Integrand size = 14, antiderivative size = 113 \[ \int e^{i \arctan (a x)} x^4 \, dx=-\frac {4 i x^2 \sqrt {1+a^2 x^2}}{15 a^3}+\frac {x^3 \sqrt {1+a^2 x^2}}{4 a^2}+\frac {i x^4 \sqrt {1+a^2 x^2}}{5 a}+\frac {(64 i-45 a x) \sqrt {1+a^2 x^2}}{120 a^5}+\frac {3 \text {arcsinh}(a x)}{8 a^5} \]
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
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {\sqrt {1+a^2 x^2} \left (64 i-45 a x-32 i a^2 x^2+30 a^3 x^3+24 i a^4 x^4\right )+45 \text {arcsinh}(a x)}{120 a^5} \]
(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)
Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {5583, 533, 27, 533, 25, 27, 533, 27, 533, 25, 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.
\(\displaystyle \int x^4 e^{i \arctan (a x)} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {x^4 (1+i a x)}{\sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {\int \frac {a x^3 (4 i-5 a x)}{\sqrt {a^2 x^2+1}}dx}{5 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {\int \frac {x^3 (4 i-5 a x)}{\sqrt {a^2 x^2+1}}dx}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {\int -\frac {a x^2 (16 i a x+15)}{\sqrt {a^2 x^2+1}}dx}{4 a^2}}{5 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\int \frac {a x^2 (16 i a x+15)}{\sqrt {a^2 x^2+1}}dx}{4 a^2}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\int \frac {x^2 (16 i a x+15)}{\sqrt {a^2 x^2+1}}dx}{4 a}}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {\int \frac {a x (32 i-45 a x)}{\sqrt {a^2 x^2+1}}dx}{3 a^2}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {\int \frac {x (32 i-45 a x)}{\sqrt {a^2 x^2+1}}dx}{3 a}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {-\frac {45 x \sqrt {a^2 x^2+1}}{2 a}-\frac {\int -\frac {a (64 i a x+45)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{3 a}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {-\frac {45 x \sqrt {a^2 x^2+1}}{2 a}+\frac {\int \frac {a (64 i a x+45)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{3 a}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {-\frac {45 x \sqrt {a^2 x^2+1}}{2 a}+\frac {\int \frac {64 i a x+45}{\sqrt {a^2 x^2+1}}dx}{2 a}}{3 a}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {-\frac {45 x \sqrt {a^2 x^2+1}}{2 a}+\frac {45 \int \frac {1}{\sqrt {a^2 x^2+1}}dx+\frac {64 i \sqrt {a^2 x^2+1}}{a}}{2 a}}{3 a}}{4 a}}{5 a}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {i x^4 \sqrt {a^2 x^2+1}}{5 a}-\frac {-\frac {5 x^3 \sqrt {a^2 x^2+1}}{4 a}+\frac {\frac {16 i x^2 \sqrt {a^2 x^2+1}}{3 a}-\frac {-\frac {45 x \sqrt {a^2 x^2+1}}{2 a}+\frac {\frac {45 \text {arcsinh}(a x)}{a}+\frac {64 i \sqrt {a^2 x^2+1}}{a}}{2 a}}{3 a}}{4 a}}{5 a}\) |
((I/5)*x^4*Sqrt[1 + a^2*x^2])/a - ((-5*x^3*Sqrt[1 + a^2*x^2])/(4*a) + (((( 16*I)/3)*x^2*Sqrt[1 + a^2*x^2])/a - ((-45*x*Sqrt[1 + a^2*x^2])/(2*a) + ((( 64*I)*Sqrt[1 + a^2*x^2])/a + (45*ArcSinh[a*x])/a)/(2*a))/(3*a))/(4*a))/(5* a)
3.1.1.3.1 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 = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {i \left (24 a^{4} x^{4}-30 i a^{3} x^{3}-32 a^{2} x^{2}+45 i a x +64\right ) \sqrt {a^{2} x^{2}+1}}{120 a^{5}}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{4} \sqrt {a^{2}}}\) | \(84\) |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right ) \sqrt {a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{4 a^{5}}}{2 a^{4} \sqrt {\pi }\, \sqrt {a^{2}}}+\frac {i \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}-8 a^{2} x^{2}+16\right ) \sqrt {a^{2} x^{2}+1}}{15}\right )}{2 a^{5} \sqrt {\pi }}\) | \(117\) |
default | \(\frac {x^{3} \sqrt {a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 \left (\frac {x \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}+i a \left (\frac {x^{4} \sqrt {a^{2} x^{2}+1}}{5 a^{2}}-\frac {4 \left (\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}}\right )}{5 a^{2}}\right )\) | \(142\) |
1/120*I*(24*a^4*x^4-30*I*a^3*x^3-32*a^2*x^2+45*I*a*x+64)*(a^2*x^2+1)^(1/2) /a^5+3/8/a^4*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.59 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {{\left (24 i \, a^{4} x^{4} + 30 \, a^{3} x^{3} - 32 i \, a^{2} x^{2} - 45 \, a x + 64 i\right )} \sqrt {a^{2} x^{2} + 1} - 45 \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{120 \, a^{5}} \]
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
Time = 0.56 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int e^{i \arctan (a x)} x^4 \, dx=\begin {cases} \sqrt {a^{2} x^{2} + 1} \left (\frac {i x^{4}}{5 a} + \frac {x^{3}}{4 a^{2}} - \frac {4 i x^{2}}{15 a^{3}} - \frac {3 x}{8 a^{4}} + \frac {8 i}{15 a^{5}}\right ) + \frac {3 \log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{8 a^{4} \sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{6}}{6} + \frac {x^{5}}{5} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a**2*x**2 + 1)*(I*x**4/(5*a) + x**3/(4*a**2) - 4*I*x**2/(1 5*a**3) - 3*x/(8*a**4) + 8*I/(15*a**5)) + 3*log(2*a**2*x + 2*sqrt(a**2*x** 2 + 1)*sqrt(a**2))/(8*a**4*sqrt(a**2)), Ne(a**2, 0)), (I*a*x**6/6 + x**5/5 , True))
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{4}}{5 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a^{2}} - \frac {4 i \, \sqrt {a^{2} x^{2} + 1} x^{2}}{15 \, a^{3}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{5}} + \frac {8 i \, \sqrt {a^{2} x^{2} + 1}}{15 \, a^{5}} \]
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*sqr t(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
Exception generated. \[ \int e^{i \arctan (a x)} x^4 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int e^{i \arctan (a x)} x^4 \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {x^3\,{\left (a^2\right )}^{3/2}}{4\,a^4}-\frac {3\,x\,\sqrt {a^2}}{8\,a^4}+\frac {a\,8{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}-\frac {a^3\,x^2\,4{}\mathrm {i}}{15\,{\left (a^2\right )}^{5/2}}+\frac {a^5\,x^4\,1{}\mathrm {i}}{5\,{\left (a^2\right )}^{5/2}}\right )}{\sqrt {a^2}}+\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{8\,a^4\,\sqrt {a^2}} \]