Integrand size = 23, antiderivative size = 91 \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {d^2 \sqrt {d+e x^2}}{5 e^{5/2}}+\frac {2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}-\frac {\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]
2/15*d*(e*x^2+d)^(3/2)/e^(5/2)-1/25*(e*x^2+d)^(5/2)/e^(5/2)+1/5*x^5*arctan h(x*e^(1/2)/(e*x^2+d)^(1/2))-1/5*d^2*(e*x^2+d)^(1/2)/e^(5/2)
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 e^{5/2}}+\frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]
-1/75*(Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4))/e^(5/2) + (x^5*Arc Tanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6775, 243, 53, 2009}
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.
\(\displaystyle \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx\) |
\(\Big \downarrow \) 6775 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{5} \sqrt {e} \int \frac {x^5}{\sqrt {e x^2+d}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {e} \int \frac {x^4}{\sqrt {e x^2+d}}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {e} \int \left (\frac {d^2}{e^2 \sqrt {e x^2+d}}-\frac {2 \sqrt {e x^2+d} d}{e^2}+\frac {\left (e x^2+d\right )^{3/2}}{e^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {e} \left (\frac {2 d^2 \sqrt {d+e x^2}}{e^3}+\frac {2 \left (d+e x^2\right )^{5/2}}{5 e^3}-\frac {4 d \left (d+e x^2\right )^{3/2}}{3 e^3}\right )\) |
-1/10*(Sqrt[e]*((2*d^2*Sqrt[d + e*x^2])/e^3 - (4*d*(d + e*x^2)^(3/2))/(3*e ^3) + (2*(d + e*x^2)^(5/2))/(5*e^3))) + (x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5
3.1.10.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_ Symbol] :> Simp[(d*x)^(m + 1)*(ArcTanh[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x] - Simp[c/(d*(m + 1)) Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; Fre eQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(67)=134\).
Time = 0.01 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.93
method | result | size |
default | \(\frac {x^{5} \operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {e^{\frac {3}{2}} \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d}\) | \(176\) |
parts | \(\frac {x^{5} \operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {e^{\frac {3}{2}} \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d}\) | \(176\) |
1/5*x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/5*e^(3/2)/d*(1/7*x^6/e*(e*x^2 +d)^(1/2)-6/7*d/e*(1/5*x^4/e*(e*x^2+d)^(1/2)-4/5*d/e*(1/3*x^2/e*(e*x^2+d)^ (1/2)-2/3*d/e^2*(e*x^2+d)^(1/2))))-1/5*e^(1/2)/d*(1/7*x^4*(e*x^2+d)^(3/2)/ e-4/7*d/e*(1/5*x^2*(e*x^2+d)^(3/2)/e-2/15*d/e^2*(e*x^2+d)^(3/2)))
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85 \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {15 \, e^{3} x^{5} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (3 \, e^{2} x^{4} - 4 \, d e x^{2} + 8 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{150 \, e^{3}} \]
1/150*(15*e^3*x^5*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*( 3*e^2*x^4 - 4*d*e*x^2 + 8*d^2)*sqrt(e*x^2 + d)*sqrt(e))/e^3
Time = 0.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.99 \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\begin {cases} - \frac {8 d^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {5}{2}}} + \frac {4 d x^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {3}{2}}} + \frac {x^{5} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {x^{4} \sqrt {d + e x^{2}}}{25 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-8*d**2*sqrt(d + e*x**2)/(75*e**(5/2)) + 4*d*x**2*sqrt(d + e*x* *2)/(75*e**(3/2)) + x**5*atanh(sqrt(e)*x/sqrt(d + e*x**2))/5 - x**4*sqrt(d + e*x**2)/(25*sqrt(e)), Ne(e, 0)), (0, True))
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.40 \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {1}{5} \, x^{5} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {15 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}}{525 \, d e^{\frac {5}{2}}} + \frac {5 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x^{2} + d} d^{3}}{175 \, d e^{\frac {5}{2}}} \]
1/5*x^5*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)) - 1/525*(15*(e*x^2 + d)^(7/2) - 42*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^2)/(d*e^(5/2)) + 1/175*(5 *(e*x^2 + d)^(7/2) - 21*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^2 - 3 5*sqrt(e*x^2 + d)*d^3)/(d*e^(5/2))
\[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int { x^{4} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) \,d x } \]
Timed out. \[ \int x^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int x^4\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]