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

Optimal. Leaf size=101 \[ -\frac{2 a^2 (3 B+i A) \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{4 a^2 (B+i A)}{f \sqrt{c-i c \tan (e+f x)}}+\frac{2 a^2 B (c-i c \tan (e+f x))^{3/2}}{3 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.168383, antiderivative size = 101, 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} \[ -\frac{2 a^2 (3 B+i A) \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{4 a^2 (B+i A)}{f \sqrt{c-i c \tan (e+f x)}}+\frac{2 a^2 B (c-i c \tan (e+f x))^{3/2}}{3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 4.86747, size = 138, normalized size = 1.37 \[ \frac{a^2 \sqrt{c-i c \tan (e+f x)} (\sin (e+3 f x)-i \cos (e+3 f x)) (A+B \tan (e+f x)) ((-7 B-3 i A) \sin (2 (e+f x))+(9 A-13 i B) \cos (2 (e+f x))+9 A-15 i B)}{3 c f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(9*A - (15*I)*B + (9*A - (13*I)*B)*Cos[2*(e + f*x)] + ((-3*I)*A - 7*B)*Sin[2*(e + f*x)])*((-I)*Cos[e + 3*
f*x] + Sin[e + 3*f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*c*f*(Cos[f*x] + I*Sin[f*x])^2*(A*Co
s[e + f*x] + B*Sin[e + f*x]))

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Maple [A]  time = 0.117, size = 93, normalized size = 0.9 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}-3\,iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\sqrt{c-ic\tan \left ( fx+e \right ) }+2\,{\frac{{c}^{2} \left ( A-iB \right ) }{\sqrt{c-ic\tan \left ( fx+e \right ) }}} \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))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.81695, size = 113, normalized size = 1.12 \begin{align*} -\frac{2 i \,{\left (\frac{3 \,{\left (2 \, A - 2 i \, B\right )} a^{2} c}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}} + \frac{i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} B a^{2} + \sqrt{-i \, c \tan \left (f x + e\right ) + c}{\left (3 \, A - 9 i \, B\right )} a^{2} c}{c}\right )}}{3 \, c 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))^(1/2),x, algorithm="maxima")

[Out]

-2/3*I*(3*(2*A - 2*I*B)*a^2*c/sqrt(-I*c*tan(f*x + e) + c) + (I*(-I*c*tan(f*x + e) + c)^(3/2)*B*a^2 + sqrt(-I*c
*tan(f*x + e) + c)*(3*A - 9*I*B)*a^2*c)/c)/(c*f)

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Fricas [A]  time = 1.11467, size = 251, normalized size = 2.49 \begin{align*} \frac{\sqrt{2}{\left ({\left (-6 i \, A - 6 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-18 i \, A - 30 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-12 i \, A - 20 \, B\right )} a^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \,{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c 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))^(1/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*((-6*I*A - 6*B)*a^2*e^(4*I*f*x + 4*I*e) + (-18*I*A - 30*B)*a^2*e^(2*I*f*x + 2*I*e) + (-12*I*A - 20
*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(2*I*f*x + 2*I*e) + c*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{A}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int - \frac{A \tan ^{2}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{B \tan{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int - \frac{B \tan ^{3}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{2 i A \tan{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{2 i B \tan ^{2}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, 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))**(1/2),x)

[Out]

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

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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}}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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