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

Optimal. Leaf size=110 \[ \frac{2 (-1)^{3/4} a \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.170372, antiderivative size = 110, 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 = {3529, 3533, 205} \[ \frac{2 (-1)^{3/4} a \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 4.53736, size = 105, normalized size = 0.95 \[ -\frac{2 a \left (\cot (e+f x) (18 \cot (2 (e+f x))+5 i)-6 \csc ^2(e+f x)+15 \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{15 d^3 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(Cot[e + f*x]*(5*I + 18*Cot[2*(e + f*x)]) - 6*Csc[e + f*x]^2 + 15*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x))
)/(1 + E^((2*I)*(e + f*x)))]]*Sqrt[I*Tan[e + f*x]]))/(15*d^3*f*Sqrt[d*Tan[e + f*x]])

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Maple [B]  time = 0.025, size = 397, normalized size = 3.6 \begin{align*} -{\frac{2\,a}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{{\frac{2\,i}{3}}a}{{d}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{a}{{d}^{3}f\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{{\frac{i}{4}}a\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{{\frac{i}{2}}a\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{{\frac{i}{2}}a\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\sqrt{2}}{4\,{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}}}}}+{\frac{a\sqrt{2}}{2\,{d}^{3}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\sqrt{2}}{2\,{d}^{3}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((a+I*a*tan(f*x+e))/(d*tan(f*x+e))^(7/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.16867, size = 1211, normalized size = 11.01 \begin{align*} \frac{15 \,{\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{4 i \, a^{2}}{d^{7} f^{2}}} \log \left (\frac{{\left (-2 i \, a 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{4 i \, a^{2}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - 15 \,{\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{4 i \, a^{2}}{d^{7} f^{2}}} \log \left (\frac{{\left (-2 i \, a 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{4 i \, a^{2}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) +{\left (184 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 8 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 88 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} + 104 i \, a\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 (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))/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/60*(15*(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(
4*I*a^2/(d^7*f^2))*log((-2*I*a*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(4*I*a^2/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a) - 15*(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(4*I*a^2/(d^7*f^
2))*log((-2*I*a*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(4*I*a^2/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a) + (184*I*a*e^(6*I*f*x + 6
*I*e) - 8*I*a*e^(4*I*f*x + 4*I*e) - 88*I*a*e^(2*I*f*x + 2*I*e) + 104*I*a)*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))/(d*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27285, size = 176, normalized size = 1.6 \begin{align*} -\frac{2}{15} \, a{\left (\frac{15 i \, \sqrt{2} \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{15 \, d^{2} \tan \left (f x + e\right )^{2} - 5 i \, d^{2} \tan \left (f x + e\right ) - 3 \, d^{2}}{\sqrt{d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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