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

Optimal. Leaf size=60 \[ \frac{2 a (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \]

[Out]

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

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Rubi [A]  time = 0.0989737, antiderivative size = 60, 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} \[ \frac{2 a (B+i A) \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.36431, size = 45, normalized size = 0.75 \[ \frac{2 a \sqrt{c-i c \tan (e+f x)} (3 i A+i B \tan (e+f x)+2 B)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.069, size = 66, normalized size = 1.1 \begin{align*}{\frac{2\,ia}{cf} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}-iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\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))*(A+B*tan(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.19481, size = 66, normalized size = 1.1 \begin{align*} \frac{2 i \,{\left (i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} B a + \sqrt{-i \, c \tan \left (f x + e\right ) + c}{\left (3 \, A - 3 i \, B\right )} a c\right )}}{3 \, c 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))*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.10666, size = 177, normalized size = 2.95 \begin{align*} \frac{\sqrt{2}{\left ({\left (6 i \, A + 6 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \,{\left (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))*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*((6*I*A + 6*B)*a*e^(2*I*f*x + 2*I*e) + (6*I*A + 2*B)*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(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 \left (\int A \sqrt{- i c \tan{\left (e + f x \right )} + c}\, dx + \int B \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int i A \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int 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))*(A+B*tan(f*x+e)),x)

[Out]

a*(Integral(A*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Inte
gral(I*A*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(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 )} \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))*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

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