3.6.70 \(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^3 \, dx\) [570]

Optimal. Leaf size=299 \[ \frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d} \]

[Out]

-1/2*(a-b)*(a^2+4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/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))/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))/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))/d*2^(1/2)+2*a*(a^2-3*b^2)*tan(d*x+c)^(1/
2)/d+2/3*b*(3*a^2-b^2)*tan(d*x+c)^(3/2)/d+32/35*a*b^2*tan(d*x+c)^(5/2)/d+2/7*b^2*tan(d*x+c)^(5/2)*(a+b*tan(d*x
+c))/d

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Rubi [A]
time = 0.27, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3647, 3711, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \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}-\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}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\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}-\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}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d} \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]*d) - ((a - b)*(a^2 + 4*a*b + b^2
)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*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]*d) - ((a + b)*(a^2 - 4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + T
an[c + d*x]])/(2*Sqrt[2]*d) + (2*a*(a^2 - 3*b^2)*Sqrt[Tan[c + d*x]])/d + (2*b*(3*a^2 - b^2)*Tan[c + d*x]^(3/2)
)/(3*d) + (32*a*b^2*Tan[c + d*x]^(5/2))/(35*d) + (2*b^2*Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x]))/(7*d)

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 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 3609

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

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 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

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

Rubi steps

\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {2}{7} \int \tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a \left (7 a^2-5 b^2\right )+\frac {7}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)+8 a b^2 \tan ^2(c+d x)\right ) \, dx\\ &=\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {2}{7} \int \tan ^{\frac {3}{2}}(c+d x) \left (\frac {7}{2} a \left (a^2-3 b^2\right )+\frac {7}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {2}{7} \int \sqrt {\tan (c+d x)} \left (-\frac {7}{2} b \left (3 a^2-b^2\right )+\frac {7}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {2}{7} \int \frac {-\frac {7}{2} a \left (a^2-3 b^2\right )-\frac {7}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}+\frac {4 \text {Subst}\left (\int \frac {-\frac {7}{2} a \left (a^2-3 b^2\right )-\frac {7}{2} b \left (3 a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{7 d}\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 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 )}{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 )}{d}\\ &=\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 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} 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} 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 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 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} 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} d}+\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 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} 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} 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} 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} 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} 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} d}+\frac {2 a \left (a^2-3 b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (3 a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {32 a b^2 \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{7 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.41, size = 144, normalized size = 0.48 \begin {gather*} \frac {105 \sqrt [4]{-1} (a-i b)^3 \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+105 \sqrt [4]{-1} (a+i b)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (105 \left (a^3-3 a b^2\right )-35 b \left (-3 a^2+b^2\right ) \tan (c+d x)+63 a b^2 \tan ^2(c+d x)+15 b^3 \tan ^3(c+d x)\right )}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(105*(-1)^(1/4)*(a - I*b)^3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 105*(-1)^(1/4)*(a + I*b)^3*ArcTanh[(-1)^(3
/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(105*(a^3 - 3*a*b^2) - 35*b*(-3*a^2 + b^2)*Tan[c + d*x] + 63*a*
b^2*Tan[c + d*x]^2 + 15*b^3*Tan[c + d*x]^3))/(105*d)

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Maple [A]
time = 0.05, size = 279, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {6 b^{2} a \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a^{2} b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {2 b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )-6 b^{2} a \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-a^{3}+3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(279\)
default	\(\frac {\frac {2 b^{3} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {6 b^{2} a \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a^{2} b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {2 b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )-6 b^{2} a \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-a^{3}+3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(279\)

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*(2/7*b^3*tan(d*x+c)^(7/2)+6/5*b^2*a*tan(d*x+c)^(5/2)+2*a^2*b*tan(d*x+c)^(3/2)-2/3*b^3*tan(d*x+c)^(3/2)+2*a
^3*tan(d*x+c)^(1/2)-6*b^2*a*tan(d*x+c)^(1/2)+1/4*(-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)))+1/4*(-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))))

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Maxima [A]
time = 0.50, size = 258, normalized size = 0.86 \begin {gather*} \frac {120 \, b^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 504 \, a b^{2} \tan \left (d x + c\right )^{\frac {5}{2}} - 210 \, \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 ) - 210 \, \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 ) - 105 \, \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 ) + 105 \, \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 ) + 280 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 840 \, {\left (a^{3} - 3 \, a b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{420 \, 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]

1/420*(120*b^3*tan(d*x + c)^(7/2) + 504*a*b^2*tan(d*x + c)^(5/2) - 210*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)))) - 210*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)))) - 105*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) + 105*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) + 280*(3*a^2*b - b^3)*tan(d*x + c)^(3/2) + 840*(a^3 - 3*a*b^2)*sqrt(tan(d*x + c)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7272 vs. \(2 (257) = 514\).
time = 3.30, size = 7272, normalized size = 24.32 \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]

