3.1 \(\int e^{i \tan ^{-1}(a x)} x^4 \, dx\)

Optimal. Leaf size=113 \[ \frac {3 \sinh ^{-1}(a x)}{8 a^5}+\frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}+\frac {x^3 \sqrt {a^2 x^2+1}}{4 a^2}+\frac {(-45 a x+64 i) \sqrt {a^2 x^2+1}}{120 a^5}-\frac {4 i x^2 \sqrt {a^2 x^2+1}}{15 a^3} \]

[Out]

3/8*arcsinh(a*x)/a^5-4/15*I*x^2*(a^2*x^2+1)^(1/2)/a^3+1/4*x^3*(a^2*x^2+1)^(1/2)/a^2+1/5*I*x^4*(a^2*x^2+1)^(1/2
)/a+1/120*(64*I-45*a*x)*(a^2*x^2+1)^(1/2)/a^5

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Rubi [A]  time = 0.09, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5060, 833, 780, 215} \[ \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}+\frac {x^3 \sqrt {a^2 x^2+1}}{4 a^2}-\frac {4 i x^2 \sqrt {a^2 x^2+1}}{15 a^3}+\frac {(-45 a x+64 i) \sqrt {a^2 x^2+1}}{120 a^5}+\frac {3 \sinh ^{-1}(a x)}{8 a^5} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a*x])*x^4,x]

[Out]

(((-4*I)/15)*x^2*Sqrt[1 + a^2*x^2])/a^3 + (x^3*Sqrt[1 + a^2*x^2])/(4*a^2) + ((I/5)*x^4*Sqrt[1 + a^2*x^2])/a +
((64*I - 45*a*x)*Sqrt[1 + a^2*x^2])/(120*a^5) + (3*ArcSinh[a*x])/(8*a^5)

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^4 \, dx &=\int \frac {x^4 (1+i a x)}{\sqrt {1+a^2 x^2}} \, dx\\ &=\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x^3 \left (-4 i a+5 a^2 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{5 a^2}\\ &=\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x^2 \left (-15 a^2-16 i a^3 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{20 a^4}\\ &=-\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {\int \frac {x \left (32 i a^3-45 a^4 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{60 a^6}\\ &=-\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{8 a^4}\\ &=-\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \sinh ^{-1}(a x)}{8 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 64, normalized size = 0.57 \[ \frac {45 \sinh ^{-1}(a x)+\sqrt {a^2 x^2+1} \left (24 i a^4 x^4+30 a^3 x^3-32 i a^2 x^2-45 a x+64 i\right )}{120 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(I*ArcTan[a*x])*x^4,x]

[Out]

(Sqrt[1 + a^2*x^2]*(64*I - 45*a*x - (32*I)*a^2*x^2 + 30*a^3*x^3 + (24*I)*a^4*x^4) + 45*ArcSinh[a*x])/(120*a^5)

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fricas [A]  time = 0.46, size = 67, normalized size = 0.59 \[ \frac {{\left (24 i \, a^{4} x^{4} + 30 \, a^{3} x^{3} - 32 i \, a^{2} x^{2} - 45 \, a x + 64 i\right )} \sqrt {a^{2} x^{2} + 1} - 45 \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{120 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^4,x, algorithm="fricas")

[Out]

1/120*((24*I*a^4*x^4 + 30*a^3*x^3 - 32*I*a^2*x^2 - 45*a*x + 64*I)*sqrt(a^2*x^2 + 1) - 45*log(-a*x + sqrt(a^2*x
^2 + 1)))/a^5

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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)*x^4,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 [A]  time = 0.16, size = 128, normalized size = 1.13 \[ \frac {i x^{4} \sqrt {a^{2} x^{2}+1}}{5 a}-\frac {4 i x^{2} \sqrt {a^{2} x^{2}+1}}{15 a^{3}}+\frac {8 i \sqrt {a^{2} x^{2}+1}}{15 a^{5}}+\frac {x^{3} \sqrt {a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 x \sqrt {a^{2} x^{2}+1}}{8 a^{4}}+\frac {3 \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{4} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^4,x)

