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

Optimal. Leaf size=312 \[ \frac{\left (\frac{5}{4}-\frac{7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}-\frac{\left (\frac{5}{4}-\frac{7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a f}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}+\frac{\left (\frac{5}{8}+\frac{7 i}{8}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{\left (\frac{5}{8}+\frac{7 i}{8}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.344064, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3550, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{5}{4}-\frac{7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}-\frac{\left (\frac{5}{4}-\frac{7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a f}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}+\frac{\left (\frac{5}{8}+\frac{7 i}{8}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{\left (\frac{5}{8}+\frac{7 i}{8}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3550

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]

Rule 3528

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 3534

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 1168

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

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

Mathematica [A]  time = 1.70613, size = 275, normalized size = 0.88 \[ -\frac{d^4 \sec ^3(e+f x) \left (54 i \sin (e+f x)+22 i \sin (3 (e+f x))-16 \cos (e+f x)+16 \cos (3 (e+f x))+(42+30 i) \sqrt{\sin (2 (e+f x))} \cos (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))+(15+21 i) \sin ^{\frac{3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt{\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt{\sin (2 (e+f x))} \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{48 a f (\tan (e+f x)-i) \sqrt{d \tan (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(d^4*Sec[e + f*x]^3*(-16*Cos[e + f*x] + 16*Cos[3*(e + f*x)] + (54*I)*Sin[e + f*x] + (21 - 15*I)*Log[Cos[e + f
*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (21 - 15*I)*Cos[2*(e + f*x)]*Log[Cos[e +
f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (42 + 30*I)*ArcSin[Cos[e + f*x] - Sin[
e + f*x]]*Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])*Sqrt[Sin[2*(e + f*x)]] + (15 + 21*I)*Log[Cos[e + f*x] +
Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (22*I)*Sin[3*(e + f*x)]))/(48*a*f*Sqrt[d*Tan[
e + f*x]]*(-I + Tan[e + f*x]))

________________________________________________________________________________________

Maple [A]  time = 0.063, size = 154, normalized size = 0.5 \begin{align*}{\frac{-{\frac{2\,i}{3}}{d}^{2}}{fa} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{d}^{3}\sqrt{d\tan \left ( fx+e \right ) }}{fa}}-{\frac{{\frac{i}{2}}{d}^{4}}{fa \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{3\,i{d}^{4}}{fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{\frac{i}{2}}{d}^{4}}{fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 2.36101, size = 1656, normalized size = 5.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((-3*I*d^4 + sqrt(9*I*d
^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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)) - 3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*
log((-3*I*d^4 - sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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)) + 3*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e
) + a*f*e^(2*I*f*x + 2*I*e))*log((-2*I*d^4*e^(2*I*f*x + 2*I*e) + 4*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x
+ 2*I*e) + a*f)*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) -
3*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((-2*I*d^4*e^(2*I*f*x + 2*
I*e) - 4*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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) + (19*d^3*e^(4*I*f*x + 4*I*e) + 38*d^3*e^(2*I*f*x + 2*I*e) +
3*d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*
I*f*x + 2*I*e))

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [A]  time = 1.19102, size = 320, normalized size = 1.03 \begin{align*} -\frac{1}{6} \, d^{2}{\left (\frac{18 \, \sqrt{2} d^{\frac{3}{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 )}{a f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 \, \sqrt{2} d^{\frac{3}{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 )}{a f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 i \, \sqrt{d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f} + \frac{4 \,{\left (i \, \sqrt{d \tan \left (f x + e\right )} a^{2} d f^{2} \tan \left (f x + e\right ) - 3 \, \sqrt{d \tan \left (f x + e\right )} a^{2} d f^{2}\right )}}{a^{3} f^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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