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

Optimal. Leaf size=58 \[ -\frac{2 i a^2 \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{4 i a^2}{f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.147839, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ -\frac{2 i a^2 \sqrt{c-i c \tan (e+f x)}}{c f}-\frac{4 i a^2}{f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

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))^2}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\left (a^2 c^2\right ) \int \frac{\sec ^4(e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{c-x}{(c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \left (\frac{2 c}{(c+x)^{3/2}}-\frac{1}{\sqrt{c+x}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=-\frac{4 i a^2}{f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i a^2 \sqrt{c-i c \tan (e+f x)}}{c f}\\ \end{align*}

Mathematica [A]  time = 1.9878, size = 91, normalized size = 1.57 \[ \frac{2 a^2 \sqrt{c-i c \tan (e+f x)} (-2 \sin (2 e)-2 i \cos (2 e)+\sin (2 f x)-i \cos (2 f x)) (\cos (e+f x)+i \sin (e+f x))^2}{c f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 45, normalized size = 0.8 \begin{align*}{\frac{-2\,i{a}^{2}}{cf} \left ( \sqrt{c-ic\tan \left ( fx+e \right ) }+2\,{\frac{c}{\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/(c-I*c*tan(f*x+e))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.44841, size = 61, normalized size = 1.05 \begin{align*} -\frac{2 i \,{\left (\sqrt{-i \, c \tan \left (f x + e\right ) + c} a^{2} + \frac{2 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.39956, size = 122, normalized size = 2.1 \begin{align*} \frac{\sqrt{2}{\left (-2 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, a^{2}\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))^2/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

sqrt(2)*(-2*I*a^2*e^(2*I*f*x + 2*I*e) - 4*I*a^2)*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^{2} \left (\int - \frac{\tan ^{2}{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{2 i \tan{\left (e + f x \right )}}{\sqrt{- i c \tan{\left (e + f x \right )} + c}}\, dx + \int \frac{1}{\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/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

a**2*(Integral(-tan(e + f*x)**2/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(2*I*tan(e + f*x)/sqrt(-I*c*tan(e +
f*x) + c), x) + Integral(1/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 (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/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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