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

Optimal. Leaf size=105 \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.181252, antiderivative size = 105, 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} \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac{4 a^2 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \]

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])^(3/2),x]

[Out]

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

Mathematica [A]  time = 5.7058, size = 116, normalized size = 1.1 \[ \frac{a^2 c \sec ^3(e+f x) \sqrt{c-i c \tan (e+f x)} (\sin (e-f x)+i \cos (e-f x)) (3 (11 B+7 i A) \sin (2 (e+f x))+(49 A-37 i B) \cos (2 (e+f x))+49 A-7 i B)}{105 f (\cos (f x)+i \sin (f x))^2} \]

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])^(3/2),x]

[Out]

(a^2*c*Sec[e + f*x]^3*(I*Cos[e - f*x] + Sin[e - f*x])*(49*A - (7*I)*B + (49*A - (37*I)*B)*Cos[2*(e + f*x)] + 3
*((7*I)*A + 11*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(105*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.066, size = 83, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({\frac{i}{7}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}+{\frac{-3\,iBc+Ac}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{ \left ( -2\,iBc+2\,Ac \right ) c}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{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))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.11722, size = 109, normalized size = 1.04 \begin{align*} -\frac{2 i \,{\left (15 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} B a^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (21 \, A - 63 i \, B\right )} a^{2} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (70 \, A - 70 i \, B\right )} a^{2} c^{2}\right )}}{105 \, 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))^(3/2),x, algorithm="maxima")

[Out]

-2/105*I*(15*I*(-I*c*tan(f*x + e) + c)^(7/2)*B*a^2 + (-I*c*tan(f*x + e) + c)^(5/2)*(21*A - 63*I*B)*a^2*c - (-I
*c*tan(f*x + e) + c)^(3/2)*(70*A - 70*I*B)*a^2*c^2)/(c^2*f)

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Fricas [A]  time = 1.30387, size = 331, normalized size = 3.15 \begin{align*} \frac{\sqrt{2}{\left ({\left (280 i \, A + 280 \, B\right )} a^{2} c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (392 i \, A + 56 \, B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (112 i \, A + 16 \, B\right )} a^{2} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \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))^(3/2),x, algorithm="fricas")

[Out]

1/105*sqrt(2)*((280*I*A + 280*B)*a^2*c*e^(4*I*f*x + 4*I*e) + (392*I*A + 56*B)*a^2*c*e^(2*I*f*x + 2*I*e) + (112
*I*A + 16*B)*a^2*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e
^(2*I*f*x + 2*I*e) + f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int A c \sqrt{- i c \tan{\left (e + f x \right )} + c}\, dx + \int A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int i A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int i A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int i B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int i B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx\right ) \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))**(3/2),x)

[Out]

a**2*(Integral(A*c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2,
x) + Integral(B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(B*c*sqrt(-I*c*tan(e + f*x) + c)*tan
(e + f*x)**3, x) + Integral(I*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(I*A*c*sqrt(-I*c*tan(
e + f*x) + c)*tan(e + f*x)**3, x) + Integral(I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(
I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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{3}{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))^(3/2),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)^(3/2), x)