3.516 \(\int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=142 \[ -\frac{2 (3 A-7 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{3 d \sqrt{\cot (c+d x)}}-\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 B \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*B*Sqrt[Cot[c + d*x]])
/d + (((2*I)/3)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c + d*x]^(3/2)) - (2*(3*A - (7*I)*B)*(I*a^3 + a^3*Cot[c +
d*x]))/(3*d*Sqrt[Cot[c + d*x]])

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Rubi [A]  time = 0.47097, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3581, 3593, 3592, 3533, 208} \[ -\frac{2 (3 A-7 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{3 d \sqrt{\cot (c+d x)}}-\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 B \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*B*Sqrt[Cot[c + d*x]])
/d + (((2*I)/3)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c + d*x]^(3/2)) - (2*(3*A - (7*I)*B)*(I*a^3 + a^3*Cot[c +
d*x]))/(3*d*Sqrt[Cot[c + d*x]])

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3593

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

Rule 3592

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

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\int \frac{(i a+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(i a+a \cot (c+d x))^2 \left (\frac{1}{2} a (3 i A+7 B)+\frac{1}{2} a (3 A+i B) \cot (c+d x)\right )}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 (3 A-7 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \sqrt{\cot (c+d x)}}+\frac{4}{3} \int \frac{(i a+a \cot (c+d x)) \left (a^2 (3 i A+5 B)+2 i a^2 B \cot (c+d x)\right )}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 B \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 (3 A-7 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \sqrt{\cot (c+d x)}}+\frac{4}{3} \int \frac{-3 a^3 (A-i B)+3 a^3 (i A+B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 B \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 (3 A-7 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \sqrt{\cot (c+d x)}}+\frac{\left (24 a^6 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{3 a^3 (A-i B)+3 a^3 (i A+B) x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 B \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 (3 A-7 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.91535, size = 132, normalized size = 0.93 \[ -\frac{a^3 \sqrt{\cot (c+d x)} \left (\sec ^2(c+d x) ((9 B+3 i A) \sin (2 (c+d x))+(3 A-i B) \cos (2 (c+d x))+3 A+i B)-24 (A-i B) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-(a^3*Sqrt[Cot[c + d*x]]*(Sec[c + d*x]^2*(3*A + I*B + (3*A - I*B)*Cos[2*(c + d*x)] + ((3*I)*A + 9*B)*Sin[2*(c
+ d*x)]) - 24*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Tan[c + d*x
]]))/(3*d)

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Maple [C]  time = 0.54, size = 1539, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/3*a^3/d*2^(1/2)*(12*I*B*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c)
)^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1
/2*2^(1/2))*cos(d*x+c)+12*I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x
+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/
2))*cos(d*x+c)-12*I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/
2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^
(1/2))*cos(d*x+c)^2-I*B*2^(1/2)*cos(d*x+c)^2-12*A*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((
cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d
*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12*I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x
+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^2-12*I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+
sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x
+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)-12*B*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((
cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d
*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12*B*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)
*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-12*A*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-12*B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12*B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2)+3*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)+I*B*2^(1/2)+3*A*2^(1/2)*cos(d*x+c)^2+9*B*2
^(1/2)*cos(d*x+c)*sin(d*x+c)+12*I*B*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/
2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2)*(cos(d*x+c)/sin(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^3

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Maxima [A]  time = 1.55794, size = 263, normalized size = 1.85 \begin{align*} \frac{3 \,{\left (\sqrt{2}{\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \sqrt{2}{\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} - \frac{6 \, A a^{3}}{\sqrt{\tan \left (d x + c\right )}} + 2 \,{\left (-i \, B a^{3} - \frac{3 \,{\left (i \, A + 3 \, B\right )} a^{3}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac{3}{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/3*(3*(sqrt(2)*(-(2*I - 2)*A - (2*I + 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + sqrt(2)*(-
(2*I - 2)*A - (2*I + 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I -
1)*B)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)/sqr
t(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 - 6*A*a^3/sqrt(tan(d*x + c)) + 2*(-I*B*a^3 - 3*(I*A + 3*B)*a^3/tan(
d*x + c))*tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.63229, size = 1185, normalized size = 8.35 \begin{align*} \frac{3 \, \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 3 \, \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) -{\left (16 \,{\left (3 \, A - 5 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \,{\left (3 \, A + i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, B a^{3}\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{12 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(3*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*lo
g(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*
c) - I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3
)) - 3*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log
(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*I*
c) + I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3
)) - (16*(3*A - 5*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 16*(3*A + I*B)*a^3*e^(2*I*d*x + 2*I*c) + 64*I*B*a^3)*sqrt((I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(3/2), x)