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)*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

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Rubi [A]  time = 0.47, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 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))^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*}

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Mathematica [A]  time = 6.92, 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]

-1/3*(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]]))/d

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fricas [B]  time = 0.55, size = 442, normalized size = 3.11 \[ \frac {3 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 3 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - {\left (16 \, {\left (3 \, A - 5 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, {\left (3 \, A + i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, B a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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

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))^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)

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maple [C]  time = 2.21, size = 1539, normalized size = 10.84 \[ \text {result too large to display} \]

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))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/3*a^3/d*(-12*I*A*cos(d*x+c)^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))/si
n(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))+12*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))+12*I*B*cos(d*x+c)^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)*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))-12*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+3*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)-12*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)*Elli
pticPi((-(-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))-12*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)*cos(d*x+c)^2+12*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)*Ell
ipticF((-(-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+I*B*2^(1/2)-12*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)-12*B*Elliptic
Pi((-(-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)+12*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)+12*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))+12*I*A*
cos(d*x+c)^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)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*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)-I*B*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*2^(1/2)

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maxima [A]  time = 1.19, size = 194, normalized size = 1.37 \[ \frac {3 \, {\left (\sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} - \frac {6 \, A a^{3}}{\sqrt {\tan \left (d x + c\right )}} + 2 \, {\left (-i \, B a^{3} + \frac {{\left (-3 i \, A - 9 \, B\right )} a^{3}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{3 \, d} \]

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))^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 - 9*B)*a^3/tan
(d*x + c))*tan(d*x + c)^(3/2))/d

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

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)^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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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