Integrand size = 10, antiderivative size = 111 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=-\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {(64+45 a x) \sqrt {1-a^2 x^2}}{120 a^5}+\frac {3 \arcsin (a x)}{8 a^5} \]
3/8*arcsin(a*x)/a^5-4/15*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*x^4*(-a^2*x^2+1)^(1/2)/a-1/120*(45*a*x+64)*(-a^2*x^2+1)^(1/2) /a^5
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.54 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=\frac {-\sqrt {1-a^2 x^2} \left (64+45 a x+32 a^2 x^2+30 a^3 x^3+24 a^4 x^4\right )+45 \arcsin (a x)}{120 a^5} \]
(-(Sqrt[1 - a^2*x^2]*(64 + 45*a*x + 32*a^2*x^2 + 30*a^3*x^3 + 24*a^4*x^4)) + 45*ArcSin[a*x])/(120*a^5)
Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.40, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {6674, 533, 27, 533, 27, 533, 27, 533, 27, 455, 223}
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.
\(\displaystyle \int x^4 e^{\text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6674 |
\(\displaystyle \int \frac {x^4 (a x+1)}{\sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int \frac {a x^3 (5 a x+4)}{\sqrt {1-a^2 x^2}}dx}{5 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^3 (5 a x+4)}{\sqrt {1-a^2 x^2}}dx}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {\int \frac {a x^2 (16 a x+15)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {x^2 (16 a x+15)}{\sqrt {1-a^2 x^2}}dx}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {\frac {\int \frac {a x (45 a x+32)}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \frac {x (45 a x+32)}{\sqrt {1-a^2 x^2}}dx}{3 a}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {a (64 a x+45)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {45 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {64 a x+45}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {45 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {\frac {\frac {45 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {64 \sqrt {1-a^2 x^2}}{a}}{2 a}-\frac {45 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {\frac {\frac {\frac {45 \arcsin (a x)}{a}-\frac {64 \sqrt {1-a^2 x^2}}{a}}{2 a}-\frac {45 x \sqrt {1-a^2 x^2}}{2 a}}{3 a}-\frac {16 x^2 \sqrt {1-a^2 x^2}}{3 a}}{4 a}-\frac {5 x^3 \sqrt {1-a^2 x^2}}{4 a}}{5 a}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}\) |
-1/5*(x^4*Sqrt[1 - a^2*x^2])/a + ((-5*x^3*Sqrt[1 - a^2*x^2])/(4*a) + ((-16 *x^2*Sqrt[1 - a^2*x^2])/(3*a) + ((-45*x*Sqrt[1 - a^2*x^2])/(2*a) + ((-64*S qrt[1 - a^2*x^2])/a + (45*ArcSin[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[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x )^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / ; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {\left (24 a^{4} x^{4}+30 a^{3} x^{3}+32 a^{2} x^{2}+45 a x +64\right ) \left (a^{2} x^{2}-1\right )}{120 a^{5} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{4} \sqrt {a^{2}}}\) | \(88\) |
meijerg | \(-\frac {-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}}{2 a^{5} \sqrt {\pi }}+\frac {-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}}{2 a^{4} \sqrt {\pi }\, \sqrt {-a^{2}}}\) | \(124\) |
default | \(-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}+a \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )\) | \(142\) |
1/120*(24*a^4*x^4+30*a^3*x^3+32*a^2*x^2+45*a*x+64)*(a^2*x^2-1)/a^5/(-a^2*x ^2+1)^(1/2)+3/8/a^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=-\frac {{\left (24 \, a^{4} x^{4} + 30 \, a^{3} x^{3} + 32 \, a^{2} x^{2} + 45 \, a x + 64\right )} \sqrt {-a^{2} x^{2} + 1} + 90 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{120 \, a^{5}} \]
-1/120*((24*a^4*x^4 + 30*a^3*x^3 + 32*a^2*x^2 + 45*a*x + 64)*sqrt(-a^2*x^2 + 1) + 90*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^5
Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (- \frac {x^{4}}{5 a} - \frac {x^{3}}{4 a^{2}} - \frac {4 x^{2}}{15 a^{3}} - \frac {3 x}{8 a^{4}} - \frac {8}{15 a^{5}}\right ) + \frac {3 \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 a^{4} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {a x^{6}}{6} + \frac {x^{5}}{5} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(-a**2*x**2 + 1)*(-x**4/(5*a) - x**3/(4*a**2) - 4*x**2/(15* a**3) - 3*x/(8*a**4) - 8/(15*a**5)) + 3*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt (-a**2*x**2 + 1))/(8*a**4*sqrt(-a**2)), Ne(a**2, 0)), (a*x**6/6 + x**5/5, True))
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=-\frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{5 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{4 \, a^{2}} - \frac {4 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{15 \, a^{3}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{4}} + \frac {3 \, \arcsin \left (a x\right )}{8 \, a^{5}} - \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, a^{5}} \]
-1/5*sqrt(-a^2*x^2 + 1)*x^4/a - 1/4*sqrt(-a^2*x^2 + 1)*x^3/a^2 - 4/15*sqrt (-a^2*x^2 + 1)*x^2/a^3 - 3/8*sqrt(-a^2*x^2 + 1)*x/a^4 + 3/8*arcsin(a*x)/a^ 5 - 8/15*sqrt(-a^2*x^2 + 1)/a^5
Exception generated. \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=\text {Exception raised: TypeError} \]
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 = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int e^{\text {arctanh}(a x)} x^4 \, dx=\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^4\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {8}{15\,a^3\,\sqrt {-a^2}}+\frac {a\,x^4}{5\,\sqrt {-a^2}}-\frac {3\,x\,\sqrt {-a^2}}{8\,a^4}+\frac {4\,x^2}{15\,a\,\sqrt {-a^2}}+\frac {x^3\,{\left (-a^2\right )}^{3/2}}{4\,a^4}\right )}{\sqrt {-a^2}} \]