3.152 \(\int \frac{(a+i a \tan (e+f x))^2}{\sqrt{d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{2 a^2 \sqrt{d \tan (e+f x)}}{d f} \]

[Out]

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

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Rubi [A]  time = 0.0982788, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3543, 3533, 205} \[ -\frac{4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{2 a^2 \sqrt{d \tan (e+f x)}}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

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

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))^2}{\sqrt{d \tan (e+f x)}} \, dx &=-\frac{2 a^2 \sqrt{d \tan (e+f x)}}{d f}+\int \frac{2 a^2+2 i a^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{d \tan (e+f x)}}{d f}+\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2 d-2 i a^2 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{d} f}-\frac{2 a^2 \sqrt{d \tan (e+f x)}}{d f}\\ \end{align*}

Mathematica [C]  time = 1.86292, size = 157, normalized size = 2.38 \[ -\frac{2 a^2 e^{-2 i (e+f x)} \sqrt{\tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (\tan (e+f x)+2 i \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{f \sqrt{-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sqrt[Tan[e + f*x]]*((2*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x
)))/(1 + E^((2*I)*(e + f*x)))]]*Sqrt[I*Tan[e + f*x]] + Tan[e + f*x]))/(E^((2*I)*(e + f*x))*Sqrt[((-I)*(-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 = 0.033, size = 362, normalized size = 5.5 \begin{align*} -2\,{\frac{{a}^{2}\sqrt{d\tan \left ( fx+e \right ) }}{df}}+{\frac{{a}^{2}\sqrt{2}}{2\,df}\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}^{2}\sqrt{2}}{df}\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}^{2}\sqrt{2}}{df}\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}{2}}{a}^{2}\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{i{a}^{2}\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{i{a}^{2}\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))^2/(d*tan(f*x+e))^(1/2),x)

[Out]

-2*a^2*(d*tan(f*x+e))^(1/2)/d/f+1/2/f*a^2/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/f*a^2/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/f*a^2/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*I/f*a^2/(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)))+I/f*a^2/(d^
2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-I/f*a^2/(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))^2/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 1.89995, size = 707, normalized size = 10.71 \begin{align*} \frac{d \sqrt{-\frac{16 i \, a^{4}}{d f^{2}}} f \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - d \sqrt{-\frac{16 i \, a^{4}}{d f^{2}}} f \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, a^{2} \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}}}{4 \, d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(d*sqrt(-16*I*a^4/(d*f^2))*f*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) + (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqr
t(-16*I*a^4/(d*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/a^
2) - d*sqrt(-16*I*a^4/(d*f^2))*f*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) - (d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqr
t(-16*I*a^4/(d*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/a^
2) - 8*a^2*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d*f)

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

[Out]

a**2*(Integral(1/sqrt(d*tan(e + f*x)), x) + Integral(-tan(e + f*x)**2/sqrt(d*tan(e + f*x)), x) + Integral(2*I*
tan(e + f*x)/sqrt(d*tan(e + f*x)), x))

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Giac [C]  time = 1.22488, size = 124, normalized size = 1.88 \begin{align*} \frac{4 i \, \sqrt{2} a^{2} \arctan \left (\frac{16 \, \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 )}} - \frac{2 \, \sqrt{d \tan \left (f x + e\right )} a^{2}}{d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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