[Out]

1/5*I*x^4*(a^2*x^2+1)^(1/2)/a-4/15*I*x^2*(a^2*x^2+1)^(1/2)/a^3+8/15*I/a^5*(a^2*x^2+1)^(1/2)+1/4*x^3*(a^2*x^2+1
)^(1/2)/a^2-3/8/a^4*x*(a^2*x^2+1)^(1/2)+3/8/a^4*ln(x*a^2/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)

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maxima [A]  time = 0.36, size = 100, normalized size = 0.88 \[ \frac {i \, \sqrt {a^{2} x^{2} + 1} x^{4}}{5 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a^{2}} - \frac {4 i \, \sqrt {a^{2} x^{2} + 1} x^{2}}{15 \, a^{3}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{5}} + \frac {8 i \, \sqrt {a^{2} x^{2} + 1}}{15 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^4,x, algorithm="maxima")

[Out]

1/5*I*sqrt(a^2*x^2 + 1)*x^4/a + 1/4*sqrt(a^2*x^2 + 1)*x^3/a^2 - 4/15*I*sqrt(a^2*x^2 + 1)*x^2/a^3 - 3/8*sqrt(a^
2*x^2 + 1)*x/a^4 + 3/8*arcsinh(a*x)/a^5 + 8/15*I*sqrt(a^2*x^2 + 1)/a^5

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mupad [B]  time = 0.10, size = 98, normalized size = 0.87 \[ \frac {\sqrt {a^2\,x^2+1}\,\left (\frac {x^3\,{\left (a^2\right )}^{3/2}}{4\,a^4}-\frac {3\,x\,\sqrt {a^2}}{8\,a^4}+\frac {a\,8{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}-\frac {a^3\,x^2\,4{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}+\frac {a^5\,x^4\,1{}\mathrm {i}}{5\,{\left (a^2\right )}^{5/2}}\right )}{\sqrt {a^2}}+\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{8\,a^4\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a*x*1i + 1))/(a^2*x^2 + 1)^(1/2),x)

[Out]

((a^2*x^2 + 1)^(1/2)*((a*8i)/(15*(a^2)^(5/2)) - (a^3*x^2*4i)/(15*(a^2)^(5/2)) + (x^3*(a^2)^(3/2))/(4*a^4) + (a
^5*x^4*1i)/(5*(a^2)^(5/2)) - (3*x*(a^2)^(1/2))/(8*a^4)))/(a^2)^(1/2) + (3*asinh(x*(a^2)^(1/2)))/(8*a^4*(a^2)^(
1/2))

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sympy [A]  time = 5.02, size = 138, normalized size = 1.22 \[ i a \left (\begin {cases} \frac {x^{4} \sqrt {a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {a^{2} x^{2} + 1}}{15 a^{4}} + \frac {8 \sqrt {a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) + \frac {x^{5}}{4 \sqrt {a^{2} x^{2} + 1}} - \frac {x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {a^{2} x^{2} + 1}} + \frac {3 \operatorname {asinh}{\left (a x \right )}}{8 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)/(a**2*x**2+1)**(1/2)*x**4,x)

[Out]

I*a*Piecewise((x**4*sqrt(a**2*x**2 + 1)/(5*a**2) - 4*x**2*sqrt(a**2*x**2 + 1)/(15*a**4) + 8*sqrt(a**2*x**2 + 1
)/(15*a**6), Ne(a, 0)), (x**6/6, True)) + x**5/(4*sqrt(a**2*x**2 + 1)) - x**3/(8*a**2*sqrt(a**2*x**2 + 1)) - 3
*x/(8*a**4*sqrt(a**2*x**2 + 1)) + 3*asinh(a*x)/(8*a**5)

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