3.143 \(\int (d \tan (e+f x))^{3/2} (a-i a \tan (e+f x)) \, dx\)

Optimal. Leaf size=82 \[ \frac{2 \sqrt [4]{-1} a d^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 a d \sqrt{d \tan (e+f x)}}{f}-\frac{2 i a (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.106302, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3528, 3533, 208} \[ \frac{2 \sqrt [4]{-1} a d^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 a d \sqrt{d \tan (e+f x)}}{f}-\frac{2 i a (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)*(a - I*a*Tan[e + f*x]),x]

[Out]

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

Mathematica [A]  time = 0.229259, size = 78, normalized size = 0.95 \[ \frac{2 a (d \tan (e+f x))^{3/2} \left (3 \sqrt [4]{-1} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (e+f x)}\right )+(3-i \tan (e+f x)) \sqrt{\tan (e+f x)}\right )}{3 f \tan ^{\frac{3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)*(a - I*a*Tan[e + f*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 367, normalized size = 4.5 \begin{align*}{\frac{-{\frac{2\,i}{3}}a}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+2\,{\frac{ad\sqrt{d\tan \left ( fx+e \right ) }}{f}}-{\frac{ad\sqrt{2}}{4\,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{ad\sqrt{2}}{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}}{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}{4}}a{d}^{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{{\frac{i}{2}}a{d}^{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{{\frac{i}{2}}a{d}^{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((d*tan(f*x+e))^(3/2)*(a-I*a*tan(f*x+e)),x)

[Out]

-2/3*I*a*(d*tan(f*x+e))^(3/2)/f+2*a*d*(d*tan(f*x+e))^(1/2)/f-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/(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/(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/(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))^(3/2)*(a-I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.09311, size = 759, normalized size = 9.26 \begin{align*} \frac{3 \, \sqrt{\frac{4 i \, a^{2} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-2 i \, a d^{2} + \sqrt{\frac{4 i \, a^{2} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) - 3 \, \sqrt{\frac{4 i \, a^{2} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-2 i \, a d^{2} - \sqrt{\frac{4 i \, a^{2} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{f}\right ) + 16 \,{\left (a d e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, a d\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}}}{12 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)*(a-I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^2 + sqrt(4*I*a^2*d^3/f^2)*(f*e^(2*I*f*
x + 2*I*e) + f)*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)/f) - 3*
sqrt(4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^2 - sqrt(4*I*a^2*d^3/f^2)*(f*e^(2*I*f*x + 2*I*
e) + f)*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)/f) + 16*(a*d*e^
(2*I*f*x + 2*I*e) + 2*a*d)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(2*I*f*x + 2
*I*e) + f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a \left (\int - \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}\, dx + \int i \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \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))**(3/2)*(a-I*a*tan(f*x+e)),x)

[Out]

-a*(Integral(-(d*tan(e + f*x))**(3/2), x) + Integral(I*(d*tan(e + f*x))**(3/2)*tan(e + f*x), x))

________________________________________________________________________________________

Giac [A]  time = 1.27915, size = 157, normalized size = 1.91 \begin{align*} -\frac{2}{3} \, a{\left (-\frac{3 i \, \sqrt{2} d^{\frac{3}{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 )}{f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{i \, \sqrt{d \tan \left (f x + e\right )} d f^{2} \tan \left (f x + e\right ) - 3 \, \sqrt{d \tan \left (f x + e\right )} d f^{2}}{f^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)*(a-I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

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