3.135 \(\int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx\)

Optimal. Leaf size=107 \[ -\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.160267, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3528, 3533, 205} \[ -\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(-2*(-1)^(3/4)*a*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((2*I)*a*d^2*Sqrt[d*Tan[e + f*
x]])/f + (2*a*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((2*I)/5)*a*(d*Tan[e + f*x])^(5/2))/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 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))^{5/2} (a+i a \tan (e+f x)) \, dx &=\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int (d \tan (e+f x))^{3/2} (-i a d+a d \tan (e+f x)) \, dx\\ &=\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \sqrt{d \tan (e+f x)} \left (-a d^2-i a d^2 \tan (e+f x)\right ) \, dx\\ &=-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \frac{i a d^3-a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}-\frac{\left (2 a^2 d^6\right ) \operatorname{Subst}\left (\int \frac{1}{i a d^4+a d^3 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}

Mathematica [A]  time = 2.45962, size = 125, normalized size = 1.17 \[ \frac{a d^2 \sqrt{d \tan (e+f x)} \left (30 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+\sqrt{i \tan (e+f x)} \sec ^2(e+f x) (5 \sin (2 (e+f x))-18 i \cos (2 (e+f x))-12 i)\right )}{15 f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x]),x]

[Out]

(a*d^2*((30*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + Sec[e + f*x]^2*(-12*I - (
18*I)*Cos[2*(e + f*x)] + 5*Sin[2*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(15*f*Sqrt[I*Tan[e +
f*x]])

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Maple [B]  time = 0.026, size = 393, normalized size = 3.7 \begin{align*}{\frac{{\frac{2\,i}{5}}a}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,ad}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2\,ia{d}^{2}}{f}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{{\frac{i}{4}}a{d}^{2}\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{d}^{2}\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{d}^{2}\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{d}^{3}\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{a{d}^{3}\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{a{d}^{3}\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))^(5/2)*(a+I*a*tan(f*x+e)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.06221, size = 1013, normalized size = 9.47 \begin{align*} -\frac{15 \, \sqrt{\frac{4 i \, a^{2} d^{5}}{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 (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 )}}{a d^{2}}\right ) - 15 \, \sqrt{\frac{4 i \, a^{2} d^{5}}{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 (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 )}}{a d^{2}}\right ) -{\left (-184 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 192 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 104 i \, a d^{2}\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))^(5/2)*(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt(4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3*e^(2*I*f
*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(I*f*e^(2*I*f*x + 2*I*e) + I*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*d^2)) - 15*sqrt(4*I*a^2*d^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*
e^(2*I*f*x + 2*I*e) + f)*log((-2*I*a*d^3*e^(2*I*f*x + 2*I*e) + sqrt(4*I*a^2*d^5/f^2)*(-I*f*e^(2*I*f*x + 2*I*e)
- I*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*d^2)) - (-18
4*I*a*d^2*e^(4*I*f*x + 4*I*e) - 192*I*a*d^2*e^(2*I*f*x + 2*I*e) - 104*I*a*d^2)*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2)*(a+I*a*tan(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.18068, size = 205, normalized size = 1.92 \begin{align*} -\frac{2}{15} \, a d{\left (\frac{15 \, \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{-3 i \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right )^{2} - 5 \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right ) + 15 i \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4}}{d^{5} f^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-2/15*a*d*(15*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)) + (-3*I*sqrt(d*tan(f*x + e))*d^6*f^4*tan(f*x + e)^2 - 5*sqrt(d*tan(f*x +
e))*d^6*f^4*tan(f*x + e) + 15*I*sqrt(d*tan(f*x + e))*d^6*f^4)/(d^5*f^5))