3.763 \(\int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac{2 a^3 (5 B+i A)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{8 a^3 (2 B+i A)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{8 a^3 (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a^3 B}{c^3 f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

(-8*a^3*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (8*a^3*(I*A + 2*B))/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2
)) - (2*a^3*(I*A + 5*B))/(3*c^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a^3*B)/(c^3*f*Sqrt[c - I*c*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.205122, antiderivative size = 142, 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} \[ -\frac{2 a^3 (5 B+i A)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{8 a^3 (2 B+i A)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{8 a^3 (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{2 a^3 B}{c^3 f \sqrt{c-i c \tan (e+f x)}} \]

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])^(7/2),x]

[Out]

(-8*a^3*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (8*a^3*(I*A + 2*B))/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2
)) - (2*a^3*(I*A + 5*B))/(3*c^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a^3*B)/(c^3*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))^3 (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+i a x)^2 (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{4 a^2 (A-i B)}{(c-i c x)^{9/2}}-\frac{4 a^2 (A-2 i B)}{c (c-i c x)^{7/2}}+\frac{a^2 (A-5 i B)}{c^2 (c-i c x)^{5/2}}+\frac{i a^2 B}{c^3 (c-i c x)^{3/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{8 a^3 (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{8 a^3 (i A+2 B)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 a^3 (i A+5 B)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{2 a^3 B}{c^3 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 13.3834, size = 141, normalized size = 0.99 \[ \frac{a^3 \cos (e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (4 e+7 f x)+i \sin (4 e+7 f x)) (i (A+13 i B) \cos (e+f x)+(89 B-23 i A) \cos (3 (e+f x))+14 \sin (e+f x) ((A-17 i B) \cos (2 (e+f x))+A-2 i B))}{105 c^4 f (\cos (f x)+i \sin (f x))^3} \]

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])^(7/2),x]

[Out]

(a^3*Cos[e + f*x]*(I*(A + (13*I)*B)*Cos[e + f*x] + ((-23*I)*A + 89*B)*Cos[3*(e + f*x)] + 14*(A - (2*I)*B + (A
- (17*I)*B)*Cos[2*(e + f*x)])*Sin[e + f*x])*(Cos[4*e + 7*f*x] + I*Sin[4*e + 7*f*x])*Sqrt[c - I*c*Tan[e + f*x]]
)/(105*c^4*f*(Cos[f*x] + I*Sin[f*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.08, size = 105, normalized size = 0.7 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ({-iB{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{4\,{c}^{3} \left ( A-iB \right ) }{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}}-{\frac{c \left ( A-5\,iB \right ) }{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,{c}^{2} \left ( A-2\,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))^(7/2),x)

[Out]

2*I/f*a^3/c^3*(-I*B/(c-I*c*tan(f*x+e))^(1/2)-4/7*c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(7/2)-1/3*c*(A-5*I*B)/(c-I*c*t
an(f*x+e))^(3/2)+4/5*c^2*(A-2*I*B)/(c-I*c*tan(f*x+e))^(5/2))

________________________________________________________________________________________

Maxima [A]  time = 1.17561, size = 143, normalized size = 1.01 \begin{align*} -\frac{2 i \,{\left (105 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} B a^{3} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2}{\left (35 \, A - 175 i \, B\right )} a^{3} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}{\left (84 \, A - 168 i \, B\right )} a^{3} c^{2} +{\left (60 \, A - 60 i \, B\right )} a^{3} c^{3}\right )}}{105 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} 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))^(7/2),x, algorithm="maxima")

[Out]

-2/105*I*(105*I*(-I*c*tan(f*x + e) + c)^3*B*a^3 + (-I*c*tan(f*x + e) + c)^2*(35*A - 175*I*B)*a^3*c - (-I*c*tan
(f*x + e) + c)*(84*A - 168*I*B)*a^3*c^2 + (60*A - 60*I*B)*a^3*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*c^3*f)

________________________________________________________________________________________

Fricas [A]  time = 1.17568, size = 333, normalized size = 2.35 \begin{align*} \frac{\sqrt{2}{\left ({\left (-15 i \, A - 15 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-18 i \, A + 24 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A - 13 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-4 i \, A + 52 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-8 i \, A + 104 \, B\right )} a^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{210 \, c^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/210*sqrt(2)*((-15*I*A - 15*B)*a^3*e^(8*I*f*x + 8*I*e) + (-18*I*A + 24*B)*a^3*e^(6*I*f*x + 6*I*e) + (I*A - 13
*B)*a^3*e^(4*I*f*x + 4*I*e) + (-4*I*A + 52*B)*a^3*e^(2*I*f*x + 2*I*e) + (-8*I*A + 104*B)*a^3)*sqrt(c/(e^(2*I*f
*x + 2*I*e) + 1))/(c^4*f)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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{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))^3*(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)^3/(-I*c*tan(f*x + e) + c)^(7/2), x)