3.80 \(\int e^{\frac {5}{2} i \tan ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=371 \[ \frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3} \]

[Out]

55/8*I*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a^3+11/4*I*(1-I*a*x)^(3/4)*(1+I*a*x)^(5/4)/a^3+2*I*(1+I*a*x)^(9/4)/a^3/
(1-I*a*x)^(1/4)+1/3*I*(1-I*a*x)^(3/4)*(1+I*a*x)^(9/4)/a^3-55/16*I*arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(
1/4))/a^3*2^(1/2)+55/16*I*arctan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^3*2^(1/2)+55/32*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)-55/32*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 = 0.24, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5062, 89, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(((5*I)/2)*ArcTan[a*x])*x^2,x]

[Out]

(((55*I)/8)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a^3 + (((11*I)/4)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(5/4))/a^3 +
((2*I)*(1 + I*a*x)^(9/4))/(a^3*(1 - I*a*x)^(1/4)) + ((I/3)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(9/4))/a^3 - (((55*I)
/8)*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) + (((55*I)/8)*ArcTan[1 + (Sqrt[2]
*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) + (((55*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) - (((55*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 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 89

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

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 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 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 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 {5}{2} i \tan ^{-1}(a x)} x^2 \, dx &=\int \frac {x^2 (1+i a x)^{5/4}}{(1-i a x)^{5/4}} \, dx\\ &=\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}-\frac {(2 i) \int \frac {(1+i a x)^{5/4} \left (\frac {5 i a}{2}-\frac {a^2 x}{2}\right )}{\sqrt [4]{1-i a x}} \, dx}{a^3}\\ &=\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {11 \int \frac {(1+i a x)^{5/4}}{\sqrt [4]{1-i a x}} \, dx}{2 a^2}\\ &=\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {55 \int \frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx}{8 a^2}\\ &=\frac {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {55 \int \frac {1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{16 a^2}\\ &=\frac {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {(55 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 {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {(55 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 {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}-\frac {(55 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 {(55 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 {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {(55 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 {(55 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 {(55 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 {(55 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 {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}+\frac {55 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 {55 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 {(55 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 {(55 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 {55 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}+\frac {11 i (1-i a x)^{3/4} (1+i a x)^{5/4}}{4 a^3}+\frac {2 i (1+i a x)^{9/4}}{a^3 \sqrt [4]{1-i a x}}+\frac {i (1-i a x)^{3/4} (1+i a x)^{9/4}}{3 a^3}-\frac {55 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 {55 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 {55 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 {55 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 = 0.04, size = 86, normalized size = 0.23 \[ \frac {44 \sqrt [4]{2} (a x+i) \, _2F_1\left (-\frac {5}{4},\frac {3}{4};\frac {7}{4};\frac {1}{2} (1-i a x)\right )-\sqrt [4]{1+i a x} (a x-i)^2 (a x+7 i)}{3 a^3 \sqrt [4]{1-i a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((5*I)/2)*ArcTan[a*x])*x^2,x]

[Out]

(-((1 + I*a*x)^(1/4)*(-I + a*x)^2*(7*I + a*x)) + 44*2^(1/4)*(I + a*x)*Hypergeometric2F1[-5/4, 3/4, 7/4, (1 - I
*a*x)/2])/(3*a^3*(1 - I*a*x)^(1/4))

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fricas [A]  time = 0.52, size = 244, normalized size = 0.66 \[ \frac {12 \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} i \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} i \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} i \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} i \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (8 \, a^{3} x^{3} - 26 i \, a^{2} x^{2} - 61 \, a x - 287 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{24 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x, algorithm="fricas")

[Out]

1/24*(12*a^3*sqrt(3025/64*I/a^6)*log(8/55*I*a^3*sqrt(3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 1
2*a^3*sqrt(3025/64*I/a^6)*log(-8/55*I*a^3*sqrt(3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 12*a^3*
sqrt(-3025/64*I/a^6)*log(8/55*I*a^3*sqrt(-3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 12*a^3*sqrt(
-3025/64*I/a^6)*log(-8/55*I*a^3*sqrt(-3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (8*a^3*x^3 - 26*
I*a^2*x^2 - 61*a*x - 287*I)*sqrt(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(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^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 Valuesym2poly/r2sym(const g
en & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2} \,d x \]

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))^(5/2),x)

[Out]

int(x^2*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)*x**2,x)

[Out]

Timed out

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