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

Optimal. Leaf size=140 \[ -\frac{2 a^3 (5 B+i A)}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{8 a^3 (2 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{8 a^3 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^3 B \sqrt{c-i c \tan (e+f x)}}{c^3 f} \]

[Out]

(-8*a^3*(I*A + B))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (8*a^3*(I*A + 2*B))/(3*c*f*(c - I*c*Tan[e + f*x])^(3/2
)) - (2*a^3*(I*A + 5*B))/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]]) - (2*a^3*B*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f)

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Rubi [A]  time = 0.201367, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.047, Rules used = {3588, 77} \[ -\frac{2 a^3 (5 B+i A)}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{8 a^3 (2 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{8 a^3 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^3 B \sqrt{c-i c \tan (e+f x)}}{c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^3*(I*A + B))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (8*a^3*(I*A + 2*B))/(3*c*f*(c - I*c*Tan[e + f*x])^(3/2
)) - (2*a^3*(I*A + 5*B))/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]]) - (2*a^3*B*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{4 a^2 (A-i B)}{(c-i c x)^{7/2}}-\frac{4 a^2 (A-2 i B)}{c (c-i c x)^{5/2}}+\frac{a^2 (A-5 i B)}{c^2 (c-i c x)^{3/2}}+\frac{i a^2 B}{c^3 \sqrt{c-i c x}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{8 a^3 (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{8 a^3 (i A+2 B)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^3 (i A+5 B)}{c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 a^3 B \sqrt{c-i c \tan (e+f x)}}{c^3 f}\\ \end{align*}

Mathematica [A]  time = 13.1025, size = 135, normalized size = 0.96 \[ \frac{a^3 \sqrt{c-i c \tan (e+f x)} (\sin (3 (e+2 f x))-i \cos (3 (e+2 f x))) (3 (A-11 i B) \cos (e+f x)+(11 A-91 i B) \cos (3 (e+f x))-10 i \sin (e+f x) ((A-17 i B) \cos (2 (e+f x))+A-14 i B))}{15 c^3 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(3*(A - (11*I)*B)*Cos[e + f*x] + (11*A - (91*I)*B)*Cos[3*(e + f*x)] - (10*I)*(A - (14*I)*B + (A - (17*I)*
B)*Cos[2*(e + f*x)])*Sin[e + f*x])*((-I)*Cos[3*(e + 2*f*x)] + Sin[3*(e + 2*f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/
(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.076, size = 105, normalized size = 0.8 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ( iB\sqrt{c-ic\tan \left ( fx+e \right ) }-{c \left ( A-5\,iB \right ){\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}+{\frac{4\,{c}^{2} \left ( A-2\,iB \right ) }{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,{c}^{3} \left ( A-iB \right ) }{5} \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+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2*I/f*a^3/c^3*(I*B*(c-I*c*tan(f*x+e))^(1/2)-c*(A-5*I*B)/(c-I*c*tan(f*x+e))^(1/2)+4/3*c^2*(A-2*I*B)/(c-I*c*tan(
f*x+e))^(3/2)-4/5*c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2))

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Maxima [A]  time = 1.23794, size = 150, normalized size = 1.07 \begin{align*} -\frac{2 i \,{\left (-\frac{15 i \, \sqrt{-i \, c \tan \left (f x + e\right ) + c} B a^{3}}{c^{2}} + \frac{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2}{\left (15 \, A - 75 i \, B\right )} a^{3} -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}{\left (20 \, A - 40 i \, B\right )} a^{3} c +{\left (12 \, A - 12 i \, B\right )} a^{3} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c}\right )}}{15 \, c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*I*(-15*I*sqrt(-I*c*tan(f*x + e) + c)*B*a^3/c^2 + ((-I*c*tan(f*x + e) + c)^2*(15*A - 75*I*B)*a^3 - (-I*c*
tan(f*x + e) + c)*(20*A - 40*I*B)*a^3*c + (12*A - 12*I*B)*a^3*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*c))/(c*f)

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Fricas [A]  time = 1.16856, size = 270, normalized size = 1.93 \begin{align*} \frac{\sqrt{2}{\left ({\left (-3 i \, A - 3 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A + 11 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-4 i \, A - 44 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-8 i \, A - 88 \, B\right )} a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*((-3*I*A - 3*B)*a^3*e^(6*I*f*x + 6*I*e) + (I*A + 11*B)*a^3*e^(4*I*f*x + 4*I*e) + (-4*I*A - 44*B)*
a^3*e^(2*I*f*x + 2*I*e) + (-8*I*A - 88*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(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+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(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{{\left (B \tan \left (f x + e\right ) + A\right )}{\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+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(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)