\(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx\) [170]

Optimal result

Integrand size = 28, antiderivative size = 314 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}+\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \]

[Out]

(7/8-5/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(5/2)/f*2^(1/2)+(-7/8+5/8*I)*arctan(1+2^(1/2)*(
d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(5/2)/f*2^(1/2)+(7/16+5/16*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)
*tan(f*x+e))/a/d^(5/2)/f*2^(1/2)-(7/16+5/16*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a/d
^(5/2)/f*2^(1/2)+5/2*I/a/d^2/f/(d*tan(f*x+e))^(1/2)-7/6/a/d/f/(d*tan(f*x+e))^(3/2)+1/2/d/f/(d*tan(f*x+e))^(3/2
)/(a+I*a*tan(f*x+e))

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3633, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a d^{5/2} f}+\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a d^{5/2} f}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (a+i a \tan (e+f x)) (d \tan (e+f x))^{3/2}}-\frac {7}{6 a d f (d \tan (e+f x))^{3/2}} \]

[In]

Int[1/((d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])),x]

[Out]

((7/4 - (5*I)/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*d^(5/2)*f) - ((7/4 - (5*I)/4)*
ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*d^(5/2)*f) + ((7/8 + (5*I)/8)*Log[Sqrt[d] + Sqr
t[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*d^(5/2)*f) - ((7/8 + (5*I)/8)*Log[Sqrt[d] + Sqrt
[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*d^(5/2)*f) - 7/(6*a*d*f*(d*Tan[e + f*x])^(3/2)) +
((5*I)/2)/(a*d^2*f*Sqrt[d*Tan[e + f*x]]) + 1/(2*d*f*(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

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 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3633

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}-\frac {\int \frac {-\frac {7 a d}{2}+\frac {5}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{2 a^2 d} \\ & = -\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}-\frac {\int \frac {\frac {5}{2} i a d^2+\frac {7}{2} a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{2 a^2 d^3} \\ & = -\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}-\frac {\int \frac {\frac {7 a d^3}{2}-\frac {5}{2} i a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d^5} \\ & = -\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}-\frac {\text {Subst}\left (\int \frac {\frac {7 a d^4}{2}-\frac {5}{2} i a d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d^5 f} \\ & = -\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}-\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d^2 f}-\frac {\left (\frac {7}{4}+\frac {5 i}{4}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d^2 f} \\ & = -\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}--\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}--\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{8}-\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d^2 f}-\frac {\left (\frac {7}{8}-\frac {5 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d^2 f} \\ & = \frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))}--\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f} \\ & = \frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{4}-\frac {5 i}{4}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{5/2} f}+\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {\left (\frac {7}{8}+\frac {5 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{5/2} f}-\frac {7}{6 a d f (d \tan (e+f x))^{3/2}}+\frac {5 i}{2 a d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.58 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.34 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\frac {i \left (-3 \cot (e+f x)+6 (i+\cot (e+f x)) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (e+f x)\right )+(i+\cot (e+f x)) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (e+f x)\right )\right )}{6 a d^2 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))} \]

[In]

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])),x]

[Out]

((I/6)*(-3*Cot[e + f*x] + 6*(I + Cot[e + f*x])*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[e + f*x]] + (I + Cot[
e + f*x])*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[e + f*x]]))/(a*d^2*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])
)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.42

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {i}{d^{4} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{4}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{4} \sqrt {i d}}\right )}{f a}\)	\(133\)
default	\(\frac {2 d^{2} \left (-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {i}{d^{4} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{4}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{4} \sqrt {i d}}\right )}{f a}\)	\(133\)

[In]

int(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f/a*d^2*(-1/3/d^3/(d*tan(f*x+e))^(3/2)+I/d^4/(d*tan(f*x+e))^(1/2)+1/4/d^4*(-(d*tan(f*x+e))^(1/2)/(I*d*tan(f*
x+e)+d)+6*I/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/4*I/d^4/(I*d)^(1/2)*arctan((d*tan(f*x+e)
)^(1/2)/(I*d)^(1/2)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 738 vs. \(2 (230) = 460\).

