\(\int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx\) [172]

Optimal result

Integrand size = 28, antiderivative size = 326 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

(-25/32+21/32*I)*d^(7/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/f*2^(1/2)+(25/32-21/32*I)*d^(7/2)*
arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/f*2^(1/2)-(25/64+21/64*I)*d^(7/2)*ln(d^(1/2)-2^(1/2)*(d*tan
(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^2/f*2^(1/2)+(25/64+21/64*I)*d^(7/2)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/
2)+d^(1/2)*tan(f*x+e))/a^2/f*2^(1/2)-25/8*d^3*(d*tan(f*x+e))^(1/2)/a^2/f+7/8*I*d^2*(d*tan(f*x+e))^(3/2)/a^2/f/
(1+I*tan(f*x+e))-1/4*d*(d*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^2

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3639, 3676, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \]

[In]

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

[Out]

((-25/16 + (21*I)/16)*d^(7/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) + ((25/16 -
(21*I)/16)*d^(7/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((25/32 + (21*I)/32)*
d^(7/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + ((25/32 + (21*I)
/32)*d^(7/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) - (25*d^3*Sqr
t[d*Tan[e + f*x]])/(8*a^2*f) + (((7*I)/8)*d^2*(d*Tan[e + f*x])^(3/2))/(a^2*f*(1 + I*Tan[e + f*x])) - (d*(d*Tan
[e + f*x])^(5/2))/(4*f*(a + I*a*Tan[e + f*x])^2)

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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

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

Rubi steps \begin{align*} \text {integral}& = -\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (-\frac {5 a d^2}{2}+\frac {9}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2} \\ & = \frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \sqrt {d \tan (e+f x)} \left (-\frac {21}{2} i a^2 d^3-\frac {25}{2} a^2 d^3 \tan (e+f x)\right ) \, dx}{8 a^4} \\ & = -\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {25 a^2 d^4}{2}-\frac {21}{2} i a^2 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4} \\ & = -\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\text {Subst}\left (\int \frac {\frac {25 a^2 d^5}{2}-\frac {21}{2} i a^2 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^4 f} \\ & = -\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\left (\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^4\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}+\frac {\left (\left (\frac {25}{16}+\frac {21 i}{16}\right ) d^4\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f} \\ & = -\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2}\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^2 f}+-\frac {\left (\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2}\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^2 f}+\frac {\left (\left (\frac {25}{32}-\frac {21 i}{32}\right ) d^4\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^2 f}+\frac {\left (\left (\frac {25}{32}-\frac {21 i}{32}\right ) d^4\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^2 f} \\ & = -\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2}\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^2 f}+\frac {\left (\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2}\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^2 f} \\ & = -\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.57 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {d^3 \sec ^2(e+f x) \left (4 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+46 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(9+41 \cos (2 (e+f x))+43 i \sin (2 (e+f x))) \sqrt {d \tan (e+f x)}\right )}{16 a^2 f (-i+\tan (e+f x))^2} \]

[In]

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

[Out]

(d^3*Sec[e + f*x]^2*(4*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*(Cos[2*(e + f*x)]
+ I*Sin[2*(e + f*x)]) + 46*(-1)^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*(Cos[2*(e + f
*x)] + I*Sin[2*(e + f*x)]) + (9 + 41*Cos[2*(e + f*x)] + (43*I)*Sin[2*(e + f*x)])*Sqrt[d*Tan[e + f*x]]))/(16*a^
2*f*(-I + Tan[e + f*x])^2)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.39

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(126\)
default	\(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(126\)

[In]

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

[Out]

2/f/a^2*d^3*(-(d*tan(f*x+e))^(1/2)-1/8*d*((11/2*I*(d*tan(f*x+e))^(3/2)+9/2*d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*
x+e)+d)^2+23/2*I/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/8*I*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. 573 vs. \(2 (240) = 480\).

Time = 0.27 (sec) , antiderivative size = 573, normalized size of antiderivative = 1.76 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {{\left (4 \, a^{2} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} 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^{3}}\right ) - 4 \, a^{2} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} 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^{3}}\right ) - 4 \, a^{2} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (23 i \, d^{4} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} 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 )}}{8 \, a^{2} f}\right ) + 4 \, a^{2} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (23 i \, d^{4} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} 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 )}}{8 \, a^{2} f}\right ) + {\left (42 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 9 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - d^{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}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]

[In]

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

[Out]

-1/16*(4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) + 4*(a^2*f*e^
(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-1/16*I*d^7/(a^4*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^3) - 4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(-2*(I*d^4*e
^(2*I*f*x + 2*I*e) - 4*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-1/16*I*d^7/(a^4*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^3) - 4*a^2*sqrt(529/64*I*d^7/(a^4*f^2))*f*e^
(4*I*f*x + 4*I*e)*log(1/8*(23*I*d^4 + 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*I*d^7/(a^4*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^2*f)) + 4*a^2*sqrt(529/64
*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(1/8*(23*I*d^4 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*
I*d^7/(a^4*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^2*f
)) + (42*d^3*e^(4*I*f*x + 4*I*e) + 9*d^3*e^(2*I*f*x + 2*I*e) - d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(
2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

[In]

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

[Out]

-Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x)/a**2

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.61 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.67 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {1}{8} \, d^{3} {\left (-\frac {2 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {23 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {16 \, \sqrt {d \tan \left (f x + e\right )}}{a^{2} f} + \frac {-11 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 9 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}\right )} \]

[In]

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

[Out]

-1/8*d^3*(-2*I*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^2*f*(I*d/sqrt(d^2) + 1)) + 23*I*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^2*f*(-I*d/sqrt(d^2) + 1)) + 16*sqrt(d*tan(f*x + e))/(a^2*f
) + (-11*I*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 9*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I*d)^2*a^2*f
))

Mupad [B] (verification not implemented)

Time = 6.56 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.62 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {9\,d^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8\,a^2\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,11{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\frac {2\,d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a^2\,f}+\mathrm {atan}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}}{23\,d^4}\right )\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i} \]

[In]

int((d*tan(e + f*x))^(7/2)/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

atan((8*a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^7*1i)/(64*a^4*f^2))^(1/2))/d^4)*(-(d^7*1i)/(64*a^4*f^2))^(1/2)*2i -
((9*d^5*(d*tan(e + f*x))^(1/2))/(8*a^2*f) + (d^4*(d*tan(e + f*x))^(3/2)*11i)/(8*a^2*f))/(d^2*tan(e + f*x)*2i +
d^2 - d^2*tan(e + f*x)^2) - atan((16*a^2*f*(d*tan(e + f*x))^(1/2)*((d^7*529i)/(256*a^4*f^2))^(1/2))/(23*d^4))
*((d^7*529i)/(256*a^4*f^2))^(1/2)*2i - (2*d^3*(d*tan(e + f*x))^(1/2))/(a^2*f)