3.786 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=311 \[ -\frac{21 (B+11 i A)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{21 (B+11 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(21*((11*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(512*Sqrt[2]*a^3*c^(5/2)*f) - (21*((
11*I)*A + B))/(640*a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Ta
n[e + f*x])^(5/2)) + ((11*I)*A + B)/(48*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + (3*((11*I
)*A + B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((11*I)*A + B))/(256*a^3*c*f*(c -
I*c*Tan[e + f*x])^(3/2)) - (21*((11*I)*A + B))/(512*a^3*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.349697, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ -\frac{21 (B+11 i A)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{21 (B+11 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(21*((11*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(512*Sqrt[2]*a^3*c^(5/2)*f) - (21*((
11*I)*A + B))/(640*a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Ta
n[e + f*x])^(5/2)) + ((11*I)*A + B)/(48*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + (3*((11*I
)*A + B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - (7*((11*I)*A + B))/(256*a^3*c*f*(c -
I*c*Tan[e + f*x])^(3/2)) - (21*((11*I)*A + B))/(512*a^3*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^4 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{((11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^3 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{(3 (11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(21 (11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{512 a^2 c f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{1024 a^2 c^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(21 (11 i A+B)) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{512 a^2 c^3 f}\\ &=\frac{21 (11 i A+B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 12.2013, size = 256, normalized size = 0.82 \[ \frac{e^{-6 i (e+f x)} \sqrt{c-i c \tan (e+f x)} \left (315 (B+11 i A) e^{6 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-i \left (1+e^{2 i (e+f x)}\right ) \left (A \left (-310 e^{2 i (e+f x)}-1335 e^{4 i (e+f x)}+2768 e^{6 i (e+f x)}+416 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}-40\right )-i B \left (190 e^{2 i (e+f x)}+315 e^{4 i (e+f x)}+688 e^{6 i (e+f x)}+256 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}+40\right )\right )\right )}{15360 a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((-I)*(1 + E^((2*I)*(e + f*x)))*((-I)*B*(40 + 190*E^((2*I)*(e + f*x)) + 315*E^((4*I)*(e + f*x)) + 688*E^((6*I
)*(e + f*x)) + 256*E^((8*I)*(e + f*x)) + 48*E^((10*I)*(e + f*x))) + A*(-40 - 310*E^((2*I)*(e + f*x)) - 1335*E^
((4*I)*(e + f*x)) + 2768*E^((6*I)*(e + f*x)) + 416*E^((8*I)*(e + f*x)) + 48*E^((10*I)*(e + f*x)))) + 315*((11*
I)*A + B)*E^((6*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I
*c*Tan[e + f*x]])/(15360*a^3*c^3*E^((6*I)*(e + f*x))*f)

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Maple [A]  time = 0.117, size = 233, normalized size = 0.8 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ( -{\frac{1}{32\,{c}^{5}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ({\frac{11\,i}{32}}B+{\frac{71\,A}{32}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ( -{\frac{59\,Ac}{6}}-{\frac{11\,i}{6}}Bc \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ({\frac{21\,i}{8}}B{c}^{2}+{\frac{89\,A{c}^{2}}{8}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( -21\,iB+231\,A \right ) \sqrt{2}}{64}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) }-{\frac{5\,A-iB}{32\,{c}^{5}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{2\,A-iB}{48\,{c}^{4}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{A-iB}{80\,{c}^{3}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2*I/f/a^3*c^3*(-1/32/c^5*(((11/32*I*B+71/32*A)*(c-I*c*tan(f*x+e))^(5/2)+(-59/6*A*c-11/6*I*B*c)*(c-I*c*tan(f*x+
e))^(3/2)+(21/8*I*B*c^2+89/8*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-21/64*(-I*B+11*A)*2^(1/2)/
c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/32/c^5*(5*A-I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/4
8/c^4*(2*A-I*B)/(c-I*c*tan(f*x+e))^(3/2)-1/80/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2))

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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((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80242, size = 1319, normalized size = 4.24 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} + 231 i \, A + 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} - 231 i \, A - 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + \sqrt{2}{\left ({\left (-48 i \, A - 48 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-464 i \, A - 304 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-3184 i \, A - 944 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-1433 i \, A - 1003 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (1645 i \, A - 505 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (350 i \, A - 230 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i \, A - 40 \, B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15360*(15*sqrt(1/2)*a^3*c^3*f*sqrt(-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*lo
g(1/256*(sqrt(2)*sqrt(1/2)*(a^3*c^2*f*e^(2*I*f*x + 2*I*e) + a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(
-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2)) + 231*I*A + 21*B)*e^(-I*f*x - I*e)/(a^3*c^2*f)) - 15*sqrt(1
/2)*a^3*c^3*f*sqrt(-(53361*A^2 - 9702*I*A*B - 441*B^2)/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/256*(sqrt(2)*
sqrt(1/2)*(a^3*c^2*f*e^(2*I*f*x + 2*I*e) + a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(53361*A^2 - 970
2*I*A*B - 441*B^2)/(a^6*c^5*f^2)) - 231*I*A - 21*B)*e^(-I*f*x - I*e)/(a^3*c^2*f)) + sqrt(2)*((-48*I*A - 48*B)*
e^(12*I*f*x + 12*I*e) + (-464*I*A - 304*B)*e^(10*I*f*x + 10*I*e) + (-3184*I*A - 944*B)*e^(8*I*f*x + 8*I*e) + (
-1433*I*A - 1003*B)*e^(6*I*f*x + 6*I*e) + (1645*I*A - 505*B)*e^(4*I*f*x + 4*I*e) + (350*I*A - 230*B)*e^(2*I*f*
x + 2*I*e) + 40*I*A - 40*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^(5/2)), x)