1/420*(420*sqrt(2)*d^5*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^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*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)/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))*((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((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)/d^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 + 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)*sq
rt((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)/d^4) - sqrt(2)*((a^3 -
3*a*b^2)*d^7*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)/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)/d^4) - (3*a^8*b + 8*a^6*b^3 +
6*a^4*b^5 - b^9)*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)
/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^5*b - 10*a^
3*b^3 + 3*a*b^5)*d^2*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)/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((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)
+ sqrt(2)*((3*a^14*b - 91*a^12*b^3 + 795*a^10*b^5 - 1611*a^8*b^7 + 1217*a^6*b^9 - 345*a^4*b^11 + 33*a^2*b^13 -
b^15)*d^3*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)/d^4)*cos(d*x +
c) - (a^21 - 30*a^19*b^2 + 249*a^17*b^4 - 280*a^15*b^6 - 1038*a^13*b^8 + 732*a^11*b^10 + 1322*a^9*b^12 - 504*a
^7*b^14 - 531*a^5*b^16 + 82*a^3*b^18 - 3*a*b^20)*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^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*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)/d^4))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 2
55*a^4*b^8 - 30*a^2*b^10 + b^12))*sqrt(sin(d*x + c)/cos(d*x + c))*((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^24 - 24*a^22*b^2 + 90*a^20*b^4 + 648*a^18*b^6 + 783*a^16*b
^8 - 624*a^14*b^10 - 1748*a^12*b^12 - 624*a^10*b^14 + 783*a^8*b^16 + 648*a^6*b^18 + 90*a^4*b^20 - 24*a^2*b^22
+ b^24)*sin(d*x + c))/cos(d*x + c))*((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^15 - 15*a^13*b^2 + 9*a^11*b^4 + 81*a^9*b^6 + 27*a^7*b^8 - 69*a^5*b^10 - 37*a^3*
b^12 + 3*a*b^14)*d^7*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)/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)/d^4) - (3*a^20*b - 28
*a^18*b^3 - 171*a^16*b^5 - 288*a^14*b^7 - 82*a^12*b^9 + 264*a^10*b^11 + 282*a^8*b^13 + 64*a^6*b^15 - 33*a^4*b^
17 - 12*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)/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^5*b
- 10*a^3*b^3 + 3*a*b^5)*d^2*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
)/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))*((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))/(a^
36 - 18*a^34*b^2 - 39*a^32*b^4 + 848*a^30*b^6 + 5556*a^28*b^8 + 15240*a^26*b^10 + 20420*a^24*b^12 + 5424*a^22*
b^14 - 25938*a^20*b^16 - 42988*a^18*b^18 - 25938*a^16*b^20 + 5424*a^14*b^22 + 20420*a^12*b^24 + 15240*a^10*b^2
6 + 5556*a^8*b^28 + 848*a^6*b^30 - 39*a^4*b^32 - 18*a^2*b^34 + b^36))*cos(d*x + c)^3 + 420*sqrt(2)*d^5*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^5*b - 10*a^3*b^3 + 3*a*b^
5)*d^2*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)/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))*((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((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 25
5*a^4*b^8 - 30*a^2*b^10 + b^12)/d^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
+ 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((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^...

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

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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

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Mupad [B]
time = 8.24, size = 1729, normalized size = 5.78 \begin {gather*} \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {2\,a^3}{d}-\frac {6\,a\,b^2}{d}\right )-{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (\frac {2\,b^3}{3\,d}-\frac {2\,a^2\,b}{d}\right )+\frac {2\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}+\frac {6\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}}{\frac {16\,\left (3\,a^8\,b+8\,a^6\,b^3+6\,a^4\,b^5-b^9\right )}{d^3}+\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}+\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}\right )\,\sqrt {-\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}}{\frac {16\,\left (3\,a^8\,b+8\,a^6\,b^3+6\,a^4\,b^5-b^9\right )}{d^3}+\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}+\left (\frac {8\,\left (4\,a^3\,d^2-12\,a\,b^2\,d^2\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}\right )\,\sqrt {-\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \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]

tan(c + d*x)^(1/2)*((2*a^3)/d - (6*a*b^2)/d) - tan(c + d*x)^(3/2)*((2*b^3)/(3*d) - (2*a^2*b)/d) - atan((((8*(4
*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d
^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b - a^
6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i - ((8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(
6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c
+ d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i
- 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i)/((16*(3*a^8*b - b^9 + 6*a^4*b^5 + 8*a^6*b^3))/d^3 + ((8*(4*a^3*
d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^
(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b - a^6*1i
+ b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2) + ((8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5
+ 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(
1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3
*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)))*(-(6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b
^2*15i)/(4*d^2))^(1/2)*2i - atan((((8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*
b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*
a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2)
*1i - ((8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b
^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 +
6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2)*1i)/((16*(3*a^8*b - b^9 + 6
*a^4*b^5 + 8*a^6*b^3))/d^3 + ((8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*1
5i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b
^2))/d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2) + ((
8*(4*a^3*d^2 - 12*a*b^2*d^2)*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/
(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(-(6*a*b^5 + 6*a^5*b
+ a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2)))*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b
^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2)*2i + (2*b^3*tan(c + d*x)^(7/2))/(7*d) + (6*a*b^
2*tan(c + d*x)^(5/2))/(5*d)

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