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

Optimal. Leaf size=40 \[ -\frac{2 \sqrt [4]{-1} a \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f} \]

[Out]

(-2*(-1)^(1/4)*a*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f)

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Rubi [A]  time = 0.0415236, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3533, 205} \[ -\frac{2 \sqrt [4]{-1} a \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(1/4)*a*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f)

Rule 3533

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

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.724999, size = 87, normalized size = 2.17 \[ -\frac{2 i a \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )}{f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.032, size = 330, normalized size = 8.3 \begin{align*}{\frac{a\sqrt{2}}{4\,fd}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{a\sqrt{2}}{2\,fd}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{a\sqrt{2}}{2\,fd}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{\frac{i}{4}}a\sqrt{2}}{f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{\frac{i}{2}}a\sqrt{2}}{f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{\frac{i}{2}}a\sqrt{2}}{f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*a/d*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*
x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/2/f*a/d*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)
^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/2/f*a/d*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+
1)+1/4*I/f*a/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan
(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/2*I/f*a/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d
^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/2*I/f*a/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/
2)+1)

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 1.97573, size = 566, normalized size = 14.15 \begin{align*} \frac{1}{4} \, \sqrt{-\frac{4 i \, a^{2}}{d f^{2}}} \log \left (\frac{{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{4 i \, a^{2}}{d f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - \frac{1}{4} \, \sqrt{-\frac{4 i \, a^{2}}{d f^{2}}} \log \left (\frac{{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} -{\left (d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{4 i \, a^{2}}{d f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.17028, size = 90, normalized size = 2.25 \begin{align*} \frac{2 \, \sqrt{2} a \arctan \left (-\frac{16 i \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{-8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{\sqrt{d} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*a*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/
(sqrt(d)*f*(-I*d/sqrt(d^2) + 1))