Time = 0.26 (sec) , antiderivative size = 738, normalized size of antiderivative = 2.35 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\frac {3 \, {\left (a d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, a d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {i}{4 \, a^{2} d^{5} f^{2}}} \log \left (-2 \, {\left (2 \, {\left (a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a d^{3} 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 {i}{4 \, a^{2} d^{5} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 3 \, {\left (a d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, a d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {i}{4 \, a^{2} d^{5} f^{2}}} \log \left (2 \, {\left (2 \, {\left (a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a d^{3} 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 {i}{4 \, a^{2} d^{5} f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 3 \, {\left (a d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, a d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {9 i}{a^{2} d^{5} f^{2}}} \log \left (-\frac {{\left ({\left (a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a d^{2} 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 {9 i}{a^{2} d^{5} f^{2}}} + 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2} f}\right ) + 3 \, {\left (a d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, a d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {9 i}{a^{2} d^{5} f^{2}}} \log \left (\frac {{\left ({\left (a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a d^{2} 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 {9 i}{a^{2} d^{5} f^{2}}} - 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d^{2} 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}} {\left (19 \, e^{\left (6 i \, f x + 6 i \, e\right )} - 19 \, e^{\left (4 i \, f x + 4 i \, e\right )} - 35 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3\right )}}{12 \, {\left (a d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, a d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} + a d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(3*(a*d^3*f*e^(6*I*f*x + 6*I*e) - 2*a*d^3*f*e^(4*I*f*x + 4*I*e) + a*d^3*f*e^(2*I*f*x + 2*I*e))*sqrt(-1/4*
I/(a^2*d^5*f^2))*log(-2*(2*(a*d^3*f*e^(2*I*f*x + 2*I*e) + a*d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2
*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I/(a^2*d^5*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 3*(a*d^3*f
*e^(6*I*f*x + 6*I*e) - 2*a*d^3*f*e^(4*I*f*x + 4*I*e) + a*d^3*f*e^(2*I*f*x + 2*I*e))*sqrt(-1/4*I/(a^2*d^5*f^2))
*log(2*(2*(a*d^3*f*e^(2*I*f*x + 2*I*e) + a*d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) +
1))*sqrt(-1/4*I/(a^2*d^5*f^2)) - I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 3*(a*d^3*f*e^(6*I*f*x + 6*I
*e) - 2*a*d^3*f*e^(4*I*f*x + 4*I*e) + a*d^3*f*e^(2*I*f*x + 2*I*e))*sqrt(9*I/(a^2*d^5*f^2))*log(-((a*d^2*f*e^(2
*I*f*x + 2*I*e) + a*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(9*I/(a^2*d^5*
f^2)) + 3*I)*e^(-2*I*f*x - 2*I*e)/(a*d^2*f)) + 3*(a*d^3*f*e^(6*I*f*x + 6*I*e) - 2*a*d^3*f*e^(4*I*f*x + 4*I*e)
+ a*d^3*f*e^(2*I*f*x + 2*I*e))*sqrt(9*I/(a^2*d^5*f^2))*log(((a*d^2*f*e^(2*I*f*x + 2*I*e) + a*d^2*f)*sqrt((-I*d
*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(9*I/(a^2*d^5*f^2)) - 3*I)*e^(-2*I*f*x - 2*I*e)/(a*
d^2*f)) - sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(19*e^(6*I*f*x + 6*I*e) - 19*e^(4*I
*f*x + 4*I*e) - 35*e^(2*I*f*x + 2*I*e) + 3))/(a*d^3*f*e^(6*I*f*x + 6*I*e) - 2*a*d^3*f*e^(4*I*f*x + 4*I*e) + a*
d^3*f*e^(2*I*f*x + 2*I*e))

Sympy [F]

\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=- \frac {i \int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} - i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \]

[In]

integrate(1/(d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-I*Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x) - I*(d*tan(e + f*x))**(5/2)), x)/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.82 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=-\frac {1}{6} \, d^{2} {\left (\frac {3 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a d^{\frac {9}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {18 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a d^{\frac {9}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {d \tan \left (f x + e\right )}}{{\left (i \, d \tan \left (f x + e\right ) + d\right )} a d^{4} f} - \frac {4 \, {\left (3 i \, d \tan \left (f x + e\right ) - d\right )}}{\sqrt {d \tan \left (f x + e\right )} a d^{5} f \tan \left (f x + e\right )}\right )} \]

[In]

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/6*d^2*(3*I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(
d)))/(a*d^(9/2)*f*(I*d/sqrt(d^2) + 1)) - 18*I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^
(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a*d^(9/2)*f*(-I*d/sqrt(d^2) + 1)) + 3*sqrt(d*tan(f*x + e))/((I*d*tan(f*
x + e) + d)*a*d^4*f) - 4*(3*I*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*a*d^5*f*tan(f*x + e)))

Mupad [B] (verification not implemented)

Time = 7.75 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx=\mathrm {atan}\left (\frac {2\,a\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {9{}\mathrm {i}}{4\,a^2\,d^5\,f^2}}}{3}\right )\,\sqrt {\frac {9{}\mathrm {i}}{4\,a^2\,d^5\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (4\,a\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d^5\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d^5\,f^2}}\,2{}\mathrm {i}-\frac {\frac {2{}\mathrm {i}}{3\,a\,f}+\frac {4\,\mathrm {tan}\left (e+f\,x\right )}{3\,a\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,5{}\mathrm {i}}{2\,a\,f}}{-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}} \]

[In]

int(1/((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x)*1i)),x)

[Out]

atan((2*a*d^2*f*(d*tan(e + f*x))^(1/2)*(9i/(4*a^2*d^5*f^2))^(1/2))/3)*(9i/(4*a^2*d^5*f^2))^(1/2)*2i - atan(4*a
*d^2*f*(d*tan(e + f*x))^(1/2)*(-1i/(16*a^2*d^5*f^2))^(1/2))*(-1i/(16*a^2*d^5*f^2))^(1/2)*2i - (2i/(3*a*f) + (4
*tan(e + f*x))/(3*a*f) + (tan(e + f*x)^2*5i)/(2*a*f))/(d*(d*tan(e + f*x))^(3/2)*1i - (d*tan(e + f*x))^(5/2))