3.1020 \(\int \frac{1}{\sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=44 \[ \frac{\tan (e+f x)}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0998342, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {3523, 39} \[ \frac{\tan (e+f x)}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

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

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

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

Mathematica [A]  time = 1.26144, size = 64, normalized size = 1.45 \[ \frac{\sin (e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (e+f x)+i \sin (e+f x))}{c f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

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

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Maple [B]  time = 0.089, size = 82, normalized size = 1.9 \begin{align*}{\frac{ \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \tan \left ( fx+e \right ) }{fac \left ( \tan \left ( fx+e \right ) +i \right ) ^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.80062, size = 22, normalized size = 0.5 \begin{align*} \frac{\sin \left (f x + e\right )}{\sqrt{a} \sqrt{c} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(f*x + e)/(sqrt(a)*sqrt(c)*f)

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Fricas [A]  time = 1.29461, size = 171, normalized size = 3.89 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-i \, e^{\left (4 i \, f x + 4 i \, e\right )} + i\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(4*I*f*x + 4*I*e) + I)*e^(-I*f*x
- I*e)/(a*c*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a*(I*tan(e + f*x) + 1))*sqrt(-c*(I*tan(e + f*x) - 1))), x)

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

[Out]

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