3.515 \(\int \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=138 \[ -\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}-\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 A \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{d \sqrt{\cot (c+d x)}} \]
[Out]
(-8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*A*Sqrt[Cot[c + d*x]])
/d + ((2*I)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Sqrt[Cot[c + d*x]]) - (2*(A + (3*I)*B)*Sqrt[Cot[c + d*x]]*(I*a^3
+ a^3*Cot[c + d*x]))/(3*d)
________________________________________________________________________________________
Rubi [A] time = 0.457495, antiderivative size = 138, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{3581, 3593, 3594, 3592, 3533, 208} \[ -\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}-\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 A \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{d \sqrt{\cot (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
[Out]
(-8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*A*Sqrt[Cot[c + d*x]])
/d + ((2*I)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Sqrt[Cot[c + d*x]]) - (2*(A + (3*I)*B)*Sqrt[Cot[c + d*x]]*(I*a^3
+ a^3*Cot[c + d*x]))/(3*d)
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 3594
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, 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{5}{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{3}{2}}(c+d x)} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{d \sqrt{\cot (c+d x)}}+2 \int \frac{(i a+a \cot (c+d x))^2 \left (\frac{1}{2} a (i A+5 B)+\frac{1}{2} a (A+3 i B) \cot (c+d x)\right )}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{d \sqrt{\cot (c+d x)}}-\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{4}{3} \int \frac{(i a+a \cot (c+d x)) \left (a^2 (A-3 i B)-2 i a^2 A \cot (c+d x)\right )}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 A \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{d \sqrt{\cot (c+d x)}}-\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{4}{3} \int \frac{3 a^3 (i A+B)+3 a^3 (A-i B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 A \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{d \sqrt{\cot (c+d x)}}-\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{\left (24 a^6 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-3 a^3 (i A+B)+3 a^3 (A-i B) x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 (i A+B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 A \sqrt{\cot (c+d x)}}{3 d}+\frac{2 i a B (i a+a \cot (c+d x))^2}{d \sqrt{\cot (c+d x)}}-\frac{2 (A+3 i B) \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}\\ \end{align*}
Mathematica [A] time = 5.467, size = 146, normalized size = 1.06 \[ -\frac{a^3 \sqrt{\cot (c+d x)} \csc (c+d x) \sec (c+d x) \left ((A-3 i B) \cos (2 (c+d x))-12 i (A-i B) \sin (2 (c+d x)) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+9 i A \sin (2 (c+d x))+A+3 B \sin (2 (c+d x))+3 i B\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
[Out]
-(a^3*Sqrt[Cot[c + d*x]]*Csc[c + d*x]*Sec[c + d*x]*(A + (3*I)*B + (A - (3*I)*B)*Cos[2*(c + d*x)] + (9*I)*A*Sin
[2*(c + d*x)] + 3*B*Sin[2*(c + d*x)] - (12*I)*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*
(c + d*x)))]]*Sin[2*(c + d*x)]*Sqrt[I*Tan[c + d*x]]))/(3*d)
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Maple [C] time = 0.526, size = 1562, normalized size = 11.3 \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)^(5/2)*(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)*sin(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))/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))-3*I*B*2^(1/2)*cos(d*x+c)^2+9*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)+12*A*cos(d*x+c)*sin(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))/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))-12*A*cos(d*x
+c)*sin(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))/sin(d*x+c))^(1/2)*((co
s(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))+3*I*2^(1/2)
*B-12*B*cos(d*x+c)*sin(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))/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))-12*I*B*((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)*(-(cos(d*
x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*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))^(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))/sin(d*x+c))^(1/2)*sin(d*x+c)*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*A*sin(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))/
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))-12*A*sin(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))
/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))-12*B*sin(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))/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))+12*I*B*sin(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))/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))-12*I*A*cos(d*x+c)*sin(d*x+c)*(-(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))*((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)+A*2^(1/2)*cos(d*x+c)^2+3*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-12*I*B*cos(d*x+c)*sin(d*x+c)*(-(co
s(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))*((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)/sin(d*x
+c))^(5/2)*sin(d*x+c)/cos(d*x+c)^3
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Maxima [A] time = 1.55885, size = 259, normalized size = 1.88 \begin{align*} \frac{-6 i \, B a^{3} \sqrt{\tan \left (d x + c\right )} + 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 \,{\left (-3 i \, A - B\right )} a^{3}}{\sqrt{\tan \left (d x + c\right )}} - \frac{2 \, A a^{3}}{\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)^(5/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
1/3*(-6*I*B*a^3*sqrt(tan(d*x + c)) + 3*(sqrt(2)*((2*I + 2)*A - (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sq
rt(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)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 6*(-3*I*A - B)*a^3/sqrt(tan(d*x + c))
- 2*A*a^3/tan(d*x + c)^(3/2))/d
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Fricas [B] time = 1.50108, size = 1077, normalized size = 7.8 \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 )} - 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 )} - 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 ({\left (-80 i \, A - 48 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-16 i \, A + 48 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, A 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 )} - d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(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) - 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)*(d*e^(2*I*d*x + 2*I*c) - 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 - 12
8*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*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)*(d*e^(2*I*d*x + 2*I*c) - 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)) - ((-80*I*A - 48*B)*a^3*e^(4*I*d*x + 4*I*c) + (-16*I*
A + 48*B)*a^3*e^(2*I*d*x + 2*I*c) + 64*I*A*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) - 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)**(5/2)*(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{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(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)^(5/2), x)