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

Optimal. Leaf size=353 \[ \frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 d^{5/2} f}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}} \]

[Out]

((49/16 - (45*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(5/2)*f) - ((49/16 - (
45*I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(5/2)*f) + ((49/32 + (45*I)/32)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(5/2)*f) - ((49/32 + (45*I)/
32)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(5/2)*f) - 49/(24*a^2*d
*f*(d*Tan[e + f*x])^(3/2)) + 9/(8*a^2*d*f*(1 + I*Tan[e + f*x])*(d*Tan[e + f*x])^(3/2)) + ((45*I)/8)/(a^2*d^2*f
*Sqrt[d*Tan[e + f*x]]) + 1/(4*d*f*(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.59132, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3559, 3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 d^{5/2} f}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((49/16 - (45*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(5/2)*f) - ((49/16 - (
45*I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(5/2)*f) + ((49/32 + (45*I)/32)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(5/2)*f) - ((49/32 + (45*I)/
32)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(5/2)*f) - 49/(24*a^2*d
*f*(d*Tan[e + f*x])^(3/2)) + 9/(8*a^2*d*f*(1 + I*Tan[e + f*x])*(d*Tan[e + f*x])^(3/2)) + ((45*I)/8)/(a^2*d^2*f
*Sqrt[d*Tan[e + f*x]]) + 1/(4*d*f*(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2)

Rule 3559

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

Rule 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
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 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{1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx &=\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{11 a d}{2}-\frac{7}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))} \, dx}{4 a^2 d}\\ &=\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{49 a^2 d^2}{2}-\frac{45}{2} i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{8 a^4 d^2}\\ &=-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\int \frac{-\frac{45}{2} i a^2 d^3-\frac{49}{2} a^2 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{8 a^4 d^4}\\ &=-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\int \frac{-\frac{49}{2} a^2 d^4+\frac{45}{2} i a^2 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4 d^6}\\ &=-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{49}{2} a^2 d^5+\frac{45}{2} i a^2 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 d^6 f}\\ &=-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d^2 f}+-\frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d^2 f}\\ &=-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+\frac{\left (\frac{49}{32}+\frac{45 i}{32}\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^2 d^{5/2} f}+\frac{\left (\frac{49}{32}+\frac{45 i}{32}\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^2 d^{5/2} f}+-\frac{\left (\frac{49}{32}-\frac{45 i}{32}\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^2 d^2 f}+-\frac{\left (\frac{49}{32}-\frac{45 i}{32}\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^2 d^2 f}\\ &=\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{49}{16}-\frac{45 i}{16}\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^2 d^{5/2} f}+\frac{\left (\frac{49}{16}-\frac{45 i}{16}\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^2 d^{5/2} f}\\ &=\frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{16}-\frac{45 i}{16}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{5/2} f}+\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{\left (\frac{49}{32}+\frac{45 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{5/2} f}-\frac{49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac{9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac{45 i}{8 a^2 d^2 f \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 2.19289, size = 346, normalized size = 0.98 \[ -\frac{\sec ^4(e+f x) \left (-142 i \sin (2 (e+f x))+199 i \sin (4 (e+f x))-64 \cos (2 (e+f x))+205 \cos (4 (e+f x))+(270+294 i) \sin (e+f x) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))-(147+135 i) \sqrt{\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(135-147 i) \sqrt{\sin (2 (e+f x))} \cos (e+f x) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-(135-147 i) \sqrt{\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(147+135 i) \sin (e+f x) \sqrt{\sin (2 (e+f x))} \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-269\right )}{192 a^2 d f (\tan (e+f x)-i)^2 (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[e + f*x]^4*(-269 - 64*Cos[2*(e + f*x)] + 205*Cos[4*(e + f*x)] + (135 - 147*I)*Cos[e + f*x]*Log[Cos[e + f
*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - (135 - 147*I)*Cos[3*(e + f*x)]*Log[Cos[e
+ f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (147 + 135*I)*Log[Cos[e + f*x] + Sin
[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[e + f*x]*Sqrt[Sin[2*(e + f*x)]] - (142*I)*Sin[2*(e + f*x)] + (270 + 29
4*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sin[e + f*x]*Sqrt[Sin[2*(e + f*x)]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e
+ f*x)]) - (147 + 135*I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]]*Sin
[3*(e + f*x)] + (199*I)*Sin[4*(e + f*x)]))/(192*a^2*d*f*(d*Tan[e + f*x])^(3/2)*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.059, size = 190, normalized size = 0.5 \begin{align*}{\frac{{\frac{13\,i}{8}}}{f{a}^{2}{d}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{15}{8\,f{a}^{2}d \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{{\frac{47\,i}{8}}}{f{a}^{2}{d}^{2}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}-{\frac{2}{3\,f{a}^{2}d} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,i}{f{a}^{2}{d}^{2}}{\frac{1}{\sqrt{d\tan \left ( fx+e \right ) }}}}-{\frac{{\frac{i}{4}}}{f{a}^{2}{d}^{2}}\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(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x)

