3.162 \(\int \frac{(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{8 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

(8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(7/2)*f) - (((8*I)/5)*a^3)/(d^2*f*(d*T
an[e + f*x])^(3/2)) + (8*a^3)/(d^3*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f*(d*Tan[e +
f*x])^(5/2))

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Rubi [A]  time = 0.261356, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3553, 3591, 3529, 3533, 205} \[ \frac{8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{8 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

(8*(-1)^(3/4)*a^3*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(d^(7/2)*f) - (((8*I)/5)*a^3)/(d^2*f*(d*T
an[e + f*x])^(3/2)) + (8*a^3)/(d^3*f*Sqrt[d*Tan[e + f*x]]) - (2*(a^3 + I*a^3*Tan[e + f*x]))/(5*d*f*(d*Tan[e +
f*x])^(5/2))

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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 \frac{(a+i a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 \int \frac{(a+i a \tan (e+f x)) \left (-6 i a^2 d+4 a^2 d \tan (e+f x)\right )}{(d \tan (e+f x))^{5/2}} \, dx}{5 d^2}\\ &=-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 \int \frac{10 a^3 d^2+10 i a^3 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{5 d^4}\\ &=-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac{8 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 \int \frac{10 i a^3 d^3-10 a^3 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{5 d^6}\\ &=-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac{8 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}+\frac{\left (80 a^6\right ) \operatorname{Subst}\left (\int \frac{1}{10 i a^3 d^4+10 a^3 d^3 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{8 (-1)^{3/4} a^3 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}-\frac{8 i a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}+\frac{8 a^3}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [B]  time = 8.38728, size = 377, normalized size = 2.86 \[ \frac{\sin ^3(e+f x) \tan (e+f x) (a+i a \tan (e+f x))^3 \left (\csc (e) \left (\frac{2}{5} \cos (3 e)-\frac{2}{5} i \sin (3 e)\right ) \sin (f x) \csc ^3(e+f x)+\csc (e) (\cos (e)+5 i \sin (e)) \left (-\frac{2}{5} \cos (3 e)+\frac{2}{5} i \sin (3 e)\right ) \csc ^2(e+f x)+\csc (e) \left (-\frac{42}{5} \cos (3 e)+\frac{42}{5} i \sin (3 e)\right ) \sin (f x) \csc (e+f x)+\csc (e) (21 \cos (e)+5 i \sin (e)) \left (\frac{2}{5} \cos (3 e)-\frac{2}{5} i \sin (3 e)\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (d \tan (e+f x))^{7/2}}-\frac{8 i e^{-3 i e} \sqrt{-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \cos ^3(e+f x) \tan ^{\frac{7}{2}}(e+f x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (a+i a \tan (e+f x))^3}{f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} (\cos (f x)+i \sin (f x))^3 (d \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]

[Out]

((Csc[e]*(21*Cos[e] + (5*I)*Sin[e])*((2*Cos[3*e])/5 - ((2*I)/5)*Sin[3*e]) + Csc[e]*Csc[e + f*x]^2*(Cos[e] + (5
*I)*Sin[e])*((-2*Cos[3*e])/5 + ((2*I)/5)*Sin[3*e]) + Csc[e]*Csc[e + f*x]^3*((2*Cos[3*e])/5 - ((2*I)/5)*Sin[3*e
])*Sin[f*x] + Csc[e]*Csc[e + f*x]*((-42*Cos[3*e])/5 + ((42*I)/5)*Sin[3*e])*Sin[f*x])*Sin[e + f*x]^3*Tan[e + f*
x]*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[f*x] + I*Sin[f*x])^3*(d*Tan[e + f*x])^(7/2)) - ((8*I)*Sqrt[((-I)*(-1 + E^
((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)
))]]*Cos[e + f*x]^3*Tan[e + f*x]^(7/2)*(a + I*a*Tan[e + f*x])^3)/(E^((3*I)*e)*Sqrt[(-1 + E^((2*I)*(e + f*x)))/
(1 + E^((2*I)*(e + f*x)))]*f*(Cos[f*x] + I*Sin[f*x])^3*(d*Tan[e + f*x])^(7/2))

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Maple [B]  time = 0.026, size = 414, normalized size = 3.1 \begin{align*}{\frac{-2\,i{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{a}^{3}}{5\,fd} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}+8\,{\frac{{a}^{3}}{{d}^{3}f\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{i{a}^{3}\sqrt{2}}{f{d}^{4}}\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{2\,i{a}^{3}\sqrt{2}}{f{d}^{4}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{2\,i{a}^{3}\sqrt{2}}{f{d}^{4}}\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}^{3}\sqrt{2}}{{d}^{3}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}}}}}+2\,{\frac{{a}^{3}\sqrt{2}}{{d}^{3}f\sqrt [4]{{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-2\,{\frac{{a}^{3}\sqrt{2}}{{d}^{3}f\sqrt [4]{{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x)

[Out]

-2*I/f*a^3/d^2/(d*tan(f*x+e))^(3/2)-2/5/f*a^3/d/(d*tan(f*x+e))^(5/2)+8*a^3/d^3/f/(d*tan(f*x+e))^(1/2)-I/f*a^3/
d^4*(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)))-2*I/f*a^3/d^4*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(
1/4)*(d*tan(f*x+e))^(1/2)+1)+2*I/f*a^3/d^4*(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^3/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)))+2/f*a^3/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)-2/f*a^3/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((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.15723, size = 1249, normalized size = 9.46 \begin{align*} \frac{5 \,{\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{4} 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}} \sqrt{\frac{64 i \, a^{6}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 5 \,{\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt{\frac{64 i \, a^{6}}{d^{7} f^{2}}} \log \left (\frac{{\left (-8 i \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{4} 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}} \sqrt{\frac{64 i \, a^{6}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) +{\left (208 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 96 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 176 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 128 i \, a^{3}\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}}}{20 \,{\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/20*(5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(6
4*I*a^6/(d^7*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (I*d^4*f*e^(2*I*f*x + 2*I*e) + I*d^4*f)*sqrt((-I*
d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(64*I*a^6/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) -
5*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(64*I*a^
6/(d^7*f^2))*log(1/4*(-8*I*a^3*d*e^(2*I*f*x + 2*I*e) + (-I*d^4*f*e^(2*I*f*x + 2*I*e) - I*d^4*f)*sqrt((-I*d*e^(
2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(64*I*a^6/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^3) + (208*
I*a^3*e^(6*I*f*x + 6*I*e) - 96*I*a^3*e^(4*I*f*x + 4*I*e) - 176*I*a^3*e^(2*I*f*x + 2*I*e) + 128*I*a^3)*sqrt((-I
*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*
I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*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((a+I*a*tan(f*x+e))**3/(d*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.26987, size = 188, normalized size = 1.42 \begin{align*} \frac{8 i \, \sqrt{2} a^{3} \arctan \left (\frac{16 i \, \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 )}{d^{\frac{7}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{40 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 10 i \, a^{3} d^{2} \tan \left (f x + e\right ) - 2 \, a^{3} d^{2}}{5 \, \sqrt{d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

8*I*sqrt(2)*a^3*arctan(16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)
))/(d^(7/2)*f*(-I*d/sqrt(d^2) + 1)) + 1/5*(40*a^3*d^2*tan(f*x + e)^2 - 10*I*a^3*d^2*tan(f*x + e) - 2*a^3*d^2)/
(sqrt(d*tan(f*x + e))*d^5*f*tan(f*x + e)^2)