3.87 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=173 \[ \frac{a^{5/2} (23 A-20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 (4 B+7 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \]
[Out]
(a^(5/2)*(23*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (4*Sqrt[2]*a^(5/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a^2*((7*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c
+ d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2))/(2*d)
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Rubi [A] time = 0.605962, antiderivative size = 173, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used =
{3593, 3600, 3480, 206, 3599, 63, 208} \[ \frac{a^{5/2} (23 A-20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 (4 B+7 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
[Out]
(a^(5/2)*(23*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (4*Sqrt[2]*a^(5/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a^2*((7*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c
+ d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2))/(2*d)
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 3600
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Rule 3480
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3599
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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]
Rubi steps
\begin{align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (7 i A+4 B)-\frac{1}{2} a (A-4 i B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a^2 (7 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (23 A-20 i B)-\frac{3}{4} a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a^2 (7 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac{1}{8} (a (23 A-20 i B)) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx-\left (4 a^2 (i A+B)\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{a^2 (7 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac{\left (8 a^3 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}-\frac{\left (a^3 (23 A-20 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 (7 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac{\left (a^2 (23 i A+20 B)\right ) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac{a^{5/2} (23 A-20 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 (7 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\\ \end{align*}
Mathematica [B] time = 8.45932, size = 427, normalized size = 2.47 \[ \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (-2 e^{-3 i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left ((23 A-20 i B) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (-2 e^{i (c+d x)} \left (1+\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+3\right )-\log \left (2 e^{i (c+d x)} \left (1+\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+3\right )\right )+64 \sqrt{2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )-\frac{8 (\cos (2 c)-i \sin (2 c)) \csc (c+d x) (2 A \csc (c+d x)+(4 B+9 i A) \sec (c+d x))}{\sec ^{\frac{3}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2}\right )}{32 d \sec ^{\frac{7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
[Out]
(((-2*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(64*Sqrt[2]*(A - I*B)*ArcS
inh[E^(I*(c + d*x))] + (23*A - (20*I)*B)*(Log[(-1 + E^(I*(c + d*x)))^2] - Log[(1 + E^(I*(c + d*x)))^2] + Log[3
+ 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E
^((2*I)*(c + d*x))])] - Log[3 + 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] + 2*E^(I*(c +
d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])))/E^((3*I)*(c + d*x)) - (8*Csc[c + d*x]*(2*A*Csc[c + d*x]
+ ((9*I)*A + 4*B)*Sec[c + d*x])*(Cos[2*c] - I*Sin[2*c]))/(Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])^2))*(a +
I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(32*d*Sec[c + d*x]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))
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Maple [B] time = 0.445, size = 1292, normalized size = 7.5 \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*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
[Out]
1/8/d*a^2*(20*I*B*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-20*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-23*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2))*sin(d*x+c)-20*I*B*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-32*I*A*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)-4*A*cos(d*x+c)^2-18*A*cos(d*x+c)+22*A*cos(d*
x+c)^3-8*I*B*cos(d*x+c)^3-8*B*cos(d*x+c)*sin(d*x+c)+8*B*cos(d*x+c)^2*sin(d*x+c)+8*I*B*cos(d*x+c)+32*A*2^(1/2)*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d
*x+c))*sin(d*x+c)-32*B*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2))*sin(d*x+c)-32*I*B*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)+23*I*A*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+23*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-
(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+32*I*A*sin(d*x+c)*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^2+
32*I*B*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*cos(d*x+c)^2-32*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+32*B*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)
+22*I*A*cos(d*x+c)^2*sin(d*x+c)-18*I*A*sin(d*x+c)*cos(d*x+c)-23*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+20*B*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c))*(a*(I*sin
(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(cos(d*x+c)-1)/(I*sin(d*x+c)+cos(d*x+c)-1)/(cos(d*x+c)+1)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.82976, size = 2133, normalized size = 12.33 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
1/8*(2*sqrt(2)*((11*A - 4*I*B)*a^2*e^(4*I*d*x + 4*I*c) + 4*A*a^2*e^(2*I*d*x + 2*I*c) - (7*A - 4*I*B)*a^2)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(d*e^(4*I*d*x +
4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((23*I*A + 20*B)*a^2*e^(2*I*d*x + 2*I*c) + (23*I*A + 20*B)*
a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 2*I*sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*d*e
^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((23*I*A + 20*B)*a^2)) - sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2
)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((23*I*A + 20*B)*a^2*e^(2*I*d*x + 2*I*c)
+ (23*I*A + 20*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*I*sqrt((529*A^2 - 920*I*A*B - 400
*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((23*I*A + 20*B)*a^2)) - 4*sqrt((32*A^2 - 64*I*A*B
- 32*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*
I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt((32*A^2 - 64*I*
A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)) + 4*sqrt((32*A^2 - 64*
I*A*B - 32*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2
*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt((32*A^2 -
64*I*A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(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*(a+I*a*tan(d*x+c))**(5/2)*(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 )}^{\frac{5}{2}} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^3, x)