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

Optimal. Leaf size=337 \[ -\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {123 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt {2} a^4}-\frac {123 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt {2} a^4}-\frac {123 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}+\frac {123 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2} \]

[Out]

-41/64*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a^4+1/4*x^2*(1-I*a*x)^(7/4)*(1+I*a*x)^(1/4)/a^2-1/32*(1-I*a*x)^(7/4)*(1
+I*a*x)^(1/4)*(11-4*I*a*x)/a^4-123/128*arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^4*2^(1/2)+123/128*a
rctan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^4*2^(1/2)+123/256*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^4*2^(1/2)-123/256*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^4*2^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 337, 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, 100, 147, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {123 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt {2} a^4}-\frac {123 \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{128 \sqrt {2} a^4}-\frac {123 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4}+\frac {123 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{64 \sqrt {2} a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-41*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(64*a^4) + (x^2*(1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(4*a^2) - ((1 -
I*a*x)^(7/4)*(1 + I*a*x)^(1/4)*(11 - (4*I)*a*x))/(32*a^4) - (123*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 +
I*a*x)^(1/4)])/(64*Sqrt[2]*a^4) + (123*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(64*Sqrt[2]*
a^4) + (123*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*Sqr
t[2]*a^4) - (123*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)])/(12
8*Sqrt[2]*a^4)

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 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 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 {3}{2} i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac {x^3 (1-i a x)^{3/4}}{(1+i a x)^{3/4}} \, dx\\ &=\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}+\frac {\int \frac {x (1-i a x)^{3/4} \left (-2+\frac {3 i a x}{2}\right )}{(1+i a x)^{3/4}} \, dx}{4 a^2}\\ &=\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {(41 i) \int \frac {(1-i a x)^{3/4}}{(1+i a x)^{3/4}} \, dx}{64 a^3}\\ &=-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {(123 i) \int \frac {1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{128 a^3}\\ &=-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}+\frac {123 \operatorname {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{32 a^4}\\ &=-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}+\frac {123 \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 )}{32 a^4}\\ &=-\frac {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {123 \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 {123 \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 {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}+\frac {123 \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 {123 \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 {123 \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 {123 \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 {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}+\frac {123 \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 {123 \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 {123 \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 {123 \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 {41 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{64 a^4}+\frac {x^2 (1-i a x)^{7/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {(1-i a x)^{7/4} \sqrt [4]{1+i a x} (11-4 i a x)}{32 a^4}-\frac {123 \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 {123 \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 {123 \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 {123 \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*}

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Mathematica [C]  time = 0.12, size = 127, normalized size = 0.38 \[ \frac {(1-i a x)^{7/4} \left (7 a^2 x^2 \sqrt [4]{1+i a x}+12 \sqrt [4]{2} \, _2F_1\left (-\frac {5}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-i a x)\right )-20 \sqrt [4]{2} \, _2F_1\left (-\frac {1}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-i a x)\right )+7 \sqrt [4]{2} \, _2F_1\left (\frac {3}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-i a x)\right )\right )}{28 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*a*x)^(7/4)*(7*a^2*x^2*(1 + I*a*x)^(1/4) + 12*2^(1/4)*Hypergeometric2F1[-5/4, 7/4, 11/4, (1 - I*a*x)/2]
- 20*2^(1/4)*Hypergeometric2F1[-1/4, 7/4, 11/4, (1 - I*a*x)/2] + 7*2^(1/4)*Hypergeometric2F1[3/4, 7/4, 11/4,
(1 - I*a*x)/2]))/(28*a^4)

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fricas [A]  time = 0.43, size = 251, normalized size = 0.74 \[ -\frac {32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} \, a^{4} \sqrt {\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (\frac {64}{123} \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 32 \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} \log \left (-\frac {64}{123} \, a^{4} \sqrt {-\frac {15129 i}{4096 \, a^{8}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (16 \, a^{4} x^{4} + 40 i \, a^{3} x^{3} - 54 \, a^{2} x^{2} - 93 i \, a x + 63\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{64 \, a^{4}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

-1/64*(32*a^4*sqrt(15129/4096*I/a^8)*log(64/123*a^4*sqrt(15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I
))) - 32*a^4*sqrt(15129/4096*I/a^8)*log(-64/123*a^4*sqrt(15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I
))) + 32*a^4*sqrt(-15129/4096*I/a^8)*log(64/123*a^4*sqrt(-15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x +
I))) - 32*a^4*sqrt(-15129/4096*I/a^8)*log(-64/123*a^4*sqrt(-15129/4096*I/a^8) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x
+ I))) + (16*a^4*x^4 + 40*I*a^3*x^3 - 54*a^2*x^2 - 93*I*a*x + 63)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a^4

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {3}{2}}}\, dx \]

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))^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

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

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