\(\int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [516]

Optimal result

Integrand size = 36, antiderivative size = 142 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {8 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {16 i a^3 B \sqrt {\cot (c+d x)}}{3 d}+\frac {2 i a B (i a+a \cot (c+d x))^2}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 A-7 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \sqrt {\cot (c+d x)}} \]

[Out]

-8*(-1)^(1/4)*a^3*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d+2/3*I*a*B*(I*a+a*cot(d*x+c))^2/d/cot(d*x+c)^(
3/2)-2/3*(3*A-7*I*B)*(I*a^3+a^3*cot(d*x+c))/d/cot(d*x+c)^(1/2)-16/3*I*a^3*B*cot(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3662, 3674, 3673, 3614, 214} \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {8 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 (3 A-7 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{3 d \sqrt {\cot (c+d x)}}-\frac {16 i a^3 B \sqrt {\cot (c+d x)}}{3 d}+\frac {2 i a B (a \cot (c+d x)+i a)^2}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[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 214

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

Rule 3614

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 3662

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 3673

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 3674

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] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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) \text {arctanh}\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 [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.95 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.80 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 a^3 \sqrt {\cot (c+d x)} \left (-9 A+12 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right )+12 \sqrt [4]{-1} B \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}+3 i A \tan (c+d x)+9 B \tan (c+d x)+i B \tan ^2(c+d x)\right )}{3 d} \]

[In]

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

[Out]

(-2*a^3*Sqrt[Cot[c + d*x]]*(-9*A + 12*A*Hypergeometric2F1[-1/2, 1, 1/2, I*Tan[c + d*x]] + 12*(-1)^(1/4)*B*ArcT
an[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + (3*I)*A*Tan[c + d*x] + 9*B*Tan[c + d*x] + I*B*Tan[c + d
*x]^2))/(3*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.66

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\left (4 i B -4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-2 A \sqrt {\cot \left (d x +c \right )}+\frac {-2 i A -6 B}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}\right )}{d}\)	\(236\)
default	\(\frac {a^{3} \left (-\frac {\left (4 i B -4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-2 A \sqrt {\cot \left (d x +c \right )}+\frac {-2 i A -6 B}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}\right )}{d}\)	\(236\)

[In]

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

[Out]

a^3/d*(-1/4*(4*I*B-4*A)*2^(1/2)*(ln((1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(
1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2)))-1/4*(4*I*A+4*B)*2^(1/2)*(ln
((1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c
)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2)))-2*A*cot(d*x+c)^(1/2)+2*(-I*A-3*B)/cot(d*x+c)^(1/2)-2/3*I*B/cot
(d*x+c)^(3/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (112) = 224\).

Time = 0.25 (sec) , antiderivative size = 442, normalized size of antiderivative = 3.11 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (3 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 3 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 2 \, {\left ({\left (3 \, A - 5 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (3 \, A + i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, B a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )}}{3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[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]

2/3*(3*sqrt(-(-I*A^2 - 2*A*B + 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(2*((A
- I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + 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)/((-I*A - B)*a^3)) - 3*sqrt(-(-I*
A^2 - 2*A*B + 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(2*((A - I*B)*a^3*e^(2*
I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + 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)/((-I*A - B)*a^3)) - 2*((3*A - 5*I*B)*a^3*e^(4*I*
d*x + 4*I*c) + (3*A + I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*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)

Sympy [F]

\[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- i a^{3} \left (\int i A \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 A \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{3}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 B \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{4}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 i A \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int i B \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 i B \tan ^{3}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

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

[Out]

-I*a**3*(Integral(I*A*cot(c + d*x)**(3/2), x) + Integral(-3*A*tan(c + d*x)*cot(c + d*x)**(3/2), x) + Integral(
A*tan(c + d*x)**3*cot(c + d*x)**(3/2), x) + Integral(-3*B*tan(c + d*x)**2*cot(c + d*x)**(3/2), x) + Integral(B
*tan(c + d*x)**4*cot(c + d*x)**(3/2), x) + Integral(-3*I*A*tan(c + d*x)**2*cot(c + d*x)**(3/2), x) + Integral(
I*B*tan(c + d*x)*cot(c + d*x)**(3/2), x) + Integral(-3*I*B*tan(c + d*x)**3*cot(c + d*x)**(3/2), x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.39 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {3 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + \frac {6 \, A a^{3}}{\sqrt {\tan \left (d x + c\right )}} - 2 \, {\left (-i \, B a^{3} - \frac {3 \, {\left (i \, A + 3 \, B\right )} a^{3}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{3 \, d} \]

[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*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((
I - 1)*A + (I + 1)*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)/sqrt(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

Giac [F]

\[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int(cot(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(cot(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)