3.1.19 \(\int e^{3 i \text {ArcTan}(a x)} x^3 \, dx\) [19]

Optimal. Leaf size=137 \[ \frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}+\frac {27 \sqrt {1+a^2 x^2}}{4 a^4}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {9 i (2 i-3 a x) \sqrt {1+a^2 x^2}}{8 a^4}-\frac {51 i \sinh ^{-1}(a x)}{8 a^4} \]

[Out]

-51/8*I*arcsinh(a*x)/a^4+(1+I*a*x)^3/a^4/(a^2*x^2+1)^(1/2)+27/4*(a^2*x^2+1)^(1/2)/a^4-x^2*(a^2*x^2+1)^(1/2)/a^
2-1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-9/8*I*(2*I-3*a*x)*(a^2*x^2+1)^(1/2)/a^4

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Rubi [A]
time = 0.44, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5168, 1647, 1607, 12, 866, 1649, 1829, 27, 757, 655, 221} \begin {gather*} -\frac {51 i \sinh ^{-1}(a x)}{8 a^4}-\frac {x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {9 i (-3 a x+2 i) \sqrt {a^2 x^2+1}}{8 a^4}+\frac {27 \sqrt {a^2 x^2+1}}{4 a^4}+\frac {(1+i a x)^3}{a^4 \sqrt {a^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + I*a*x)^3/(a^4*Sqrt[1 + a^2*x^2]) + (27*Sqrt[1 + a^2*x^2])/(4*a^4) - (x^2*Sqrt[1 + a^2*x^2])/a^2 - ((I/4)*
x^3*Sqrt[1 + a^2*x^2])/a - (((9*I)/8)*(2*I - 3*a*x)*Sqrt[1 + a^2*x^2])/a^4 - (((51*I)/8)*ArcSinh[a*x])/a^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 655

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

Rule 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rule 1829

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*(q + 2*p + 1))), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 5168

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rubi steps

\begin {align*} \int e^{3 i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac {x^3 (1+i a x)^2}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx\\ &=-\left ((i a) \int \frac {\sqrt {1+a^2 x^2} \left (\frac {i x^3}{a}-x^4\right )}{(1-i a x)^2} \, dx\right )\\ &=-\left ((i a) \int \frac {\left (\frac {i}{a}-x\right ) x^3 \sqrt {1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=a^2 \int \frac {x^3 \left (1+a^2 x^2\right )^{3/2}}{a^2 (1-i a x)^3} \, dx\\ &=\int \frac {x^3 \left (1+a^2 x^2\right )^{3/2}}{(1-i a x)^3} \, dx\\ &=\int \frac {x^3 (1+i a x)^3}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\int \frac {(1+i a x)^2 \left (\frac {3 i}{a^3}-\frac {x}{a^2}-\frac {i x^2}{a}\right )}{\sqrt {1+a^2 x^2}} \, dx\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {\int \frac {\frac {12 i}{a}-28 x-27 i a x^2+12 a^2 x^3}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {\int \frac {36 i a-108 a^2 x-81 i a^3 x^2}{\sqrt {1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {\int -\frac {9 i a (-2 i+3 a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}+\frac {(3 i) \int \frac {(-2 i+3 a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{4 a^3}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {9 i (2 i-3 a x) \sqrt {1+a^2 x^2}}{8 a^4}+\frac {(3 i) \int \frac {-17 a^2-18 i a^3 x}{\sqrt {1+a^2 x^2}} \, dx}{8 a^5}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}+\frac {27 \sqrt {1+a^2 x^2}}{4 a^4}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {9 i (2 i-3 a x) \sqrt {1+a^2 x^2}}{8 a^4}-\frac {(51 i) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{8 a^3}\\ &=\frac {(1+i a x)^3}{a^4 \sqrt {1+a^2 x^2}}+\frac {27 \sqrt {1+a^2 x^2}}{4 a^4}-\frac {x^2 \sqrt {1+a^2 x^2}}{a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {9 i (2 i-3 a x) \sqrt {1+a^2 x^2}}{8 a^4}-\frac {51 i \sinh ^{-1}(a x)}{8 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 80, normalized size = 0.58 \begin {gather*} \sqrt {1+a^2 x^2} \left (\frac {6}{a^4}+\frac {19 i x}{8 a^3}-\frac {x^2}{a^2}-\frac {i x^3}{4 a}+\frac {4 i}{a^4 (i+a x)}\right )-\frac {51 i \sinh ^{-1}(a x)}{8 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + a^2*x^2]*(6/a^4 + (((19*I)/8)*x)/a^3 - x^2/a^2 - ((I/4)*x^3)/a + (4*I)/(a^4*(I + a*x))) - (((51*I)/8)
*ArcSinh[a*x])/a^4

