3.179 \(\int \frac{(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=370 \[ \frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 f}+\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3} \]

[Out]

((7/4 + (15*I)/8)*d^(9/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/4 + (15*I)
/8)*d^(9/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/8 - (15*I)/16)*d^(9/2)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + ((7/8 - (15*I)/16)*d^(9/2
)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + (((15*I)/4)*d^4*Sqrt[d
*Tan[e + f*x]])/(a^3*f) - (d*(d*Tan[e + f*x])^(7/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + (((5*I)/12)*d^2*(d*Tan[e
+ f*x])^(5/2))/(a*f*(a + I*a*Tan[e + f*x])^2) + (7*d^3*(d*Tan[e + f*x])^(3/2))/(6*f*(a^3 + I*a^3*Tan[e + f*x]
))

________________________________________________________________________________________

Rubi [A]  time = 0.674508, antiderivative size = 370, 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 = {3558, 3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 f}+\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((7/4 + (15*I)/8)*d^(9/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/4 + (15*I)
/8)*d^(9/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/8 - (15*I)/16)*d^(9/2)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + ((7/8 - (15*I)/16)*d^(9/2
)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + (((15*I)/4)*d^4*Sqrt[d
*Tan[e + f*x]])/(a^3*f) - (d*(d*Tan[e + f*x])^(7/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + (((5*I)/12)*d^2*(d*Tan[e
+ f*x])^(5/2))/(a*f*(a + I*a*Tan[e + f*x])^2) + (7*d^3*(d*Tan[e + f*x])^(3/2))/(6*f*(a^3 + I*a^3*Tan[e + f*x]
))

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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]

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))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{(d \tan (e+f x))^{5/2} \left (-\frac{7 a d^2}{2}+\frac{13}{2} i a d^2 \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{\int \frac{(d \tan (e+f x))^{3/2} \left (-25 i a^2 d^3-31 a^2 d^3 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{24 a^4}\\ &=-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\int \sqrt{d \tan (e+f x)} \left (84 a^3 d^4-90 i a^3 d^4 \tan (e+f x)\right ) \, dx}{48 a^6}\\ &=\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\int \frac{90 i a^3 d^5+84 a^3 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{48 a^6}\\ &=\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{90 i a^3 d^6+84 a^3 d^5 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{24 a^6 f}\\ &=\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac{\left (\left (\frac{7}{4}-\frac{15 i}{8}\right ) d^5\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^3 f}-\frac{\left (\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^5\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^3 f}\\ &=\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2}\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^3 f}-\frac{\left (\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2}\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^3 f}-\frac{\left (\left (\frac{7}{8}+\frac{15 i}{16}\right ) d^5\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^3 f}-\frac{\left (\left (\frac{7}{8}+\frac{15 i}{16}\right ) d^5\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^3 f}\\ &=-\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 f}+\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 f}+\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac{\left (\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2}\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^3 f}-\frac{\left (\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2}\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^3 f}\\ &=\frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{7}{4}+\frac{15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 f}+\frac{\left (\frac{7}{8}-\frac{15 i}{16}\right ) d^{9/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 f}+\frac{15 i d^4 \sqrt{d \tan (e+f x)}}{4 a^3 f}-\frac{d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac{7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.49007, size = 236, normalized size = 0.64 \[ \frac{i d^4 e^{-6 i (e+f x)} \sqrt{d \tan (e+f x)} \left (9 e^{2 i (e+f x)}-49 e^{4 i (e+f x)}-105 e^{6 i (e+f x)}+146 e^{8 i (e+f x)}-87 e^{6 i (e+f x)} \sqrt{-1+e^{4 i (e+f x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (e+f x)}}\right )-6 e^{6 i (e+f x)} \sqrt{-1+e^{2 i (e+f x)}} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )-1\right )}{48 a^3 f \left (-1+e^{2 i (e+f x)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((I/48)*d^4*(-1 + 9*E^((2*I)*(e + f*x)) - 49*E^((4*I)*(e + f*x)) - 105*E^((6*I)*(e + f*x)) + 146*E^((8*I)*(e +
f*x)) - 87*E^((6*I)*(e + f*x))*Sqrt[-1 + E^((4*I)*(e + f*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(e + f*x))]] - 6*E^((
6*I)*(e + f*x))*Sqrt[-1 + E^((2*I)*(e + f*x))]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(e +
f*x)))/(1 + E^((2*I)*(e + f*x)))]])*Sqrt[d*Tan[e + f*x]])/(a^3*E^((6*I)*(e + f*x))*(-1 + E^((2*I)*(e + f*x)))*
f)

________________________________________________________________________________________

Maple [A]  time = 0.06, size = 203, normalized size = 0.6 \begin{align*}{\frac{2\,i{d}^{4}}{f{a}^{3}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{5\,{d}^{5}}{2\,f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{{\frac{49\,i}{12}}{d}^{6}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{d}^{7}}{4\,f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{29\,{d}^{5}}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{d}^{5}}{8\,f{a}^{3}}\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))^(9/2)/(a+I*a*tan(f*x+e))^3,x)

[Out]

2*I/f/a^3*d^4*(d*tan(f*x+e))^(1/2)+5/2/f/a^3*d^5/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(5/2)-49/12*I/f/a^3*d^6/
(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(3/2)-7/4/f/a^3*d^7/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(1/2)-29/8/f/a^3
*d^5/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/8/f/a^3*d^5/(I*d)^(1/2)*arctan((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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 2.62108, size = 1631, normalized size = 4.41 \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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/48*(12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log((-2*I*d^5*e^(2*I*f*x + 2*I*e) + (16*I*a^3*f*
e^(2*I*f*x + 2*I*e) + 16*I*a^3*f)*sqrt(1/64*I*d^9/(a^6*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^4) - 12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log((-2*I
*d^5*e^(2*I*f*x + 2*I*e) + (-16*I*a^3*f*e^(2*I*f*x + 2*I*e) - 16*I*a^3*f)*sqrt(1/64*I*d^9/(a^6*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^4) + 12*a^3*sqrt(-841/64*I*d^9
/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64*I*d^9/(
a^6*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^3*f)) - 12
*a^3*sqrt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*
f)*sqrt(-841/64*I*d^9/(a^6*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^3*f)) + (146*I*d^4*e^(6*I*f*x + 6*I*e) + 41*I*d^4*e^(4*I*f*x + 4*I*e) - 8*I*d^4*e^(2*I*f*x + 2*I*
e) + I*d^4)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

________________________________________________________________________________________

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))**(9/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [A]  time = 1.19397, size = 339, normalized size = 0.92 \begin{align*} -\frac{1}{24} \, d^{4}{\left (\frac{87 i \, \sqrt{2} \sqrt{d} \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^{3} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 i \, \sqrt{2} \sqrt{d} \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^{3} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{48 i \, \sqrt{d \tan \left (f x + e\right )}}{a^{3} f} - \frac{2 \,{\left (30 \, \sqrt{d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - 49 i \, \sqrt{d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 21 \, \sqrt{d \tan \left (f x + e\right )} d^{3}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/24*d^4*(87*I*sqrt(2)*sqrt(d)*arctan(16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sq
rt(d^2)*sqrt(d)))/(a^3*f*(I*d/sqrt(d^2) + 1)) + 3*I*sqrt(2)*sqrt(d)*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^3*f*(-I*d/sqrt(d^2) + 1)) - 48*I*sqrt(d*tan(f*x + e
))/(a^3*f) - 2*(30*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)^2 - 49*I*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e) - 21*s
qrt(d*tan(f*x + e))*d^3)/((d*tan(f*x + e) - I*d)^3*a^3*f))