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

Optimal. Leaf size=140 \[ -\frac{4 (-1)^{3/4} a^2 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{4 i a^2 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

(-4*(-1)^(3/4)*a^2*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((4*I)*a^2*d^2*Sqrt[d*Tan[e
+ f*x]])/f + (4*a^2*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((4*I)/5)*a^2*(d*Tan[e + f*x])^(5/2))/f - (2*a^2*(d*Tan
[e + f*x])^(7/2))/(7*d*f)

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Rubi [A]  time = 0.217504, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3543, 3528, 3533, 205} \[ -\frac{4 (-1)^{3/4} a^2 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{4 i a^2 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(-1)^(3/4)*a^2*d^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ((4*I)*a^2*d^2*Sqrt[d*Tan[e
+ f*x]])/f + (4*a^2*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (((4*I)/5)*a^2*(d*Tan[e + f*x])^(5/2))/f - (2*a^2*(d*Tan
[e + f*x])^(7/2))/(7*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))^{5/2} (a+i a \tan (e+f x))^2 \, dx &=-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\int (d \tan (e+f x))^{5/2} \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \, dx\\ &=\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\int (d \tan (e+f x))^{3/2} \left (-2 i a^2 d+2 a^2 d \tan (e+f x)\right ) \, dx\\ &=\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\int \sqrt{d \tan (e+f x)} \left (-2 a^2 d^2-2 i a^2 d^2 \tan (e+f x)\right ) \, dx\\ &=-\frac{4 i a^2 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}+\int \frac{2 i a^2 d^3-2 a^2 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=-\frac{4 i a^2 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}-\frac{\left (8 a^4 d^6\right ) \operatorname{Subst}\left (\int \frac{1}{2 i a^2 d^4+2 a^2 d^3 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{4 (-1)^{3/4} a^2 d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{4 i a^2 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{4 a^2 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (d \tan (e+f x))^{7/2}}{7 d f}\\ \end{align*}

Mathematica [A]  time = 3.01511, size = 145, normalized size = 1.04 \[ \frac{a^2 d^2 \sqrt{d \tan (e+f x)} \left (840 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )-i \sqrt{i \tan (e+f x)} \sec ^3(e+f x) (25 i \sin (e+f x)+85 i \sin (3 (e+f x))+588 \cos (e+f x)+252 \cos (3 (e+f x)))\right )}{210 f \sqrt{i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*d^2*((840*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] - I*Sec[e + f*x]^3*(588*
Cos[e + f*x] + 252*Cos[3*(e + f*x)] + (25*I)*Sin[e + f*x] + (85*I)*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]])*Sqr
t[d*Tan[e + f*x]])/(210*f*Sqrt[I*Tan[e + f*x]])

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Maple [B]  time = 0.019, size = 431, normalized size = 3.1 \begin{align*} -{\frac{2\,{a}^{2}}{7\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{{\frac{4\,i}{5}}{a}^{2}}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{4\,{a}^{2}d}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{4\,i{a}^{2}{d}^{2}}{f}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{{\frac{i}{2}}{a}^{2}{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{i{a}^{2}{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{i{a}^{2}{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}^{2}{d}^{3}\sqrt{2}}{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{{a}^{2}{d}^{3}\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{{a}^{2}{d}^{3}\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))^(5/2)*(a+I*a*tan(f*x+e))^2,x)

[Out]

-2/7*a^2*(d*tan(f*x+e))^(7/2)/d/f+4/5*I*a^2*(d*tan(f*x+e))^(5/2)/f+4/3*a^2*d*(d*tan(f*x+e))^(3/2)/f-4*I*a^2*d^
2*(d*tan(f*x+e))^(1/2)/f+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)-1/2/f*a^2*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/f*a^2*
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/f*a^2*d^3/(d^2)^(1/4)*2^(1/2)*arc
tan(-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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.38118, size = 1211, normalized size = 8.65 \begin{align*} -\frac{105 \, \sqrt{\frac{16 i \, a^{4} d^{5}}{f^{2}}}{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{16 i \, a^{4} 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 )}}{2 \, a^{2} d^{2}}\right ) - 105 \, \sqrt{\frac{16 i \, a^{4} d^{5}}{f^{2}}}{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-4 i \, a^{2} d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{16 i \, a^{4} 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 )}}{2 \, a^{2} d^{2}}\right ) -{\left (-2696 i \, a^{2} d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 4904 i \, a^{2} d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 4504 i \, a^{2} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 1336 i \, a^{2} 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}}}{420 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, 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))^2,x, algorithm="fricas")

[Out]

-1/420*(105*sqrt(16*I*a^4*d^5/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e)
+ f)*log(1/2*(-4*I*a^2*d^3*e^(2*I*f*x + 2*I*e) + sqrt(16*I*a^4*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^2*d^2)) - 105*sqrt(16*I*a^
4*d^5/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d
^3*e^(2*I*f*x + 2*I*e) + sqrt(16*I*a^4*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^2*d^2)) - (-2696*I*a^2*d^2*e^(6*I*f*x + 6*I*e) -
4904*I*a^2*d^2*e^(4*I*f*x + 4*I*e) - 4504*I*a^2*d^2*e^(2*I*f*x + 2*I*e) - 1336*I*a^2*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^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2848, size = 259, normalized size = 1.85 \begin{align*} -\frac{4 \, \sqrt{2} a^{2} d^{\frac{5}{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{30 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{9} f^{6} \tan \left (f x + e\right )^{3} - 84 i \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{9} f^{6} \tan \left (f x + e\right )^{2} - 140 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{9} f^{6} \tan \left (f x + e\right ) + 420 i \, \sqrt{d \tan \left (f x + e\right )} a^{2} d^{9} f^{6}}{105 \, d^{7} f^{7}} \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))^2,x, algorithm="giac")

[Out]

-4*sqrt(2)*a^2*d^(5/2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqr
t(d)))/(f*(I*d/sqrt(d^2) + 1)) - 1/105*(30*sqrt(d*tan(f*x + e))*a^2*d^9*f^6*tan(f*x + e)^3 - 84*I*sqrt(d*tan(f
*x + e))*a^2*d^9*f^6*tan(f*x + e)^2 - 140*sqrt(d*tan(f*x + e))*a^2*d^9*f^6*tan(f*x + e) + 420*I*sqrt(d*tan(f*x
+ e))*a^2*d^9*f^6)/(d^7*f^7)