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

Optimal. Leaf size=99 \[ -\frac {2 a^2 (A+i B) \sqrt {\cot (c+d x)}}{d}-\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{d \sqrt {\cot (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {2 a^2 (A+i B) \sqrt {\cot (c+d x)}}{d}-\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 4.99, size = 163, normalized size = 1.65 \[ -\frac {2 a^2 e^{-2 i c} \cos (c+d x) \sqrt {\cot (c+d x)} (\cos (2 (c+d x))+i \sin (2 (c+d x))) (A+B \tan (c+d x)) \left (-2 (A-i B) \sqrt {i \tan (c+d x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+A+B \tan (c+d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Cos[c + d*x]*Sqrt[Cot[c + d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(A + B*Tan[c + d*x])*(A - 2*(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]] + B*Tan[c + d
*x]))/(d*E^((2*I)*c)*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.55, size = 384, normalized size = 3.88 \[ \frac {\sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A + i \, B\right )} a^{2}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-(4*(A - I*B)*a^2*e^(2*I*d*x
+ 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2)) - sqrt((16*I*A^2 + 32*A*B -
16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32
*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2)) - 8*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + (A + I*B)*a^2)*sq
rt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.00, size = 1440, normalized size = 14.55 \[ \text {result too large to display} \]

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))^2*(A+B*tan(d*x+c)),x)

[Out]

-a^2/d*(2*I*A*cos(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))-2*I*A*cos(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/
2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*
2^(1/2))+2*I*B*cos(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c
))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*
I,1/2*2^(1/2))-2*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*
x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*cos(d*x+c)+2*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))-2*I*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+2*
B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)-2*B*Ellip
ticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)+2*I*B
*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))
/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*A*Ellipt
icPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*B*(-(-sin(d*x+c
)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-2*B*EllipticPi((-(-sin(d*x+c)-1+cos
(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+
c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+A*2^(1/2)*cos(d*x+c)+B*sin(d*x+c)*2^(1/2))
*(cos(d*x+c)/sin(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^2*2^(1/2)

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maxima [B]  time = 1.04, size = 174, normalized size = 1.76 \[ -\frac {4 \, B a^{2} \sqrt {\tan \left (d x + c\right )} - {\left (2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2} + \frac {4 \, A a^{2}}{\sqrt {\tan \left (d x + c\right )}}}{2 \, d} \]

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))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(4*B*a^2*sqrt(tan(d*x + c)) - (2*sqrt(2)*(-(I - 1)*A - (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(ta
n(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)))) - sqr
t(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^2 + 4*A*a^2/sqrt(tan(d*x + c)))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- A \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- B \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{3}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

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))**2*(A+B*tan(d*x+c)),x)

[Out]

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

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