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

Optimal. Leaf size=61 \[ \frac{2 (-1)^{3/4} a \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 i a \sqrt{d \tan (e+f x)}}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0727427, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3528, 3533, 205} \[ \frac{2 (-1)^{3/4} a \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 i a \sqrt{d \tan (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x)) \, dx &=\frac{2 i a \sqrt{d \tan (e+f x)}}{f}+\int \frac{-i a d+a d \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{2 i a \sqrt{d \tan (e+f x)}}{f}-\frac{\left (2 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-i a d^2-a d x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{2 (-1)^{3/4} a \sqrt{d} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 i a \sqrt{d \tan (e+f x)}}{f}\\ \end{align*}

Mathematica [A]  time = 0.782633, size = 85, normalized size = 1.39 \[ \frac{2 i a \sqrt{d \tan (e+f x)} \left (\sqrt{i \tan (e+f x)}-\tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.016, size = 341, normalized size = 5.6 \begin{align*}{\frac{2\,ia}{f}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{i}{4}}a\sqrt{2}}{f}\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{{\frac{i}{2}}a\sqrt{2}}{f}\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\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{ad\sqrt{2}}{4\,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{ad\sqrt{2}}{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{ad\sqrt{2}}{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((d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.01241, size = 655, normalized size = 10.74 \begin{align*} \frac{\sqrt{\frac{4 i \, a^{2} d}{f^{2}}} f \log \left (\frac{{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt{\frac{4 i \, a^{2} 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 )}}{a}\right ) - \sqrt{\frac{4 i \, a^{2} d}{f^{2}}} f \log \left (\frac{{\left (-2 i \, a d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt{\frac{4 i \, a^{2} 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 )}}{a}\right ) + 8 i \, a \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 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.1426, size = 115, normalized size = 1.89 \begin{align*} 2 \, a{\left (\frac{\sqrt{2} \sqrt{d} \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 )}{f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{i \, \sqrt{d \tan \left (f x + e\right )}}{f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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