Integrand size = 14, antiderivative size = 90 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}+\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16+9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}+\frac {3 i \text {arcsinh}(a x)}{8 a^4} \]
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
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {\sqrt {1+a^2 x^2} \left (-16-9 i a x+8 a^2 x^2+6 i a^3 x^3\right )+9 i \text {arcsinh}(a x)}{24 a^4} \]
(Sqrt[1 + a^2*x^2]*(-16 - (9*I)*a*x + 8*a^2*x^2 + (6*I)*a^3*x^3) + (9*I)*A rcSinh[a*x])/(24*a^4)
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.39, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5583, 533, 27, 533, 25, 27, 533, 27, 455, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 e^{i \arctan (a x)} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {x^3 (1+i a x)}{\sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {\int \frac {a x^2 (3 i-4 a x)}{\sqrt {a^2 x^2+1}}dx}{4 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {\int \frac {x^2 (3 i-4 a x)}{\sqrt {a^2 x^2+1}}dx}{4 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {\int -\frac {a x (9 i a x+8)}{\sqrt {a^2 x^2+1}}dx}{3 a^2}}{4 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\int \frac {a x (9 i a x+8)}{\sqrt {a^2 x^2+1}}dx}{3 a^2}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\int \frac {x (9 i a x+8)}{\sqrt {a^2 x^2+1}}dx}{3 a}}{4 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\frac {9 i x \sqrt {a^2 x^2+1}}{2 a}-\frac {\int \frac {a (9 i-16 a x)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\frac {9 i x \sqrt {a^2 x^2+1}}{2 a}-\frac {\int \frac {9 i-16 a x}{\sqrt {a^2 x^2+1}}dx}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\frac {9 i x \sqrt {a^2 x^2+1}}{2 a}-\frac {-\frac {16 \sqrt {a^2 x^2+1}}{a}+9 i \int \frac {1}{\sqrt {a^2 x^2+1}}dx}{2 a}}{3 a}}{4 a}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {-\frac {4 x^2 \sqrt {a^2 x^2+1}}{3 a}+\frac {\frac {9 i x \sqrt {a^2 x^2+1}}{2 a}-\frac {-\frac {16 \sqrt {a^2 x^2+1}}{a}+\frac {9 i \text {arcsinh}(a x)}{a}}{2 a}}{3 a}}{4 a}\) |
((I/4)*x^3*Sqrt[1 + a^2*x^2])/a - ((-4*x^2*Sqrt[1 + a^2*x^2])/(3*a) + (((( 9*I)/2)*x*Sqrt[1 + a^2*x^2])/a - ((-16*Sqrt[1 + a^2*x^2])/a + ((9*I)*ArcSi nh[a*x])/a)/(2*a))/(3*a))/(4*a)
3.1.2.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
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] /; Free Q[{a, m}, x] && IntegerQ[(I*n - 1)/2]
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {i \left (6 a^{3} x^{3}-8 i a^{2} x^{2}-9 a x +16 i\right ) \sqrt {a^{2} x^{2}+1}}{24 a^{4}}+\frac {3 i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{3} \sqrt {a^{2}}}\) | \(77\) |
meijerg | \(\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right ) \sqrt {a^{2} x^{2}+1}}{6}}{2 a^{4} \sqrt {\pi }}+\frac {i \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right ) \sqrt {a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{4 a^{5}}\right )}{2 a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}\) | \(109\) |
default | \(\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}}+i a \left (\frac {x^{3} \sqrt {a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 \left (\frac {x \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}\right )\) | \(117\) |
1/24*I*(6*a^3*x^3-8*I*a^2*x^2-9*a*x+16*I)*(a^2*x^2+1)^(1/2)/a^4+3/8*I/a^3* ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {{\left (6 i \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 9 i \, a x - 16\right )} \sqrt {a^{2} x^{2} + 1} - 9 i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{24 \, a^{4}} \]
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
Time = 0.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16 \[ \int e^{i \arctan (a x)} x^3 \, dx=\begin {cases} \sqrt {a^{2} x^{2} + 1} \left (\frac {i x^{3}}{4 a} + \frac {x^{2}}{3 a^{2}} - \frac {3 i x}{8 a^{3}} - \frac {2}{3 a^{4}}\right ) + \frac {3 i \log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{8 a^{3} \sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{5}}{5} + \frac {x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a**2*x**2 + 1)*(I*x**3/(4*a) + x**2/(3*a**2) - 3*I*x/(8*a* *3) - 2/(3*a**4)) + 3*I*log(2*a**2*x + 2*sqrt(a**2*x**2 + 1)*sqrt(a**2))/( 8*a**3*sqrt(a**2)), Ne(a**2, 0)), (I*a*x**5/5 + x**4/4, True))
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac {3 i \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {3 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{3 \, a^{4}} \]
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
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int e^{i \arctan (a x)} x^3 \, dx=-\frac {1}{24} \, \sqrt {a^{2} x^{2} + 1} {\left ({\left (2 \, x {\left (-\frac {3 i \, x}{a} - \frac {4}{a^{2}}\right )} + \frac {9 i}{a^{3}}\right )} x + \frac {16}{a^{4}}\right )} - \frac {3 i \, \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{8 \, a^{3} {\left | a \right |}} \]
-1/24*sqrt(a^2*x^2 + 1)*((2*x*(-3*I*x/a - 4/a^2) + 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))
Time = 0.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,3{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {2}{3\,{\left (a^2\right )}^{3/2}}-\frac {a^2\,x^2}{3\,{\left (a^2\right )}^{3/2}}-\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}+\frac {x\,\sqrt {a^2}\,3{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}} \]