3.747 \(\int \frac{(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 a B}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

(-2*a*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (2*a*B)/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.103932, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 43} \[ \frac{2 a B}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(-2*a*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (2*a*B)/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2))

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{A-i B}{(c-i c x)^{9/2}}+\frac{i B}{c (c-i c x)^{7/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 a (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a B}{5 c f (c-i c \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 11.2774, size = 100, normalized size = 1.61 \[ \frac{2 a \cos ^3(e+f x) (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\cos (4 e+5 f x)+i \sin (4 e+5 f x)) ((2 B-5 i A) \cos (e+f x)-7 i B \sin (e+f x))}{35 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(2*a*Cos[e + f*x]^3*(Cos[f*x] - I*Sin[f*x])*(((-5*I)*A + 2*B)*Cos[e + f*x] - (7*I)*B*Sin[e + f*x])*(Cos[4*e +
5*f*x] + I*Sin[4*e + 5*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(35*c^4*f)

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Maple [A]  time = 0.071, size = 53, normalized size = 0.9 \begin{align*}{\frac{2\,ia}{cf} \left ( -{\frac{c \left ( A-iB \right ) }{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}}-{{\frac{i}{5}}B \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))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

2*I/f*a/c*(-1/7*c*(A-I*B)/(c-I*c*tan(f*x+e))^(7/2)-1/5*I*B/(c-I*c*tan(f*x+e))^(5/2))

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Maxima [A]  time = 1.12567, size = 63, normalized size = 1.02 \begin{align*} -\frac{2 i \,{\left (7 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a +{\left (5 \, A - 5 i \, B\right )} a c\right )}}{35 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-2/35*I*(7*I*(-I*c*tan(f*x + e) + c)*B*a + (5*A - 5*I*B)*a*c)/((-I*c*tan(f*x + e) + c)^(7/2)*c*f)

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Fricas [B]  time = 1.40384, size = 320, normalized size = 5.16 \begin{align*} \frac{\sqrt{2}{\left ({\left (-5 i \, A - 5 \, B\right )} a e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-20 i \, A - 6 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-30 i \, A + 12 \, B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-20 i \, A + 22 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-5 i \, A + 9 \, B\right )} a\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{280 \, c^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/280*sqrt(2)*((-5*I*A - 5*B)*a*e^(8*I*f*x + 8*I*e) + (-20*I*A - 6*B)*a*e^(6*I*f*x + 6*I*e) + (-30*I*A + 12*B)
*a*e^(4*I*f*x + 4*I*e) + (-20*I*A + 22*B)*a*e^(2*I*f*x + 2*I*e) + (-5*I*A + 9*B)*a)*sqrt(c/(e^(2*I*f*x + 2*I*e
) + 1))/(c^4*f)

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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((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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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 )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)/(-I*c*tan(f*x + e) + c)^(7/2), x)