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (114 ) = 228\).
time = 0.13, size = 286, normalized size = 2.09

method	result	size
risch	\(-\frac {i \left (2 a^{3} x^{3}-8 i a^{2} x^{2}-19 a x +48 i\right ) \sqrt {a^{2} x^{2}+1}}{8 a^{4}}-\frac {51 i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{3} \sqrt {a^{2}}}+\frac {4 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{5} \left (x +\frac {i}{a}\right )}\)	\(122\)
meijerg	\(\frac {-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}}{a^{4} \sqrt {\pi }}+\frac {3 i \left (\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \arcsinh \left (a x \right )}{2 a^{5}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}-\frac {3 \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}+8 a^{2} x^{2}+16\right )}{6 \sqrt {a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {\pi }}-\frac {i \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{56 a^{6} \sqrt {a^{2} x^{2}+1}}+\frac {15 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {7}{2}} \arcsinh \left (a x \right )}{8 a^{7}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}\)	\(231\)
default	\(-i a^{3} \left (\frac {x^{5}}{4 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {5 \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )}{4 a^{2}}\right )-3 a^{2} \left (\frac {x^{4}}{3 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {4 \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )}{3 a^{2}}\right )+3 i a \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )+\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a^3*(1/4*x^5/a^2/(a^2*x^2+1)^(1/2)-5/4/a^2*(1/2*x^3/a^2/(a^2*x^2+1)^(1/2)-3/2/a^2*(-x/a^2/(a^2*x^2+1)^(1/2)
+1/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2))))-3*a^2*(1/3*x^4/a^2/(a^2*x^2+1)^(1/2)-4/3/a^2*(x^
2/a^2/(a^2*x^2+1)^(1/2)+2/a^4/(a^2*x^2+1)^(1/2)))+3*I*a*(1/2*x^3/a^2/(a^2*x^2+1)^(1/2)-3/2/a^2*(-x/a^2/(a^2*x^
2+1)^(1/2)+1/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)))+x^2/a^2/(a^2*x^2+1)^(1/2)+2/a^4/(a^2*x^
2+1)^(1/2)

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Maxima [A]
time = 0.29, size = 114, normalized size = 0.83 \begin {gather*} -\frac {i \, a x^{5}}{4 \, \sqrt {a^{2} x^{2} + 1}} - \frac {x^{4}}{\sqrt {a^{2} x^{2} + 1}} + \frac {17 i \, x^{3}}{8 \, \sqrt {a^{2} x^{2} + 1} a} + \frac {5 \, x^{2}}{\sqrt {a^{2} x^{2} + 1} a^{2}} + \frac {51 i \, x}{8 \, \sqrt {a^{2} x^{2} + 1} a^{3}} - \frac {51 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} + \frac {10}{\sqrt {a^{2} x^{2} + 1} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*a*x^5/sqrt(a^2*x^2 + 1) - x^4/sqrt(a^2*x^2 + 1) + 17/8*I*x^3/(sqrt(a^2*x^2 + 1)*a) + 5*x^2/(sqrt(a^2*x^
2 + 1)*a^2) + 51/8*I*x/(sqrt(a^2*x^2 + 1)*a^3) - 51/8*I*arcsinh(a*x)/a^4 + 10/(sqrt(a^2*x^2 + 1)*a^4)

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Fricas [A]
time = 2.38, size = 88, normalized size = 0.64 \begin {gather*} \frac {32 i \, a x - 51 \, {\left (-i \, a x + 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (-2 i \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 11 i \, a^{2} x^{2} + 29 \, a x + 80 i\right )} \sqrt {a^{2} x^{2} + 1} - 32}{8 \, {\left (a^{5} x + i \, a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(32*I*a*x - 51*(-I*a*x + 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + (-2*I*a^4*x^4 - 6*a^3*x^3 + 11*I*a^2*x^2 + 29*
a*x + 80*I)*sqrt(a^2*x^2 + 1) - 32)/(a^5*x + I*a^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \left (\int \frac {i x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x^{4}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{6}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{5}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(I*x**3/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x**4/(a**2*x**2*
sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**6/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2
*x**2 + 1)), x) + Integral(-3*I*a**2*x**5/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [B]
time = 0.46, size = 137, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a^2\,x^2+1}\,\left (\frac {4}{{\left (a^2\right )}^{3/2}}+\frac {2\,\sqrt {a^2}}{a^4}-\frac {x^2\,\sqrt {a^2}}{a^2}-\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}+\frac {x\,\sqrt {a^2}\,19{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,51{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}+\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a^3\,\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*x^2 + 1)^(1/2)*(4/(a^2)^(3/2) + (2*(a^2)^(1/2))/a^4 - (x^2*(a^2)^(1/2))/a^2 - (x^3*(a^2)^(3/2)*1i)/(4*a^
3) + (x*(a^2)^(1/2)*19i)/(8*a^3)))/(a^2)^(1/2) - (asinh(x*(a^2)^(1/2))*51i)/(8*a^3*(a^2)^(1/2)) + ((a^2*x^2 +
1)^(1/2)*4i)/(a^3*(((a^2)^(1/2)*1i)/a + x*(a^2)^(1/2))*(a^2)^(1/2))

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