*d^4 + 67200*a^10*b^12*d^4 + 53760*
a^12*b^10*d^4 + 25344*a^14*b^8*d^4 + 5952*a^16*b^6*d^4 + 192*a^18*b^4*d^4 - 128*a^20*b^2*d^4)/(2*(a^16*d^5 + b
^16*d^5 + 8*a^2*b^14*d^5 + 28*a^4*b^12*d^5 + 56*a^6*b^10*d^5 + 70*a^8*b^8*d^5 + 56*a^10*b^6*d^5 + 28*a^12*b^4*
d^5 + 8*a^14*b^2*d^5)) - (tan(c + d*x)^(1/2)*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4
*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2)*(512*b^25*d^4 + 4608*a^2*b^23*d^4 + 17920*a^4*b^21*d^4 + 3
8400*a^6*b^19*d^4 + 46080*a^8*b^17*d^4 + 21504*a^10*b^15*d^4 - 21504*a^12*b^13*d^4 - 46080*a^14*b^11*d^4 - 384
00*a^16*b^9*d^4 - 17920*a^18*b^7*d^4 - 4608*a^20*b^5*d^4 - 512*a^22*b^3*d^4))/(4*(a^16*d^4 + b^16*d^4 + 8*a^2*
b^14*d^4 + 28*a^4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28*a^12*b^4*d^4 + 8*a^14*b^2
*d^4)))*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d
^2*15i))^(1/2))/2 + (tan(c + d*x)^(1/2)*(1544*a*b^18*d^2 + 64*a^3*b^16*d^2 - 7456*a^5*b^14*d^2 - 576*a^7*b^12*
d^2 + 19504*a^9*b^10*d^2 + 18880*a^11*b^8*d^2 + 3808*a^13*b^6*d^2 - 960*a^15*b^4*d^2 + 8*a^17*b^2*d^2))/(2*(a^
16*d^4 + b^16*d^4 + 8*a^2*b^14*d^4 + 28*a^4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28
*a^12*b^4*d^4 + 8*a^14*b^2*d^4)))*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i -
20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2))/2)*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4
*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2))/2 - (tan(c + d*x)^(1/2)*(9*a^12*b + 41*b^13 - 82*a^2*b^11
+ 1831*a^4*b^9 - 4348*a^6*b^7 + 1671*a^8*b^5 - 210*a^10*b^3))/(2*(a^16*d^4 + b^16*d^4 + 8*a^2*b^14*d^4 + 28*a
^4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28*a^12*b^4*d^4 + 8*a^14*b^2*d^4)))*(1/(b^6
*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2)*
1i - ((((518*a*b^15*d^2 - 18*a^15*b*d^2 - 4494*a^3*b^13*d^2 + 3022*a^5*b^11*d^2 + 17194*a^7*b^9*d^2 + 5298*a^9
*b^7*d^2 - 3338*a^11*b^5*d^2 + 506*a^13*b^3*d^2)/(2*(a^16*d^5 + b^16*d^5 + 8*a^2*b^14*d^5 + 28*a^4*b^12*d^5 +
56*a^6*b^10*d^5 + 70*a^8*b^8*d^5 + 56*a^10*b^6*d^5 + 28*a^12*b^4*d^5 + 8*a^14*b^2*d^5)) + (((((4224*a^4*b^18*d
^4 - 320*a^2*b^20*d^4 - 192*b^22*d^4 + 22272*a^6*b^16*d^4 + 51072*a^8*b^14*d^4 + 67200*a^10*b^12*d^4 + 53760*a
^12*b^10*d^4 + 25344*a^14*b^8*d^4 + 5952*a^16*b^6*d^4 + 192*a^18*b^4*d^4 - 128*a^20*b^2*d^4)/(2*(a^16*d^5 + b^
16*d^5 + 8*a^2*b^14*d^5 + 28*a^4*b^12*d^5 + 56*a^6*b^10*d^5 + 70*a^8*b^8*d^5 + 56*a^10*b^6*d^5 + 28*a^12*b^4*d
^5 + 8*a^14*b^2*d^5)) + (tan(c + d*x)^(1/2)*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*
d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2)*(512*b^25*d^4 + 4608*a^2*b^23*d^4 + 17920*a^4*b^21*d^4 + 38
400*a^6*b^19*d^4 + 46080*a^8*b^17*d^4 + 21504*a^10*b^15*d^4 - 21504*a^12*b^13*d^4 - 46080*a^14*b^11*d^4 - 3840
0*a^16*b^9*d^4 - 17920*a^18*b^7*d^4 - 4608*a^20*b^5*d^4 - 512*a^22*b^3*d^4))/(4*(a^16*d^4 + b^16*d^4 + 8*a^2*b
^14*d^4 + 28*a^4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28*a^12*b^4*d^4 + 8*a^14*b^2*
d^4)))*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^
2*15i))^(1/2))/2 - (tan(c + d*x)^(1/2)*(1544*a*b^18*d^2 + 64*a^3*b^16*d^2 - 7456*a^5*b^14*d^2 - 576*a^7*b^12*d
^2 + 19504*a^9*b^10*d^2 + 18880*a^11*b^8*d^2 + 3808*a^13*b^6*d^2 - 960*a^15*b^4*d^2 + 8*a^17*b^2*d^2))/(2*(a^1
6*d^4 + b^16*d^4 + 8*a^2*b^14*d^4 + 28*a^4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28*
a^12*b^4*d^4 + 8*a^14*b^2*d^4)))*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 2
0*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2))/2)*(1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*
d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2))/2 + (tan(c + d*x)^(1/2)*(9*a^12*b + 41*b^13 - 82*a^2*b^11
+ 1831*a^4*b^9 - 4348*a^6*b^7 + 1671*a^8*b^5 - 210*a^10*b^3))/(2*(a^16*d^4 + b^16*d^4 + 8*a^2*b^14*d^4 + 28*a^
4*b^12*d^4 + 56*a^6*b^10*d^4 + 70*a^8*b^8*d^4 + 56*a^10*b^6*d^4 + 28*a^12*b^4*d^4 + 8*a^14*b^2*d^4)))*(1/(b^6*
d^2*1i - a^6*d^2*1i + 6*a*b^5*d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^(1/2)*1
i)/(((((518*a*b^15*d^2 - 18*a^15*b*d^2 - 4494*a^3*b^13*d^2 + 3022*a^5*b^11*d^2 + 17194*a^7*b^9*d^2 + 5298*a^9*
b^7*d^2 - 3338*a^11*b^5*d^2 + 506*a^13*b^3*d^2)/(2*(a^16*d^5 + b^16*d^5 + 8*a^2*b^14*d^5 + 28*a^4*b^12*d^5 + 5
6*a^6*b^10*d^5 + 70*a^8*b^8*d^5 + 56*a^10*b^6*d^5 + 28*a^12*b^4*d^5 + 8*a^14*b^2*d^5)) + (((((4224*a^4*b^18*d^
4 - 320*a^2*b^20*d^4 - 192*b^22*d^4 + 22272*a^6...

________________________________________________________________________________________