3.181 \(\int \frac{(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=329 \[ \frac{d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 \sqrt{2} a^3 f}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{8 \sqrt{2} a^3 f}-\frac{i d^2 \sqrt{d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}+\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}+\frac{i d^2 \sqrt{d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.599929, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {3558, 3595, 3596, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 \sqrt{2} a^3 f}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{8 \sqrt{2} a^3 f}-\frac{i d^2 \sqrt{d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}+\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}+\frac{i d^2 \sqrt{d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

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

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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

Mathematica [A]  time = 1.0254, size = 232, normalized size = 0.71 \[ \frac{d^3 \sec ^4(e+f x) \left (-6 \sin (2 (e+f x))+3 \sin (4 (e+f x))-i \cos (4 (e+f x))+6 i \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-6 \sqrt{\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+6 i \sqrt{\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+i\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(d^3*Sec[e + f*x]^4*(I - I*Cos[4*(e + f*x)] + (6*I)*Cos[3*(e + f*x)]*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Si
n[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - 6*Sin[2*(e + f*x)] + (6*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sqrt[
Sin[2*(e + f*x)]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) - 6*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)] + 3*Sin[4*(e + f*x)]))/(96*a^3*f*Sqrt[d*Tan[e + f*x]]*(-I +
Tan[e + f*x])^3)

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Maple [A]  time = 0.058, size = 145, normalized size = 0.4 \begin{align*} -{\frac{{d}^{3}}{4\,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{i}{12}}{d}^{4}}{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{{d}^{3}}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}-{\frac{{d}^{3}}{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))^(5/2)/(a+I*a*tan(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.52808, size = 1608, normalized size = 4.89 \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))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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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((d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.21139, size = 296, normalized size = 0.9 \begin{align*} -\frac{1}{24} \, d^{4}{\left (\frac{3 \, \sqrt{2} \arctan \left (\frac{16 \, \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} d^{\frac{3}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 \, \sqrt{2} \arctan \left (\frac{16 \, \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} d^{\frac{3}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{2 \,{\left (3 \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} - i \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/24*d^4*(3*sqrt(2)*arctan(16*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*d^(3/2)*f*(I*d/sqrt(d^2) + 1)) + 3*sqrt(2)*arctan(16*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*d^(3/2)*f*(-I*d/sqrt(d^2) + 1)) + 2*(3*sqrt(d*tan(f*x + e))*d^2*tan
(f*x + e)^2 - I*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e))/((d*tan(f*x + e) - I*d)^3*a^3*d*f))