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

Optimal. Leaf size=103 \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac{4 a^2 (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.163603, antiderivative size = 103, 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} \[ -\frac{2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac{4 a^2 (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}+\frac{2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \]

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

Mathematica [A]  time = 4.49815, size = 83, normalized size = 0.81 \[ \frac{a^2 \sec ^2(e+f x) \sqrt{c-i c \tan (e+f x)} ((-5 A+9 i B) \sin (2 (e+f x))+(21 B+25 i A) \cos (2 (e+f x))+5 (3 B+5 i A))}{15 f} \]

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*Sec[e + f*x]^2*(5*((5*I)*A + 3*B) + ((25*I)*A + 21*B)*Cos[2*(e + f*x)] + (-5*A + (9*I)*B)*Sin[2*(e + f*x)
])*Sqrt[c - I*c*Tan[e + f*x]])/(15*f)

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Maple [A]  time = 0.072, size = 83, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{f{c}^{2}} \left ({\frac{i}{5}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+{\frac{-3\,iBc+Ac}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-2\, \left ( -iBc+Ac \right ) c\sqrt{c-ic\tan \left ( fx+e \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16914, size = 109, normalized size = 1.06 \begin{align*} -\frac{2 i \,{\left (3 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} B a^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (5 \, A - 15 i \, B\right )} a^{2} c - \sqrt{-i \, c \tan \left (f x + e\right ) + c}{\left (30 \, A - 30 i \, B\right )} a^{2} c^{2}\right )}}{15 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

-2/15*I*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*a^2 + (-I*c*tan(f*x + e) + c)^(3/2)*(5*A - 15*I*B)*a^2*c - sqrt(-
I*c*tan(f*x + e) + c)*(30*A - 30*I*B)*a^2*c^2)/(c^2*f)

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Fricas [A]  time = 1.15521, size = 282, normalized size = 2.74 \begin{align*} \frac{\sqrt{2}{\left ({\left (60 i \, A + 60 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (100 i \, A + 60 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (40 i \, A + 24 \, B\right )} a^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*((60*I*A + 60*B)*a^2*e^(4*I*f*x + 4*I*e) + (100*I*A + 60*B)*a^2*e^(2*I*f*x + 2*I*e) + (40*I*A + 2
4*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4*I*e) + 2*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 \sqrt{- i c \tan{\left (e + f x \right )} + c}\, dx + \int - A \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int B \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int - B \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int 2 i A \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int 2 i B \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x)
+ Integral(B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f
*x)**3, x) + Integral(2*I*A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(2*I*B*sqrt(-I*c*tan(e + f*
x) + c)*tan(e + f*x)**2, 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} \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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e)),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)