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

Optimal. Leaf size=58 \[ -\frac{2 a (B+i A)}{f \sqrt{c-i c \tan (e+f x)}}-\frac{2 a B \sqrt{c-i c \tan (e+f x)}}{c f} \]

[Out]

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

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Rubi [A]  time = 0.100029, antiderivative size = 58, 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} \[ -\frac{2 a (B+i A)}{f \sqrt{c-i c \tan (e+f x)}}-\frac{2 a B \sqrt{c-i c \tan (e+f x)}}{c f} \]

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

Mathematica [A]  time = 2.52525, size = 82, normalized size = 1.41 \[ \frac{2 a (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\sin (e+2 f x)-i \cos (e+2 f x)) (-B \sin (e+f x)+(A-2 i B) \cos (e+f x))}{c f} \]

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

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Maple [A]  time = 0.133, size = 53, normalized size = 0.9 \begin{align*}{\frac{2\,ia}{cf} \left ( iB\sqrt{c-ic\tan \left ( fx+e \right ) }-{c \left ( A-iB \right ){\frac{1}{\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))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.1618, size = 65, normalized size = 1.12 \begin{align*} \frac{2 i \,{\left (i \, \sqrt{-i \, c \tan \left (f x + e\right ) + c} B a - \frac{{\left (A - i \, B\right )} a c}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\right )}}{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.06398, size = 136, normalized size = 2.34 \begin{align*} \frac{\sqrt{2}{\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, A - 3 \, B\right )} a\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{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))^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

a*(Integral(A/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(B*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Inte
gral(I*A*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(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 )}}{\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))*(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)/sqrt(-I*c*tan(f*x + e) + c), x)