3.1.27 \(\int e^{4 i \text {ArcTan}(a x)} x^3 \, dx\) [27]

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

[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 = {5170, 90} \begin {gather*} \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} \end {gather*}

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 90

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 5170

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 = 0.03, size = 65, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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Maple [A]
time = 0.10, size = 67, normalized size = 1.03

method	result	size
default	\(-\frac {-\frac {1}{4} a^{3} x^{4}+\frac {4}{3} i a^{2} x^{3}+4 a \,x^{2}-12 i x}{a^{3}}-\frac {4 \left (-\frac {i}{a \left (a x +i\right )}-\frac {4 \ln \left (a x +i\right )}{a}\right )}{a^{3}}\)	\(67\)
risch	\(\frac {x^{4}}{4}-\frac {4 i x^{3}}{3 a}-\frac {4 x^{2}}{a^{2}}+\frac {12 i x}{a^{3}}+\frac {4 i}{a^{4} \left (a x +i\right )}+\frac {8 \ln \left (a^{2} x^{2}+1\right )}{a^{4}}-\frac {16 i \arctan \left (a x \right )}{a^{4}}\)	\(70\)
meijerg	\(\frac {-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}+\frac {2 i \left (\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a^{3} \sqrt {a^{2}}}-\frac {3 \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{3 a^{2} x^{2}+3}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{a^{4}}-\frac {2 i \left (-\frac {x \left (a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}+70 a^{2} x^{2}+105\right )}{21 a^{6} \left (a^{2} x^{2}+1\right )}+\frac {5 \left (a^{2}\right )^{\frac {7}{2}} \arctan \left (a x \right )}{a^{7}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {-\frac {x^{2} a^{2} \left (-2 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{4 \left (a^{2} x^{2}+1\right )}+3 \ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}\)	\(263\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 77, normalized size = 1.18 \begin {gather*} -\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}} \end {gather*}

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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Fricas [A]
time = 2.20, size = 70, normalized size = 1.08 \begin {gather*} \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 + 192 \, {\left (a x + i\right )} \log \left (\frac {a x + i}{a}\right ) + 48 i}{12 \, {\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)^4/(a^2*x^2+1)^2*x^3,x, algorithm="fricas")

[Out]

1/12*(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 + I)*log((a*x + I)/a) + 48*I)/
(a^5*x + I*a^4)

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Sympy [A]
time = 0.11, size = 56, normalized size = 0.86 \begin {gather*} \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}} \end {gather*}

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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Giac [A]
time = 0.41, size = 60, normalized size = 0.92 \begin {gather*} \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 i \, a^{7} x^{3} - 48 \, a^{6} x^{2} + 144 i \, a^{5} x}{12 \, a^{8}} \end {gather*}

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*I*a^7*x^3 - 48*a^6*x^2 + 144*I*a^5*x)/a^8

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Mupad [B]
time = 0.43, size = 60, normalized size = 0.92 \begin {gather*} \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} \end {gather*}

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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