3.88 \(\int e^{-\frac{1}{2} i \tan ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=337 \[ \frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}-\frac{11 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt{2} a^4}+\frac{11 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt{2} a^4}-\frac{11 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}+\frac{11 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4} \]

[Out]

(-11*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(64*a^4) + (x^2*(1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(4*a^2) - ((1 -
I*a*x)^(5/4)*(1 + I*a*x)^(3/4)*(25 - (4*I)*a*x))/(96*a^4) - (11*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I
*a*x)^(1/4)])/(64*Sqrt[2]*a^4) + (11*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(64*Sqrt[2]*a^
4) - (11*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(128*Sqrt[2
]*a^4) + (11*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(128*Sq
rt[2]*a^4)

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Rubi [A]  time = 0.213259, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5062, 100, 147, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}-\frac{11 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt{2} a^4}+\frac{11 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt{2} a^4}-\frac{11 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}+\frac{11 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^((I/2)*ArcTan[a*x]),x]

[Out]

(-11*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(64*a^4) + (x^2*(1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(4*a^2) - ((1 -
I*a*x)^(5/4)*(1 + I*a*x)^(3/4)*(25 - (4*I)*a*x))/(96*a^4) - (11*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I
*a*x)^(1/4)])/(64*Sqrt[2]*a^4) + (11*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(64*Sqrt[2]*a^
4) - (11*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(128*Sqrt[2
]*a^4) + (11*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(128*Sq
rt[2]*a^4)

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 100

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

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 e^{-\frac{1}{2} i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac{x^3 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx\\ &=\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}+\frac{\int \frac{x \left (-2+\frac{i a x}{2}\right ) \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{4 a^2}\\ &=\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{(11 i) \int \frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{64 a^3}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{(11 i) \int \frac{1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{128 a^3}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{32 a^4}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{32 a^4}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 a^4}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 a^4}-\frac{11 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}-\frac{11 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{11 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}+\frac{11 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}\\ &=-\frac{11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{64 a^4}+\frac{x^2 (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^2}-\frac{(1-i a x)^{5/4} (1+i a x)^{3/4} (25-4 i a x)}{96 a^4}-\frac{11 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}+\frac{11 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt{2} a^4}-\frac{11 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}+\frac{11 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{128 \sqrt{2} a^4}\\ \end{align*}

Mathematica [C]  time = 0.100572, size = 127, normalized size = 0.38 \[ \frac{(1-i a x)^{5/4} \left (4\ 2^{3/4} \text{Hypergeometric2F1}\left (-\frac{7}{4},\frac{5}{4},\frac{9}{4},\frac{1}{2} (1-i a x)\right )-12\ 2^{3/4} \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{5}{4},\frac{9}{4},\frac{1}{2} (1-i a x)\right )+5\ 2^{3/4} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{5}{4},\frac{9}{4},\frac{1}{2} (1-i a x)\right )+5 a^2 x^2 (1+i a x)^{3/4}\right )}{20 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/E^((I/2)*ArcTan[a*x]),x]

[Out]

((1 - I*a*x)^(5/4)*(5*a^2*x^2*(1 + I*a*x)^(3/4) + 4*2^(3/4)*Hypergeometric2F1[-7/4, 5/4, 9/4, (1 - I*a*x)/2] -
12*2^(3/4)*Hypergeometric2F1[-3/4, 5/4, 9/4, (1 - I*a*x)/2] + 5*2^(3/4)*Hypergeometric2F1[1/4, 5/4, 9/4, (1 -
I*a*x)/2]))/(20*a^4)

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Maple [F]  time = 0.146, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3}{\frac{1}{\sqrt{{(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(x^3/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.16295, size = 733, normalized size = 2.18 \begin{align*} -\frac{96 \, a^{4} \sqrt{\frac{121 i}{4096 \, a^{8}}} \log \left (\frac{64}{11} i \, a^{4} \sqrt{\frac{121 i}{4096 \, a^{8}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 96 \, a^{4} \sqrt{\frac{121 i}{4096 \, a^{8}}} \log \left (-\frac{64}{11} i \, a^{4} \sqrt{\frac{121 i}{4096 \, a^{8}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 96 \, a^{4} \sqrt{-\frac{121 i}{4096 \, a^{8}}} \log \left (\frac{64}{11} i \, a^{4} \sqrt{-\frac{121 i}{4096 \, a^{8}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + 96 \, a^{4} \sqrt{-\frac{121 i}{4096 \, a^{8}}} \log \left (-\frac{64}{11} i \, a^{4} \sqrt{-\frac{121 i}{4096 \, a^{8}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) -{\left (-48 i \, a^{3} x^{3} + 56 \, a^{2} x^{2} + 58 i \, a x - 83\right )} \sqrt{a^{2} x^{2} + 1} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{192 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(96*a^4*sqrt(121/4096*I/a^8)*log(64/11*I*a^4*sqrt(121/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))
) - 96*a^4*sqrt(121/4096*I/a^8)*log(-64/11*I*a^4*sqrt(121/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) -
96*a^4*sqrt(-121/4096*I/a^8)*log(64/11*I*a^4*sqrt(-121/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 9
6*a^4*sqrt(-121/4096*I/a^8)*log(-64/11*I*a^4*sqrt(-121/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (-
48*I*a^3*x^3 + 56*a^2*x^2 + 58*I*a*x - 83)*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^4

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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