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

Optimal. Leaf size=113 \[ \frac{4 \sqrt [4]{-1} a^2 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^2 d \sqrt{d \tan (e+f x)}}{f} \]

[Out]

(4*(-1)^(1/4)*a^2*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + (4*a^2*d*Sqrt[d*Tan[e + f*x]]
)/f + (((4*I)/3)*a^2*(d*Tan[e + f*x])^(3/2))/f - (2*a^2*(d*Tan[e + f*x])^(5/2))/(5*d*f)

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Rubi [A]  time = 0.173103, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3543, 3528, 3533, 205} \[ \frac{4 \sqrt [4]{-1} a^2 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^2 d \sqrt{d \tan (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + (4*a^2*d*Sqrt[d*Tan[e + f*x]]
)/f + (((4*I)/3)*a^2*(d*Tan[e + f*x])^(3/2))/f - (2*a^2*(d*Tan[e + f*x])^(5/2))/(5*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 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 (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2 \, dx &=-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int (d \tan (e+f x))^{3/2} \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \, dx\\ &=\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int \sqrt{d \tan (e+f x)} \left (-2 i a^2 d+2 a^2 d \tan (e+f x)\right ) \, dx\\ &=\frac{4 a^2 d \sqrt{d \tan (e+f x)}}{f}+\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int \frac{-2 a^2 d^2-2 i a^2 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{4 a^2 d \sqrt{d \tan (e+f x)}}{f}+\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac{\left (8 a^4 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{-2 a^2 d^3+2 i a^2 d^2 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{4 \sqrt [4]{-1} a^2 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{4 a^2 d \sqrt{d \tan (e+f x)}}{f}+\frac{4 i a^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 3.47719, size = 127, normalized size = 1.12 \[ \frac{a^2 d^2 \left (120 i \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+\sec ^2(e+f x) (-20 i \cos (2 (e+f x))+21 \tan (e+f x)+33 \sin (3 (e+f x)) \sec (e+f x)+20 i)\right )}{30 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*d^2*((120*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]]*Sqrt[I*Tan[e + f*x]] + S
ec[e + f*x]^2*(20*I - (20*I)*Cos[2*(e + f*x)] + 33*Sec[e + f*x]*Sin[3*(e + f*x)] + 21*Tan[e + f*x])))/(30*f*Sq
rt[d*Tan[e + f*x]])

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Maple [B]  time = 0.02, size = 403, normalized size = 3.6 \begin{align*} -{\frac{2\,{a}^{2}}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{{\frac{4\,i}{3}}{a}^{2}}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+4\,{\frac{{a}^{2}d\sqrt{d\tan \left ( fx+e \right ) }}{f}}-{\frac{{a}^{2}d\sqrt{2}}{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{{a}^{2}d\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{{a}^{2}d\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}^{2}{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{i{a}^{2}{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{i{a}^{2}{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))^2,x)

[Out]

-2/5*a^2*(d*tan(f*x+e))^(5/2)/d/f+4/3*I*a^2*(d*tan(f*x+e))^(3/2)/f+4*a^2*d*(d*tan(f*x+e))^(1/2)/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/(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/(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/(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((d*tan(f*x+e))^(3/2)*(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32855, size = 1017, normalized size = 9. \begin{align*} -\frac{15 \, \sqrt{-\frac{16 i \, a^{4} d^{3}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{-\frac{16 i \, a^{4} 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 )}}{2 \, a^{2} d}\right ) - 15 \, \sqrt{-\frac{16 i \, a^{4} d^{3}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt{-\frac{16 i \, a^{4} 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 )}}{2 \, a^{2} d}\right ) - 8 \,{\left (43 \, a^{2} d e^{\left (4 i \, f x + 4 i \, e\right )} + 54 \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + 23 \, a^{2} 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}}}{60 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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))^2,x, algorithm="fricas")

[Out]

-1/60*(15*sqrt(-16*I*a^4*d^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d^2*
e^(2*I*f*x + 2*I*e) + sqrt(-16*I*a^4*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)/(a^2*d)) - 15*sqrt(-16*I*a^4*d^3/f^2)*(f*e^(4*I*f*x + 4*I*e
) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d^2*e^(2*I*f*x + 2*I*e) - sqrt(-16*I*a^4*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)/(a^2
*d)) - 8*(43*a^2*d*e^(4*I*f*x + 4*I*e) + 54*a^2*d*e^(2*I*f*x + 2*I*e) + 23*a^2*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^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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

[Out]

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

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Giac [A]  time = 1.18859, size = 220, normalized size = 1.95 \begin{align*} -\frac{1}{15} \,{\left (\frac{60 i \, \sqrt{2} a^{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{6 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )^{2} - 20 i \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right ) - 60 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{10} f^{4}}{d^{10} f^{5}}\right )} d \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))^2,x, algorithm="giac")

[Out]

-1/15*(60*I*sqrt(2)*a^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)) + (6*sqrt(d*tan(f*x + e))*a^2*d^10*f^4*tan(f*x + e)^2 - 20*I*sqrt(d*ta
n(f*x + e))*a^2*d^10*f^4*tan(f*x + e) - 60*sqrt(d*tan(f*x + e))*a^2*d^10*f^4)/(d^10*f^5))*d