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

Optimal. Leaf size=62 \[ \frac{2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.105502, 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} \[ \frac{2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}} \]

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])^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 7.85796, size = 100, normalized size = 1.61 \[ \frac{2 a \cos ^2(e+f x) (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\cos (3 e+4 f x)+i \sin (3 e+4 f x)) ((2 B-3 i A) \cos (e+f x)-5 i B \sin (e+f x))}{15 c^3 f} \]

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])^(5/2),x]

[Out]

(2*a*Cos[e + f*x]^2*(Cos[f*x] - I*Sin[f*x])*(((-3*I)*A + 2*B)*Cos[e + f*x] - (5*I)*B*Sin[e + f*x])*(Cos[3*e +
4*f*x] + I*Sin[3*e + 4*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f)

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 53, normalized size = 0.9 \begin{align*}{\frac{2\,ia}{cf} \left ({-{\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{c \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))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.14671, size = 63, normalized size = 1.02 \begin{align*} -\frac{2 i \,{\left (5 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a +{\left (3 \, A - 3 i \, B\right )} a c\right )}}{15 \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{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))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.17192, size = 258, normalized size = 4.16 \begin{align*} \frac{\sqrt{2}{\left ({\left (-3 i \, A - 3 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-9 i \, A + B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-9 i \, A + 11 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-3 i \, A + 7 \, B\right )} a\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/60*sqrt(2)*((-3*I*A - 3*B)*a*e^(6*I*f*x + 6*I*e) + (-9*I*A + B)*a*e^(4*I*f*x + 4*I*e) + (-9*I*A + 11*B)*a*e^
(2*I*f*x + 2*I*e) + (-3*I*A + 7*B)*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

________________________________________________________________________________________

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

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{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))*(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)/(-I*c*tan(f*x + e) + c)^(5/2), x)