\(\int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx\) [165]

Optimal result

Integrand size = 28, antiderivative size = 287 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=-\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {3}{8}-\frac {5 i}{8}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}-\frac {\left (\frac {3}{8}-\frac {5 i}{8}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}-\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3631, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=-\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) d^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a f}+\frac {\left (\frac {3}{8}-\frac {5 i}{8}\right ) d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {\left (\frac {3}{8}-\frac {5 i}{8}\right ) d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]

[In]

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

[Out]

((-3/4 - (5*I)/4)*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*f) + ((3/4 + (5*I)/4)
*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*f) + ((3/8 - (5*I)/8)*d^(5/2)*Log[Sqrt
[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*f) - ((3/8 - (5*I)/8)*d^(5/2)*Log[Sqrt[
d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*f) - (((5*I)/2)*d^2*Sqrt[d*Tan[e + f*x]]
)/(a*f) - (d*(d*Tan[e + f*x])^(3/2))/(2*f*(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 3609

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 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 3631

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

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

Mathematica [A] (verified)

Time = 2.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.43 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\frac {d^2 \left (-(-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-4 (-1)^{3/4} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {(-5-4 i \tan (e+f x)) \sqrt {d \tan (e+f x)}}{-i+\tan (e+f x)}\right )}{2 a f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.38

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-i \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{d \tan \left (f x +e \right )-i d}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{4}-\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}\right )}{f a}\)	\(109\)
default	\(\frac {2 d^{2} \left (-i \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{d \tan \left (f x +e \right )-i d}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{4}-\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 \sqrt {i d}}\right )}{f a}\)	\(109\)

[In]

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

[Out]

2/f/a*d^2*(-I*(d*tan(f*x+e))^(1/2)-1/4*d*((d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)-I*d)-4/(-I*d)^(1/2)*arctan((d*tan
(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/4*d/(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. 536 vs. \(2 (207) = 414\).

Time = 0.25 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.87 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\frac {{\left (a \sqrt {\frac {i \, d^{5}}{4 \, a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {i \, d^{5}}{4 \, a^{2} f^{2}}} \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 )}}{d^{2}}\right ) - a \sqrt {\frac {i \, d^{5}}{4 \, a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {i \, d^{5}}{4 \, a^{2} f^{2}}} \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 )}}{d^{2}}\right ) - a \sqrt {-\frac {4 i \, d^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (2 \, d^{3} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {4 i \, d^{5}}{a^{2} f^{2}}} \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 )}}{a f}\right ) + a \sqrt {-\frac {4 i \, d^{5}}{a^{2} f^{2}}} f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (2 \, d^{3} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {4 i \, d^{5}}{a^{2} f^{2}}} \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 )}}{a f}\right ) + {\left (-9 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((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 = 0.54 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.67 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=-\frac {1}{2} \, d^{2} {\left (\frac {\sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {4 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {4 i \, \sqrt {d \tan \left (f x + e\right )}}{a f} + \frac {\sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f}\right )} \]

[In]

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

[Out]

-1/2*d^2*(sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*s
qrt(d)))/(a*f*(I*d/sqrt(d^2) + 1)) - 4*sqrt(2)*sqrt(d)*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*f*(-I*d/sqrt(d^2) + 1)) + 4*I*sqrt(d*tan(f*x + e))/(a*f) + sqrt(d*ta
n(f*x + e))*d/((d*tan(f*x + e) - I*d)*a*f))

Mupad [B] (verification not implemented)

Time = 7.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.56 \[ \int \frac {(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\mathrm {atan}\left (\frac {a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^5\,1{}\mathrm {i}}{a^2\,f^2}}\,1{}\mathrm {i}}{d^3}\right )\,\sqrt {-\frac {d^5\,1{}\mathrm {i}}{a^2\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,4{}\mathrm {i}}{d^3}\right )\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,2{}\mathrm {i}-\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{a\,f}+\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

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