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

[Out]

(49/32-45/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(5/2)/f*2^(1/2)+(-49/32+45/32*I)*arctan(1
+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(5/2)/f*2^(1/2)+(49/64+45/64*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))
^(1/2)+d^(1/2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)-(49/64+45/64*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/
2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)+45/8*I/a^2/d^2/f/(d*tan(f*x+e))^(1/2)-49/24/a^2/d/f/(d*tan(f*x+e))^(3/2)+
9/8/a^2/d/f/(1+I*tan(f*x+e))/(d*tan(f*x+e))^(3/2)+1/4/d/f/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2

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Rubi [A]  time = 0.59, antiderivative size = 353, normalized size of antiderivative = 1.00, 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 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 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.37, 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]

-1/192*(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)] + (27
0 + 294*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)] + Si
n[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)]))/(a^2*d*f*(d*Tan[e + f*x])^(3/2)*(-I + Tan[e + f*x])^2)

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fricas [B]  time = 0.64, size = 797, normalized size = 2.26 \[ \frac {12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} \log \left (-{\left (8 \, {\left (a^{2} d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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}{16 \, a^{4} d^{5} f^{2}}} + 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 ) - 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} \log \left ({\left (8 \, {\left (a^{2} d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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}{16 \, a^{4} d^{5} f^{2}}} - 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 ) - 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} \log \left (-\frac {{\left (8 \, {\left (a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {2209 i}{64 \, a^{4} d^{5} f^{2}}} + 47 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} d^{2} f}\right ) + 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} \log \left (\frac {{\left (8 \, {\left (a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {2209 i}{64 \, a^{4} d^{5} f^{2}}} - 47 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} 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 (202 \, e^{\left (8 i \, f x + 8 i \, e\right )} - 103 \, e^{\left (6 i \, f x + 6 i \, e\right )} - 269 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 39 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3\right )}}{48 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]

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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giac [A]  time = 1.85, size = 246, normalized size = 0.70 \[ \frac {47 \, \sqrt {2} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{8 \, a^{2} d^{\frac {5}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {\sqrt {2} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{4 \, a^{2} d^{\frac {5}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {13 i \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 15 \, \sqrt {d \tan \left (f x + e\right )} d}{8 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} d^{2} f} + \frac {2 \, {\left (6 i \, d \tan \left (f x + e\right ) - d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f \tan \left (f x + e\right )} \]

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]

47/8*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)*sqrt(d)))/
(a^2*d^(5/2)*f*(I*d/sqrt(d^2) + 1)) - 1/4*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)*sqrt(d)))/(a^2*d^(5/2)*f*(-I*d/sqrt(d^2) + 1)) + 1/8*(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^2*f) + 2/3*(6*I*d*tan(f*x + e) - d)/
(sqrt(d*tan(f*x + e))*a^2*d^3*f*tan(f*x + e))

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maple [A]  time = 0.34, size = 190, normalized size = 0.54 \[ \frac {13 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{2} d^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}+\frac {15 \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{2} d \left (d \tan \left (f x +e \right )-i d \right )^{2}}+\frac {47 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 f \,a^{2} d^{2} \sqrt {-i d}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 f \,a^{2} d^{2} \sqrt {i d}}-\frac {2}{3 a^{2} d f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 i}{f \,a^{2} d^{2} \sqrt {d \tan \left (f x +e \right )}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.28, size = 209, normalized size = 0.59 \[ -\mathrm {atan}\left (8\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}}{47}\right )\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}-\frac {\frac {2\,d}{3\,a^2\,f}+\frac {221\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{24\,a^2\,f}+\frac {d\,{\mathrm {tan}\left (e+f\,x\right )}^3\,45{}\mathrm {i}}{8\,a^2\,f}-\frac {d\,\mathrm {tan}\left (e+f\,x\right )\,8{}\mathrm {i}}{3\,a^2\,f}}{d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((16*a^2*d^2*f*(d*tan(e + f*x))^(1/2)*(2209i/(256*a^4*d^5*f^2))^(1/2))/47)*(2209i/(256*a^4*d^5*f^2))^(1/2)
*2i - atan(8*a^2*d^2*f*(d*tan(e + f*x))^(1/2)*(-1i/(64*a^4*d^5*f^2))^(1/2))*(-1i/(64*a^4*d^5*f^2))^(1/2)*2i -
((2*d)/(3*a^2*f) + (221*d*tan(e + f*x)^2)/(24*a^2*f) + (d*tan(e + f*x)^3*45i)/(8*a^2*f) - (d*tan(e + f*x)*8i)/
(3*a^2*f))/(d*(d*tan(e + f*x))^(5/2)*2i - (d*tan(e + f*x))^(7/2) + d^2*(d*tan(e + f*x))^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} - \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{2}} \]

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]

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

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