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

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

[Out]

(2*(-1)^(3/4)*a*d^(5/2)*ArcTanh[((-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 = 0.147094, 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, 208} \[ \frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 (-1)^{3/4} a d^{5/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}-\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f} \]

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)*ArcTanh[((-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 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))^{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} \tanh ^{-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 = 0.533973, size = 90, normalized size = 0.84 \[ \frac{2 a (d \tan (e+f x))^{5/2} \left (15 (-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (e+f x)}\right )+\sqrt{\tan (e+f x)} \left (-3 i \tan ^2(e+f x)+5 \tan (e+f x)+15 i\right )\right )}{15 f \tan ^{\frac{5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(d*Tan[e + f*x])^(5/2)*(15*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + Sqrt[Tan[e + f*x]]*(15*I +
5*Tan[e + f*x] - (3*I)*Tan[e + f*x]^2)))/(15*f*Tan[e + f*x]^(5/2))

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Maple [B]  time = 0.018, 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 = 2.15146, size = 925, normalized size = 8.64 \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 \, a d^{3} + \sqrt{-\frac{4 i \, a^{2} d^{5}}{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 ) - 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 \, a d^{3} - \sqrt{-\frac{4 i \, a^{2} d^{5}}{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 ) +{\left (104 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 )} + 184 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*a*d^3 + sqrt(-4*I
*a^2*d^5/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) - 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
*a*d^3 - sqrt(-4*I*a^2*d^5/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) + (104*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) +
184*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 = 1.20038, 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))