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

Optimal. Leaf size=90 \[ \frac {3 i \sinh ^{-1}(a x)}{8 a^4}+\frac {x^2 \sqrt {a^2 x^2+1}}{3 a^2}+\frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {(16+9 i a x) \sqrt {a^2 x^2+1}}{24 a^4} \]

[Out]

3/8*I*arcsinh(a*x)/a^4+1/3*x^2*(a^2*x^2+1)^(1/2)/a^2+1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-1/24*(16+9*I*a*x)*(a^2*x^2+
1)^(1/2)/a^4

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Rubi [A]  time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5060, 833, 780, 215} \[ \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {x^2 \sqrt {a^2 x^2+1}}{3 a^2}-\frac {(16+9 i a x) \sqrt {a^2 x^2+1}}{24 a^4}+\frac {3 i \sinh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[1 + a^2*x^2])/(3*a^2) + ((I/4)*x^3*Sqrt[1 + a^2*x^2])/a - ((16 + (9*I)*a*x)*Sqrt[1 + a^2*x^2])/(24*a
^4) + (((3*I)/8)*ArcSinh[a*x])/a^4

Rule 215

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

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

Rule 5060

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^{i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac {x^3 (1+i a x)}{\sqrt {1+a^2 x^2}} \, dx\\ &=\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}+\frac {\int \frac {x^2 \left (-3 i a+4 a^2 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}+\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}+\frac {\int \frac {x \left (-8 a^2-9 i a^3 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}+\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16+9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}+\frac {(3 i) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{8 a^3}\\ &=\frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}+\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16+9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}+\frac {3 i \sinh ^{-1}(a x)}{8 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.62 \[ \frac {\sqrt {a^2 x^2+1} \left (6 i a^3 x^3+8 a^2 x^2-9 i a x-16\right )+9 i \sinh ^{-1}(a x)}{24 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*(-16 - (9*I)*a*x + 8*a^2*x^2 + (6*I)*a^3*x^3) + (9*I)*ArcSinh[a*x])/(24*a^4)

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fricas [A]  time = 0.47, size = 59, normalized size = 0.66 \[ \frac {{\left (6 i \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 9 i \, a x - 16\right )} \sqrt {a^{2} x^{2} + 1} - 9 i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{24 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((6*I*a^3*x^3 + 8*a^2*x^2 - 9*I*a*x - 16)*sqrt(a^2*x^2 + 1) - 9*I*log(-a*x + sqrt(a^2*x^2 + 1)))/a^4

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giac [A]  time = 2.01, size = 73, normalized size = 0.81 \[ \frac {1}{24} \, \sqrt {a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (\frac {3 \, i x}{a} + \frac {4}{a^{2}}\right )} x - \frac {9 \, i}{a^{3}}\right )} x - \frac {16}{a^{4}}\right )} - \frac {3 \, i \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{8 \, a^{3} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(a^2*x^2 + 1)*((2*(3*i*x/a + 4/a^2)*x - 9*i/a^3)*x - 16/a^4) - 3/8*i*log(-x*abs(a) + sqrt(a^2*x^2 + 1
))/(a^3*abs(a))

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maple [A]  time = 0.18, size = 109, normalized size = 1.21 \[ \frac {i x^{3} \sqrt {a^{2} x^{2}+1}}{4 a}-\frac {3 i x \sqrt {a^{2} x^{2}+1}}{8 a^{3}}+\frac {3 i \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{3} \sqrt {a^{2}}}+\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-3/8*I/a^3*x*(a^2*x^2+1)^(1/2)+3/8*I/a^3*ln(x*a^2/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/
(a^2)^(1/2)+1/3*x^2*(a^2*x^2+1)^(1/2)/a^2-2/3*(a^2*x^2+1)^(1/2)/a^4

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maxima [A]  time = 0.32, size = 81, normalized size = 0.90 \[ \frac {i \, \sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac {3 i \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {3 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{3 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.43, size = 85, normalized size = 0.94 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,3{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {2}{3\,{\left (a^2\right )}^{3/2}}-\frac {a^2\,x^2}{3\,{\left (a^2\right )}^{3/2}}-\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}+\frac {x\,\sqrt {a^2}\,3{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}} \]

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),x)

[Out]

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

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sympy [A]  time = 4.49, size = 119, normalized size = 1.32 \[ \frac {i a x^{5}}{4 \sqrt {a^{2} x^{2} + 1}} + \begin {cases} \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} - \frac {i x^{3}}{8 a \sqrt {a^{2} x^{2} + 1}} - \frac {3 i x}{8 a^{3} \sqrt {a^{2} x^{2} + 1}} + \frac {3 i \operatorname {asinh}{\left (a x \right )}}{8 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*x**5/(4*sqrt(a**2*x**2 + 1)) + Piecewise((x**2*sqrt(a**2*x**2 + 1)/(3*a**2) - 2*sqrt(a**2*x**2 + 1)/(3*a**
4), Ne(a, 0)), (x**4/4, True)) - I*x**3/(8*a*sqrt(a**2*x**2 + 1)) - 3*I*x/(8*a**3*sqrt(a**2*x**2 + 1)) + 3*I*a
sinh(a*x)/(8*a**4)

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