[Out]

13/8*I/f/a^2/d^2/(-I*d+d*tan(f*x+e))^2*(d*tan(f*x+e))^(3/2)+15/8/f/a^2/d/(-I*d+d*tan(f*x+e))^2*(d*tan(f*x+e))^
(1/2)+47/8*I/f/a^2/d^2/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))-2/3/a^2/d/f/(d*tan(f*x+e))^(3/2)
+4*I/f/a^2/d^2/(d*tan(f*x+e))^(1/2)-1/4*I/f/a^2/d^2/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2))

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.66993, size = 2136, normalized size = 6.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqr
t(-1/16*I/(a^4*d^5*f^2))*log(-(8*(a^2*d^3*f*e^(2*I*f*x + 2*I*e) + a^2*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(-1/16*I/(a^4*d^5*f^2)) + 2*I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e))
- 12*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(-
1/16*I/(a^4*d^5*f^2))*log((8*(a^2*d^3*f*e^(2*I*f*x + 2*I*e) + a^2*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(-1/16*I/(a^4*d^5*f^2)) - 2*I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 1
2*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(2209/
64*I/(a^4*d^5*f^2))*log(-1/8*(8*(a^2*d^2*f*e^(2*I*f*x + 2*I*e) + a^2*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(2209/64*I/(a^4*d^5*f^2)) + 47*I)*e^(-2*I*f*x - 2*I*e)/(a^2*d^2*f)) + 12*(a
^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(2209/64*I
/(a^4*d^5*f^2))*log(1/8*(8*(a^2*d^2*f*e^(2*I*f*x + 2*I*e) + a^2*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(2209/64*I/(a^4*d^5*f^2)) - 47*I)*e^(-2*I*f*x - 2*I*e)/(a^2*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))*(202*e^(8*I*f*x + 8*I*e) - 103*e^(6*I*f*x + 6*I*e) - 26
9*e^(4*I*f*x + 4*I*e) + 39*e^(2*I*f*x + 2*I*e) + 3))/(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x +
6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.19036, size = 336, normalized size = 0.95 \begin{align*} \frac{1}{24} \, d^{3}{\left (\frac{141 \, \sqrt{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^{2} d^{\frac{11}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{6 \, \sqrt{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^{2} d^{\frac{11}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{3 \,{\left (-13 i \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 15 \, \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} d^{5} f} - \frac{16 \,{\left (-6 i \, d \tan \left (f x + e\right ) + d\right )}}{\sqrt{d \tan \left (f x + e\right )} a^{2} d^{6} f \tan \left (f x + e\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*d^3*(141*sqrt(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)*s
qrt(d)))/(a^2*d^(11/2)*f*(I*d/sqrt(d^2) + 1)) - 6*sqrt(2)*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*sq
rt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*d^(11/2)*f*(-I*d/sqrt(d^2) + 1)) - 3*(-13*I*sqrt(d*tan(f*x
+ e))*d*tan(f*x + e) - 15*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^2*a^2*d^5*f) - 16*(-6*I*d*tan(f*x +
e) + d)/(sqrt(d*tan(f*x + e))*a^2*d^6*f*tan(f*x + e)))