3.573 \(\int \frac{(a+b \tan (c+d x))^3}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=245 \[ \frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b \left (a^2+b^2\right ) \sqrt{\tan (c+d x)}}{d}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 (a+b \tan (c+d x))}{d \sqrt{\tan (c+d x)}} \]

[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*b*(a^2 + b^2)*Sqrt[Tan[c + d*x]])/d - (2*a^2*(a + b*Tan[c + d*x]))/(d*Sqrt[Ta
n[c + d*x]])

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Rubi [A]  time = 0.307961, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3565, 3630, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b \left (a^2+b^2\right ) \sqrt{\tan (c+d x)}}{d}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 (a+b \tan (c+d x))}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3630

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]

Rule 3534

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 1168

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

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

Mathematica [C]  time = 1.90866, size = 192, normalized size = 0.78 \[ \frac{-8 a \left (a^2-3 b^2\right ) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(c+d x)\right )+\sqrt{2} b \left (b^2-3 a^2\right ) \sqrt{\tan (c+d x)} \left (2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )+\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )+8 b^2 (a+b \tan (c+d x))-32 a b^2}{4 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*a*b^2 - 8*a*(a^2 - 3*b^2)*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[c + d*x]^2] + Sqrt[2]*b*(-3*a^2 + b^2)*(2*
ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])*Sqrt[Tan[c + d*x]] + 8*b^2*(a +
b*Tan[c + d*x]))/(4*d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.014, size = 472, normalized size = 1.9 \begin{align*} 2\,{\frac{{b}^{3}\sqrt{\tan \left ( dx+c \right ) }}{d}}-2\,{\frac{{a}^{3}}{d\sqrt{\tan \left ( dx+c \right ) }}}+{\frac{3\,\sqrt{2}{a}^{2}b}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{b}^{3}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{3\,\sqrt{2}{a}^{2}b}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}{b}^{3}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}{a}^{2}b}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{b}^{3}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{a}^{3}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}a{b}^{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}{a}^{3}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{3\,\sqrt{2}a{b}^{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{a}^{3}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{3\,\sqrt{2}a{b}^{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63071, size = 289, normalized size = 1.18 \begin{align*} \frac{8 \, b^{3} \sqrt{\tan \left (d x + c\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 ) - 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 ) - \frac{8 \, a^{3}}{\sqrt{\tan \left (d x + c\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*b^3*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)))) - 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) - 8*a^3/sqrt(tan(d*x + c)))/
d

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Fricas [B]  time = 15.2081, size = 17420, normalized size = 71.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(2)*(d^5*cos(d*x + c)^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^1
0 + 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)*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(2)*((3*a^2*b - b^3)*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) + (a
^9 - 6*a^5*b^4 - 8*a^3*b^6 - 3*a*b^8)*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^1
4 - 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)*((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^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) + (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*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 + 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)^(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)*((3*a^14*b - 37*a^12*b^3 - 69*a^10*b^5 + 27*a^8*b^7 + 81*a^6*b^
9 + 9*a^4*b^11 - 15*a^2*b^13 + b^15)*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) + (a^21 - 12*a^19*b^2 - 33*a^17*b^4 + 64*a^15*b^6 + 282*a^13*b^8 + 264*a^11*b^10 - 82*a^9*b^12 - 288*a^7*b
^14 - 171*a^5*b^16 - 28*a^3*b^18 + 3*a*b^20)*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^26 + 5556*a^8*b^28 + 848*a^6*b^30 - 39*a^4*b^32 - 18*a^2*b^34 + b^36)) + 4*sqrt(2)*(d^5*cos
(d*x + c)^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)*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(2)*((3*a^2*b -
b^3)*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) + (a^9 - 6*a^5*b^4 - 8*a^3*b
^6 - 3*a*b^8)*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) - sq
rt(2)*((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^1
4)*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) +
(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*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 + 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)^(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)*((3*a^14*b - 37*a^12*b^3 - 69*a^10*b^5 + 27*a^8*b^7 + 81*a^6*b^9 + 9*a^4*b^11 - 15*a^2*
b^13 + b^15)*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) + (a^21 - 12*a^19*b^
2 - 33*a^17*b^4 + 64*a^15*b^6 + 282*a^13*b^8 + 264*a^11*b^10 - 82*a^9*b^12 - 288*a^7*b^14 - 171*a^5*b^16 - 28*
a^3*b^18 + 3*a*b^20)*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^26 +
5556*a^8*b^28 + 848*a^6*b^30 - 39*a^4*b^32 - 18*a^2*b^34 + b^36)) + sqrt(2)*((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*cos(d*x + c)^2 - (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 + 2*((3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d^3*cos(d*x + c)^2 - (3*a^5*b - 10*a
^3*b^3 + 3*a*b^5)*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
))*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)^(1/4)*log(((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)*((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^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) + (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*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 + 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)^(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 - 174
8*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))/co
s(d*x + c)) - sqrt(2)*((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*cos(d*
x + c)^2 - (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 + 2*((3*a^5*b - 10
*a^3*b^3 + 3*a*b^5)*d^3*cos(d*x + c)^2 - (3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*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))*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)^(1/4)*log(((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)*((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^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) + (3*a^20*b - 82*a^18*b^3 + 531*a^16*b^5 + 504*a^14*b^7 - 1322*a^12*b^9 - 7
32*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*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 + 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)^(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)) - 8*(a^12*b^3 + 6*a^10*b^5 + 15*a^8*b^7 +
20*a^6*b^9 + 15*a^4*b^11 + 6*a^2*b^13 + b^15 - (a^12*b^3 + 6*a^10*b^5 + 15*a^8*b^7 + 20*a^6*b^9 + 15*a^4*b^11
+ 6*a^2*b^13 + b^15)*cos(d*x + c)^2 - (a^15 + 6*a^13*b^2 + 15*a^11*b^4 + 20*a^9*b^6 + 15*a^7*b^8 + 6*a^5*b^10
+ a^3*b^12)*cos(d*x + c)*sin(d*x + c))*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*cos(d*x + c)^2 - (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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.95206, size = 365, normalized size = 1.49 \begin{align*} \frac{2 \, b^{3} \sqrt{\tan \left (d x + c\right )}}{d} - \frac{2 \, a^{3}}{d \sqrt{\tan \left (d x + c\right )}} - \frac{{\left (\sqrt{2} a^{3} - 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} + \sqrt{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 \, d} - \frac{{\left (\sqrt{2} a^{3} - 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} + \sqrt{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 \, d} + \frac{{\left (\sqrt{2} a^{3} + 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} - \sqrt{2} b^{3}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} - \frac{{\left (\sqrt{2} a^{3} + 3 \, \sqrt{2} a^{2} b - 3 \, \sqrt{2} a b^{2} - \sqrt{2} b^{3}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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