3.1157 \(\int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=82 \[ -\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

[Out]

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

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Rubi [A]  time = 0.111601, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3544, 208} \[ -\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

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

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 2.47423, size = 147, normalized size = 1.79 \[ -\frac{i e^{-i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \sqrt{a+i a \tan (e+f x)} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]],x]

[Out]

((-I)*Sqrt[1 + E^((2*I)*(e + f*x))]*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[
c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[c - I*d]*E
^(I*(e + f*x))*f)

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Maple [B]  time = 0.115, size = 209, normalized size = 2.6 \begin{align*} -{\frac{ \left ( id\tan \left ( fx+e \right ) -ic+c\tan \left ( fx+e \right ) +d \right ) a\sqrt{2}}{2\,f \left ( -\tan \left ( fx+e \right ) +i \right ) \left ( ic-d \right ) }\ln \left ({\frac{1}{\tan \left ( fx+e \right ) +i} \left ( 3\,ac+ia\tan \left ( fx+e \right ) c-iad+3\,a\tan \left ( fx+e \right ) d+2\,\sqrt{2}\sqrt{-a \left ( id-c \right ) }\sqrt{a \left ( c+d\tan \left ( fx+e \right ) \right ) \left ( 1+i\tan \left ( fx+e \right ) \right ) } \right ) } \right ) \sqrt{c+d\tan \left ( fx+e \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{-a \left ( id-c \right ) }}}{\frac{1}{\sqrt{a \left ( c+d\tan \left ( fx+e \right ) \right ) \left ( 1+i\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))^(1/2)/(c+d*tan(f*x+e))^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.42481, size = 767, normalized size = 9.35 \begin{align*} \frac{1}{2} \, \sqrt{-\frac{2 i \, a}{{\left (i \, c + d\right )} f^{2}}} \log \left ({\left ({\left (i \, c + d\right )} f \sqrt{-\frac{2 i \, a}{{\left (i \, c + d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{2} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - \frac{1}{2} \, \sqrt{-\frac{2 i \, a}{{\left (i \, c + d\right )} f^{2}}} \log \left ({\left ({\left (-i \, c - d\right )} f \sqrt{-\frac{2 i \, a}{{\left (i \, c + d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{2} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-2*I*a/((I*c + d)*f^2))*log(((I*c + d)*f*sqrt(-2*I*a/((I*c + d)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*s
qrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^
(2*I*f*x + 2*I*e) + 1)*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)) - 1/2*sqrt(-2*I*a/((I*c + d)*f^2))*log(((-I*c -
d)*f*sqrt(-2*I*a/((I*c + d)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)
/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)*e^(I*f*x + I*e))*e^(-2
*I*f*x - 2*I*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out