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

Optimal. Leaf size=339 \[ \frac {i (1+i a x)^{3/4} (1-i a x)^{5/4}}{12 a^3}+\frac {3 i (1+i a x)^{3/4} \sqrt [4]{1-i a x}}{8 a^3}+\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}+\frac {3 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {x (1+i a x)^{3/4} (1-i a x)^{5/4}}{3 a^2} \]

[Out]

3/8*I*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a^3+1/12*I*(1-I*a*x)^(5/4)*(1+I*a*x)^(3/4)/a^3+1/3*x*(1-I*a*x)^(5/4)*(1+
I*a*x)^(3/4)/a^2+3/16*I*arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^3*2^(1/2)-3/16*I*arctan(1+(1-I*a*x
)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^3*2^(1/2)+3/32*I*ln(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/
2)/(1+I*a*x)^(1/2))/a^3*2^(1/2)-3/32*I*ln(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)^
(1/2))/a^3*2^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5062, 90, 80, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac {x (1+i a x)^{3/4} (1-i a x)^{5/4}}{3 a^2}+\frac {i (1+i a x)^{3/4} (1-i a x)^{5/4}}{12 a^3}+\frac {3 i (1+i a x)^{3/4} \sqrt [4]{1-i a x}}{8 a^3}+\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}+\frac {3 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/8)*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/a^3 + ((I/12)*(1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/a^3 + (x*(1
- I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(3*a^2) + (((3*I)/8)*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/
4)])/(Sqrt[2]*a^3) - (((3*I)/8)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) + (((
3*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) - (((3*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)])/(Sq
rt[2]*a^3)

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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int e^{-\frac {1}{2} i \tan ^{-1}(a x)} x^2 \, dx &=\int \frac {x^2 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx\\ &=\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}+\frac {\int \frac {\left (-1+\frac {i a x}{2}\right ) \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{3 a^2}\\ &=\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {3 \int \frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{8 a^2}\\ &=\frac {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {3 \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{16 a^2}\\ &=\frac {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{4 a^3}\\ &=\frac {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {(3 i) \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 )}{4 a^3}\\ &=\frac {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {(3 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 {(3 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 {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}-\frac {(3 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 {(3 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 {(3 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 {(3 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 {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}+\frac {3 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 {3 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 {(3 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 {(3 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 {3 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}+\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{12 a^3}+\frac {x (1-i a x)^{5/4} (1+i a x)^{3/4}}{3 a^2}+\frac {3 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 {3 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 {3 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 {3 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*}

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.22 \[ \frac {(1-i a x)^{5/4} \left (5 (1+i a x)^{3/4} (4 a x+i)-9 i 2^{3/4} \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-i a x)\right )\right )}{60 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*a*x)^(5/4)*(5*(1 + I*a*x)^(3/4)*(I + 4*a*x) - (9*I)*2^(3/4)*Hypergeometric2F1[1/4, 5/4, 9/4, (1 - I*a*
x)/2]))/(60*a^3)

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fricas [A]  time = 0.45, size = 247, normalized size = 0.73 \[ -\frac {12 \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} \log \left (\frac {8}{3} \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} \log \left (-\frac {8}{3} \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} \log \left (\frac {8}{3} \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} \log \left (-\frac {8}{3} \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \sqrt {a^{2} x^{2} + 1} {\left (-8 i \, a^{2} x^{2} + 10 \, a x + 11 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{24 \, a^{3}} \]

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))^(1/2),x, algorithm="fricas")

[Out]

-1/24*(12*a^3*sqrt(9/64*I/a^6)*log(8/3*a^3*sqrt(9/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12*a^3*sq
rt(9/64*I/a^6)*log(-8/3*a^3*sqrt(9/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12*a^3*sqrt(-9/64*I/a^6)
*log(8/3*a^3*sqrt(-9/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 12*a^3*sqrt(-9/64*I/a^6)*log(-8/3*a^3*
sqrt(-9/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - sqrt(a^2*x^2 + 1)*(-8*I*a^2*x^2 + 10*a*x + 11*I)*sq
rt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for
the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc in
dex_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with p
arameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argu
ment ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.T
he choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choos
e a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0
,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a pol
ynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + E
rror: Bad Argument ValueDone

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}}\, dx \]

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))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}}}\,{d x} \]

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))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}\, dx \]

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))**(1/2),x)

[Out]

Integral(x**2/sqrt(I*(a*x - I)/sqrt(a**2*x**2 + 1)), x)

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