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

Optimal. Leaf size=339 \[ \frac{x \sqrt [4]{1+i a x} (1-i a x)^{7/4}}{3 a^2}+\frac{i \sqrt [4]{1+i a x} (1-i a x)^{7/4}}{4 a^3}+\frac{17 i \sqrt [4]{1+i a x} (1-i a x)^{3/4}}{24 a^3}-\frac{17 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt{2} a^3}+\frac{17 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt{2} a^3}+\frac{17 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3}-\frac{17 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3} \]

[Out]

(((17*I)/24)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a^3 + ((I/4)*(1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/a^3 + (x*(
1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(3*a^2) + (((17*I)/8)*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(
1/4)])/(Sqrt[2]*a^3) - (((17*I)/8)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) -
(((17*I)/16)*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)])/(Sqrt[2
]*a^3) + (((17*I)/16)*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)]
)/(Sqrt[2]*a^3)

________________________________________________________________________________________

Rubi [A]  time = 0.219318, antiderivative size = 339, 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, 90, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac{x \sqrt [4]{1+i a x} (1-i a x)^{7/4}}{3 a^2}+\frac{i \sqrt [4]{1+i a x} (1-i a x)^{7/4}}{4 a^3}+\frac{17 i \sqrt [4]{1+i a x} (1-i a x)^{3/4}}{24 a^3}-\frac{17 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt{2} a^3}+\frac{17 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt{2} a^3}+\frac{17 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3}-\frac{17 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((17*I)/24)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a^3 + ((I/4)*(1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/a^3 + (x*(
1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(3*a^2) + (((17*I)/8)*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(
1/4)])/(Sqrt[2]*a^3) - (((17*I)/8)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) -
(((17*I)/16)*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)])/(Sqrt[2
]*a^3) + (((17*I)/16)*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)]
)/(Sqrt[2]*a^3)

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

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 e^{-\frac{3}{2} i \tan ^{-1}(a x)} x^2 \, dx &=\int \frac{x^2 (1-i a x)^{3/4}}{(1+i a x)^{3/4}} \, dx\\ &=\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}+\frac{\int \frac{(1-i a x)^{3/4} \left (-1+\frac{3 i a x}{2}\right )}{(1+i a x)^{3/4}} \, dx}{3 a^2}\\ &=\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{17 \int \frac{(1-i a x)^{3/4}}{(1+i a x)^{3/4}} \, dx}{24 a^2}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{17 \int \frac{1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{16 a^2}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{(17 i) \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{4 a^3}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{(17 i) \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^3}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}+\frac{(17 i) \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 )}{8 a^3}-\frac{(17 i) \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 )}{8 a^3}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{(17 i) \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 )}{16 a^3}-\frac{(17 i) \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 )}{16 a^3}-\frac{(17 i) \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 )}{16 \sqrt{2} a^3}-\frac{(17 i) \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 )}{16 \sqrt{2} a^3}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}-\frac{17 i \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 )}{16 \sqrt{2} a^3}+\frac{17 i \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 )}{16 \sqrt{2} a^3}-\frac{(17 i) \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 )}{8 \sqrt{2} a^3}+\frac{(17 i) \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 )}{8 \sqrt{2} a^3}\\ &=\frac{17 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 a^3}+\frac{i (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^3}+\frac{x (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{3 a^2}+\frac{17 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3}-\frac{17 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^3}-\frac{17 i \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 )}{16 \sqrt{2} a^3}+\frac{17 i \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 )}{16 \sqrt{2} a^3}\\ \end{align*}

Mathematica [C]  time = 0.0285041, size = 73, normalized size = 0.22 \[ \frac{(1-i a x)^{7/4} \left (7 \sqrt [4]{1+i a x} (4 a x+3 i)-17 i \sqrt [4]{2} \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{7}{4},\frac{11}{4},\frac{1}{2} (1-i a x)\right )\right )}{84 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*a*x)^(7/4)*(7*(1 + I*a*x)^(1/4)*(3*I + 4*a*x) - (17*I)*2^(1/4)*Hypergeometric2F1[3/4, 7/4, 11/4, (1 -
I*a*x)/2]))/(84*a^3)

________________________________________________________________________________________

Maple [F]  time = 0.158, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.05278, size = 678, normalized size = 2. \begin{align*} -\frac{12 \, a^{3} \sqrt{\frac{289 i}{64 \, a^{6}}} \log \left (\frac{8}{17} i \, a^{3} \sqrt{\frac{289 i}{64 \, a^{6}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt{\frac{289 i}{64 \, a^{6}}} \log \left (-\frac{8}{17} i \, a^{3} \sqrt{\frac{289 i}{64 \, a^{6}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, a^{3} \sqrt{-\frac{289 i}{64 \, a^{6}}} \log \left (\frac{8}{17} i \, a^{3} \sqrt{-\frac{289 i}{64 \, a^{6}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt{-\frac{289 i}{64 \, a^{6}}} \log \left (-\frac{8}{17} i \, a^{3} \sqrt{-\frac{289 i}{64 \, a^{6}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) +{\left (8 \, a^{3} x^{3} + 22 i \, a^{2} x^{2} - 37 \, a x - 23 i\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{24 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*a^3*sqrt(289/64*I/a^6)*log(8/17*I*a^3*sqrt(289/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12
*a^3*sqrt(289/64*I/a^6)*log(-8/17*I*a^3*sqrt(289/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 12*a^3*sqr
t(-289/64*I/a^6)*log(8/17*I*a^3*sqrt(-289/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12*a^3*sqrt(-289/
64*I/a^6)*log(-8/17*I*a^3*sqrt(-289/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + (8*a^3*x^3 + 22*I*a^2*x
^2 - 37*a*x - 23*I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^3

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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