3.27 \(\int e^{4 i \tan ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=65 \[ \frac {4 i}{a^4 (a x+i)}+\frac {16 \log (a x+i)}{a^4}+\frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4} \]

[Out]

12*I*x/a^3-4*x^2/a^2-4/3*I*x^3/a+1/4*x^4+4*I/a^4/(I+a*x)+16*ln(I+a*x)/a^4

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5062, 88} \[ -\frac {4 x^2}{a^2}+\frac {12 i x}{a^3}+\frac {4 i}{a^4 (a x+i)}+\frac {16 \log (a x+i)}{a^4}-\frac {4 i x^3}{3 a}+\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])*x^3,x]

[Out]

((12*I)*x)/a^3 - (4*x^2)/a^2 - (((4*I)/3)*x^3)/a + x^4/4 + (4*I)/(a^4*(I + a*x)) + (16*Log[I + a*x])/a^4

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

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Mathematica [A]  time = 0.05, size = 65, normalized size = 1.00 \[ \frac {4 i}{a^4 (a x+i)}+\frac {16 \log (a x+i)}{a^4}+\frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])*x^3,x]

[Out]

((12*I)*x)/a^3 - (4*x^2)/a^2 - (((4*I)/3)*x^3)/a + x^4/4 + (4*I)/(a^4*(I + a*x)) + (16*Log[I + a*x])/a^4

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fricas [A]  time = 0.42, size = 70, normalized size = 1.08 \[ \frac {3 \, a^{5} x^{5} - 13 i \, a^{4} x^{4} - 32 \, a^{3} x^{3} + 96 i \, a^{2} x^{2} - 144 \, a x + {\left (192 \, a x + 192 i\right )} \log \left (\frac {a x + i}{a}\right ) + 48 i}{12 \, a^{5} x + 12 i \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^3,x, algorithm="fricas")

[Out]

(3*a^5*x^5 - 13*I*a^4*x^4 - 32*a^3*x^3 + 96*I*a^2*x^2 - 144*a*x + (192*a*x + 192*I)*log((a*x + I)/a) + 48*I)/(
12*a^5*x + 12*I*a^4)

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giac [A]  time = 0.12, size = 63, normalized size = 0.97 \[ \frac {16 \, \log \left (a x + i\right )}{a^{4}} + \frac {4 \, i}{{\left (a x + i\right )} a^{4}} + \frac {3 \, a^{8} x^{4} - 16 \, a^{7} i x^{3} - 48 \, a^{6} x^{2} + 144 \, a^{5} i x}{12 \, a^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^3,x, algorithm="giac")

[Out]

16*log(a*x + i)/a^4 + 4*i/((a*x + i)*a^4) + 1/12*(3*a^8*x^4 - 16*a^7*i*x^3 - 48*a^6*x^2 + 144*a^5*i*x)/a^8

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maple [A]  time = 0.06, size = 70, normalized size = 1.08 \[ \frac {x^{4}}{4}-\frac {4 i x^{3}}{3 a}-\frac {4 x^{2}}{a^{2}}+\frac {12 i x}{a^{3}}+\frac {4 i}{a^{4} \left (a x +i\right )}+\frac {8 \ln \left (a^{2} x^{2}+1\right )}{a^{4}}-\frac {16 i \arctan \left (a x \right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2*x^3,x)

[Out]

1/4*x^4-4/3*I*x^3/a-4*x^2/a^2+12*I*x/a^3+4*I/a^4/(I+a*x)+8/a^4*ln(a^2*x^2+1)-16*I/a^4*arctan(a*x)

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maxima [A]  time = 0.42, size = 77, normalized size = 1.18 \[ -\frac {4 \, {\left (-i \, a x - 1\right )}}{a^{6} x^{2} + a^{4}} + \frac {3 \, a^{3} x^{4} - 16 i \, a^{2} x^{3} - 48 \, a x^{2} + 144 i \, x}{12 \, a^{3}} - \frac {16 i \, \arctan \left (a x\right )}{a^{4}} + \frac {8 \, \log \left (a^{2} x^{2} + 1\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^3,x, algorithm="maxima")

[Out]

-4*(-I*a*x - 1)/(a^6*x^2 + a^4) + 1/12*(3*a^3*x^4 - 16*I*a^2*x^3 - 48*a*x^2 + 144*I*x)/a^3 - 16*I*arctan(a*x)/
a^4 + 8*log(a^2*x^2 + 1)/a^4

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mupad [B]  time = 0.43, size = 60, normalized size = 0.92 \[ \frac {16\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^4}+\frac {x^4}{4}-\frac {4\,x^2}{a^2}+\frac {4{}\mathrm {i}}{a^5\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}+\frac {x\,12{}\mathrm {i}}{a^3}-\frac {x^3\,4{}\mathrm {i}}{3\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a*x*1i + 1)^4)/(a^2*x^2 + 1)^2,x)

[Out]

4i/(a^5*(x + 1i/a)) + (16*log(x + 1i/a))/a^4 + (x*12i)/a^3 + x^4/4 - (x^3*4i)/(3*a) - (4*x^2)/a^2

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sympy [A]  time = 0.24, size = 56, normalized size = 0.86 \[ \frac {x^{4}}{4} + \frac {4 i}{a^{5} x + i a^{4}} - \frac {4 i x^{3}}{3 a} - \frac {4 x^{2}}{a^{2}} + \frac {12 i x}{a^{3}} + \frac {16 \log {\left (a x + i \right )}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2*x**3,x)

[Out]

x**4/4 + 4*I/(a**5*x + I*a**4) - 4*I*x**3/(3*a) - 4*x**2/a**2 + 12*I*x/a**3 + 16*log(a*x + I)/a**4

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