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

Optimal. Leaf size=103 \[ \frac{2 a^2 (3 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 a^2 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B}{c^2 f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

(-4*a^2*(I*A + B))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (2*a^2*(I*A + 3*B))/(3*c*f*(c - I*c*Tan[e + f*x])^(3/2
)) - (2*a^2*B)/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.179448, antiderivative size = 103, 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} \[ \frac{2 a^2 (3 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 a^2 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^2 B}{c^2 f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(I*A + B))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (2*a^2*(I*A + 3*B))/(3*c*f*(c - I*c*Tan[e + f*x])^(3/2
)) - (2*a^2*B)/(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 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))^2 (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) (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{2 a (A-i B)}{(c-i c x)^{7/2}}-\frac{a (A-3 i B)}{c (c-i c x)^{5/2}}-\frac{i a B}{c^2 (c-i c x)^{3/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{4 a^2 (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a^2 (i A+3 B)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^2 B}{c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 11.9614, size = 118, normalized size = 1.15 \[ \frac{a^2 \cos (e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (3 e+5 f x)+i \sin (3 e+5 f x)) (5 (A+3 i B) \sin (2 (e+f x))+(-21 B-i A) \cos (2 (e+f x))-i A+9 B)}{15 c^3 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[e + f*x]*((-I)*A + 9*B + ((-I)*A - 21*B)*Cos[2*(e + f*x)] + 5*(A + (3*I)*B)*Sin[2*(e + f*x)])*(Cos[3*
e + 5*f*x] + I*Sin[3*e + 5*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.081, size = 80, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({-iB{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{c \left ( A-3\,iB \right ) }{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{c}^{2} \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))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.15955, size = 107, normalized size = 1.04 \begin{align*} \frac{2 i \,{\left (15 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} B a^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}{\left (5 \, A - 15 i \, B\right )} a^{2} c -{\left (6 \, A - 6 i \, B\right )} a^{2} c^{2}\right )}}{15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*I*(15*I*(-I*c*tan(f*x + e) + c)^2*B*a^2 + (-I*c*tan(f*x + e) + c)*(5*A - 15*I*B)*a^2*c - (6*A - 6*I*B)*a^
2*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*c^2*f)

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Fricas [A]  time = 1.18839, size = 266, normalized size = 2.58 \begin{align*} \frac{\sqrt{2}{\left ({\left (-3 i \, A - 3 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-4 i \, A + 6 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (i \, A - 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 18 \, B\right )} a^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(2)*((-3*I*A - 3*B)*a^2*e^(6*I*f*x + 6*I*e) + (-4*I*A + 6*B)*a^2*e^(4*I*f*x + 4*I*e) + (I*A - 9*B)*a^
2*e^(2*I*f*x + 2*I*e) + (2*I*A - 18*B)*a^2)*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))**2*(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 )}^{2}}{{\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))^2*(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)^2/(-I*c*tan(f*x + e) + c)^(5/2), x)