3.133 \(\int \frac{e^{\frac{1}{4} i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=364 \[ -\frac{a^2 \log \left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt{2}}+\frac{a^2 \log \left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt{2}}+\frac{1}{16} a^2 \tan ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{a^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{1}{16} a^2 \tanh ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x} \]

[Out]

((-I/8)*a*(1 - I*a*x)^(7/8)*(1 + I*a*x)^(1/8))/x - ((1 - I*a*x)^(7/8)*(1 + I*a*x)^(9/8))/(2*x^2) + (a^2*ArcTan
[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*ArcTan[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/(1
6*Sqrt[2]) + (a^2*ArcTan[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/(16*Sqrt[2]) + (a^2*ArcTanh[(1 +
I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*Log[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*a*x)
^(1/4)/(1 - I*a*x)^(1/4)])/(32*Sqrt[2]) + (a^2*Log[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*
a*x)^(1/4)/(1 - I*a*x)^(1/4)])/(32*Sqrt[2])

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Rubi [A]  time = 0.161089, antiderivative size = 364, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {5062, 96, 94, 93, 214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \[ -\frac{a^2 \log \left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt{2}}+\frac{a^2 \log \left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{32 \sqrt{2}}+\frac{1}{16} a^2 \tan ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{a^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{1}{16} a^2 \tanh ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x} \]

Antiderivative was successfully verified.

[In]

Int[E^((I/4)*ArcTan[a*x])/x^3,x]

[Out]

((-I/8)*a*(1 - I*a*x)^(7/8)*(1 + I*a*x)^(1/8))/x - ((1 - I*a*x)^(7/8)*(1 + I*a*x)^(9/8))/(2*x^2) + (a^2*ArcTan
[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*ArcTan[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/(1
6*Sqrt[2]) + (a^2*ArcTan[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)])/(16*Sqrt[2]) + (a^2*ArcTanh[(1 +
I*a*x)^(1/8)/(1 - I*a*x)^(1/8)])/16 - (a^2*Log[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*a*x)
^(1/4)/(1 - I*a*x)^(1/4)])/(32*Sqrt[2]) + (a^2*Log[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*
a*x)^(1/4)/(1 - I*a*x)^(1/4)])/(32*Sqrt[2])

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 214

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[r/(2*a), Int[1/(r - s*x^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a,
b}, x] && IGtQ[n/4, 1] &&  !GtQ[a/b, 0]

Rule 212

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

Rule 206

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

Rule 203

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

Rule 211

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

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]

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])

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{4} i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac{\sqrt [8]{1+i a x}}{x^3 \sqrt [8]{1-i a x}} \, dx\\ &=-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{8} (i a) \int \frac{\sqrt [8]{1+i a x}}{x^2 \sqrt [8]{1-i a x}} \, dx\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac{1}{32} a^2 \int \frac{1}{x \sqrt [8]{1-i a x} (1+i a x)^{7/8}} \, dx\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}-\frac{1}{4} a^2 \operatorname{Subst}\left (\int \frac{1}{-1+x^8} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{8} a^2 \operatorname{Subst}\left (\int \frac{1}{1-x^4} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{8} a^2 \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{16} a^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{16} a^2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{16} a^2 \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{16} a^2 \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{16} a^2 \tan ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{16} a^2 \tanh ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{32} a^2 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{32} a^2 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{32 \sqrt{2}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{32 \sqrt{2}}\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{16} a^2 \tan ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac{1}{16} a^2 \tanh ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \log \left (1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt{2}}+\frac{a^2 \log \left (1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt{2}}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}\\ &=-\frac{i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac{(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac{1}{16} a^2 \tan ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{a^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt{2}}+\frac{1}{16} a^2 \tanh ^{-1}\left (\frac{\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac{a^2 \log \left (1-\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt{2}}+\frac{a^2 \log \left (1+\frac{\sqrt{2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}{32 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0215134, size = 84, normalized size = 0.23 \[ \frac{(1-i a x)^{7/8} \left (2 a^2 x^2 \text{Hypergeometric2F1}\left (\frac{7}{8},1,\frac{15}{8},\frac{a x+i}{-a x+i}\right )+7 \left (5 a^2 x^2-9 i a x-4\right )\right )}{56 x^2 (1+i a x)^{7/8}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((I/4)*ArcTan[a*x])/x^3,x]

[Out]

((1 - I*a*x)^(7/8)*(7*(-4 - (9*I)*a*x + 5*a^2*x^2) + 2*a^2*x^2*Hypergeometric2F1[7/8, 1, 15/8, (I + a*x)/(I -
a*x)]))/(56*x^2*(1 + I*a*x)^(7/8))

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Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}}\sqrt [4]{{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{1}{4}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(1/4)/x^3, x)

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Fricas [A]  time = 1.83024, size = 883, normalized size = 2.43 \begin{align*} \frac{a^{2} x^{2} \log \left (\left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} + 1\right ) + i \, a^{2} x^{2} \log \left (\left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} + i\right ) - i \, a^{2} x^{2} \log \left (\left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} - i\right ) - a^{2} x^{2} \log \left (\left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} - 1\right ) + \sqrt{i \, a^{4}} x^{2} \log \left (\frac{a^{2} \left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} + \sqrt{i \, a^{4}}}{a^{2}}\right ) - \sqrt{i \, a^{4}} x^{2} \log \left (\frac{a^{2} \left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} - \sqrt{i \, a^{4}}}{a^{2}}\right ) + \sqrt{-i \, a^{4}} x^{2} \log \left (\frac{a^{2} \left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} + \sqrt{-i \, a^{4}}}{a^{2}}\right ) - \sqrt{-i \, a^{4}} x^{2} \log \left (\frac{a^{2} \left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}} - \sqrt{-i \, a^{4}}}{a^{2}}\right ) -{\left (20 \, a^{2} x^{2} + 4 i \, a x + 16\right )} \left (\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}\right )^{\frac{1}{4}}}{32 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="fricas")

[Out]

1/32*(a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + 1) + I*a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(
1/4) + I) - I*a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) - I) - a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x +
I))^(1/4) - 1) + sqrt(I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + sqrt(I*a^4))/a^2) - sqrt(I*
a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) - sqrt(I*a^4))/a^2) + sqrt(-I*a^4)*x^2*log((a^2*(I*sqr
t(a^2*x^2 + 1)/(a*x + I))^(1/4) + sqrt(-I*a^4))/a^2) - sqrt(-I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I
))^(1/4) - sqrt(-I*a^4))/a^2) - (20*a^2*x^2 + 4*I*a*x + 16)*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4))/x^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/4)/x**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="giac")

[Out]

Exception raised: TypeError