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

Optimal. Leaf size=118 \[ \frac{4 (-1)^{3/4} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{4 a^2}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.203235, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3542, 3529, 3533, 205} \[ \frac{4 (-1)^{3/4} a^2 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}+\frac{4 a^2}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3542

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
((b*c - a*d)^2*(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*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, 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))^2}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}}+\frac{\int \frac{2 i a^2 d-2 a^2 d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{\int \frac{-2 a^2 d^2-2 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{4 a^2}{d^3 f \sqrt{d \tan (e+f x)}}+\frac{\int \frac{-2 i a^2 d^3+2 a^2 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{4 a^2}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{\left (8 a^4\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 \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}-\frac{2 a^2}{5 d f (d \tan (e+f x))^{5/2}}-\frac{4 i a^2}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{4 a^2}{d^3 f \sqrt{d \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 7.7974, size = 381, normalized size = 3.23 \[ \frac{\sin ^2(e+f x) \tan ^2(e+f x) (a+i a \tan (e+f x))^2 \left (\csc (e) \left (\frac{2}{5} \cos (2 e)-\frac{2}{5} i \sin (2 e)\right ) \sin (f x) \csc ^3(e+f x)+\csc (e) (3 \cos (e)+10 i \sin (e)) \left (-\frac{2}{15} \cos (2 e)+\frac{2}{15} i \sin (2 e)\right ) \csc ^2(e+f x)+\csc (e) \left (-\frac{22}{5} \cos (2 e)+\frac{22}{5} i \sin (2 e)\right ) \sin (f x) \csc (e+f x)+\csc (e) (33 \cos (e)+10 i \sin (e)) \left (\frac{2}{15} \cos (2 e)-\frac{2}{15} i \sin (2 e)\right )\right )}{f (\cos (f x)+i \sin (f x))^2 (d \tan (e+f x))^{7/2}}-\frac{4 i e^{-2 i e} \sqrt{-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}} \cos ^2(e+f x) \tan ^{\frac{7}{2}}(e+f x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (a+i a \tan (e+f x))^2}{f \sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} (\cos (f x)+i \sin (f x))^2 (d \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Csc[e]*(33*Cos[e] + (10*I)*Sin[e])*((2*Cos[2*e])/15 - ((2*I)/15)*Sin[2*e]) + Csc[e]*Csc[e + f*x]^2*(3*Cos[e]
+ (10*I)*Sin[e])*((-2*Cos[2*e])/15 + ((2*I)/15)*Sin[2*e]) + Csc[e]*Csc[e + f*x]^3*((2*Cos[2*e])/5 - ((2*I)/5)
*Sin[2*e])*Sin[f*x] + Csc[e]*Csc[e + f*x]*((-22*Cos[2*e])/5 + ((22*I)/5)*Sin[2*e])*Sin[f*x])*Sin[e + f*x]^2*Ta
n[e + f*x]^2*(a + I*a*Tan[e + f*x])^2)/(f*(Cos[f*x] + I*Sin[f*x])^2*(d*Tan[e + f*x])^(7/2)) - ((4*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]^2*Tan[e + f*x]^(7/2)*(a + I*a*Tan[e + f*x])^2)/(E^((2*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])^2*(d*Tan[e + f*x])^(7/2))

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 414, normalized size = 3.5 \begin{align*} -{\frac{2\,{a}^{2}}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{{\frac{4\,i}{3}}{a}^{2}}{{d}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+4\,{\frac{{a}^{2}}{{d}^{3}f\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{{\frac{i}{2}}{a}^{2}\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{i{a}^{2}\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{i{a}^{2}\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}^{2}\sqrt{2}}{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}}}}}+{\frac{{a}^{2}\sqrt{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}^{2}\sqrt{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))^2/(d*tan(f*x+e))^(7/2),x)

[Out]

-2/5*a^2/d/f/(d*tan(f*x+e))^(5/2)-4/3*I*a^2/d^2/f/(d*tan(f*x+e))^(3/2)+4*a^2/d^3/f/(d*tan(f*x+e))^(1/2)-1/2*I/
f*a^2/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)))-I/f*a^2/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)+I/f*a^2/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/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)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan
(f*x+e))^(1/2)+1)

________________________________________________________________________________________

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))^2/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.22045, size = 1251, normalized size = 10.6 \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{16 i \, a^{4}}{d^{7} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\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{16 i \, a^{4}}{d^{7} f^{2}}} \log \left (\frac{{\left (-4 i \, a^{2} 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{16 i \, a^{4}}{d^{7} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) +{\left (344 i \, a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 88 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 248 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 184 i \, a^{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 (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))^2/(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(
16*I*a^4/(d^7*f^2))*log(1/2*(-4*I*a^2*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(16*I*a^4/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) -
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(16*I*
a^4/(d^7*f^2))*log(1/2*(-4*I*a^2*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(16*I*a^4/(d^7*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) + (34
4*I*a^2*e^(6*I*f*x + 6*I*e) - 88*I*a^2*e^(4*I*f*x + 4*I*e) - 248*I*a^2*e^(2*I*f*x + 2*I*e) + 184*I*a^2)*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)

________________________________________________________________________________________

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))**2/(d*tan(f*x+e))**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21342, size = 188, normalized size = 1.59 \begin{align*} -\frac{4 i \, \sqrt{2} a^{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{60 \, a^{2} d^{2} \tan \left (f x + e\right )^{2} - 20 i \, a^{2} d^{2} \tan \left (f x + e\right ) - 6 \, a^{2} d^{2}}{15 \, \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))^2/(d*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-4*I*sqrt(2)*a^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)) + 1/15*(60*a^2*d^2*tan(f*x + e)^2 - 20*I*a^2*d^2*tan(f*x + e) - 6*a^2*d^
2)/(sqrt(d*tan(f*x + e))*d^5*f*tan(f*x + e)^2)