Integrand size = 12, antiderivative size = 57 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {b x^2}{10 c^3}+\frac {b x^4}{20 c}+\frac {1}{5} x^5 (a+b \text {arctanh}(c x))+\frac {b \log \left (1-c^2 x^2\right )}{10 c^5} \]
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {b x^2}{10 c^3}+\frac {b x^4}{20 c}+\frac {a x^5}{5}+\frac {1}{5} b x^5 \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{10 c^5} \]
(b*x^2)/(10*c^3) + (b*x^4)/(20*c) + (a*x^5)/5 + (b*x^5*ArcTanh[c*x])/5 + ( b*Log[1 - c^2*x^2])/(10*c^5)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6452, 243, 49, 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.
\(\displaystyle \int x^4 (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{5} x^5 (a+b \text {arctanh}(c x))-\frac {1}{5} b c \int \frac {x^5}{1-c^2 x^2}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{5} x^5 (a+b \text {arctanh}(c x))-\frac {1}{10} b c \int \frac {x^4}{1-c^2 x^2}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{5} x^5 (a+b \text {arctanh}(c x))-\frac {1}{10} b c \int \left (-\frac {x^2}{c^2}-\frac {1}{c^4 \left (c^2 x^2-1\right )}-\frac {1}{c^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 (a+b \text {arctanh}(c x))-\frac {1}{10} b c \left (-\frac {x^2}{c^4}-\frac {x^4}{2 c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^6}\right )\) |
3.1.2.3.1 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}, x] && IGtQ[m, 0] && IGtQ[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[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02
method | result | size |
parts | \(\frac {a \,x^{5}}{5}+\frac {b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}+\frac {\ln \left (c x -1\right )}{10}+\frac {\ln \left (c x +1\right )}{10}\right )}{c^{5}}\) | \(58\) |
derivativedivides | \(\frac {\frac {c^{5} x^{5} a}{5}+b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}+\frac {\ln \left (c x -1\right )}{10}+\frac {\ln \left (c x +1\right )}{10}\right )}{c^{5}}\) | \(62\) |
default | \(\frac {\frac {c^{5} x^{5} a}{5}+b \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}+\frac {\ln \left (c x -1\right )}{10}+\frac {\ln \left (c x +1\right )}{10}\right )}{c^{5}}\) | \(62\) |
parallelrisch | \(\frac {4 b \,\operatorname {arctanh}\left (c x \right ) x^{5} c^{5}+4 c^{5} x^{5} a +b \,c^{4} x^{4}+2 b \,c^{2} x^{2}+4 \ln \left (c x -1\right ) b +4 b \,\operatorname {arctanh}\left (c x \right )+2 b}{20 c^{5}}\) | \(65\) |
risch | \(\frac {b \,x^{5} \ln \left (c x +1\right )}{10}-\frac {b \,x^{5} \ln \left (-c x +1\right )}{10}+\frac {a \,x^{5}}{5}+\frac {b \,x^{4}}{20 c}+\frac {b \,x^{2}}{10 c^{3}}+\frac {b \ln \left (c^{2} x^{2}-1\right )}{10 c^{5}}\) | \(67\) |
1/5*a*x^5+b/c^5*(1/5*c^5*x^5*arctanh(c*x)+1/20*c^4*x^4+1/10*c^2*x^2+1/10*l n(c*x-1)+1/10*ln(c*x+1))
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {2 \, b c^{5} x^{5} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 4 \, a c^{5} x^{5} + b c^{4} x^{4} + 2 \, b c^{2} x^{2} + 2 \, b \log \left (c^{2} x^{2} - 1\right )}{20 \, c^{5}} \]
1/20*(2*b*c^5*x^5*log(-(c*x + 1)/(c*x - 1)) + 4*a*c^5*x^5 + b*c^4*x^4 + 2* b*c^2*x^2 + 2*b*log(c^2*x^2 - 1))/c^5
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a x^{5}}{5} + \frac {b x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b x^{4}}{20 c} + \frac {b x^{2}}{10 c^{3}} + \frac {b \log {\left (x - \frac {1}{c} \right )}}{5 c^{5}} + \frac {b \operatorname {atanh}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\\frac {a x^{5}}{5} & \text {otherwise} \end {cases} \]
Piecewise((a*x**5/5 + b*x**5*atanh(c*x)/5 + b*x**4/(20*c) + b*x**2/(10*c** 3) + b*log(x - 1/c)/(5*c**5) + b*atanh(c*x)/(5*c**5), Ne(c, 0)), (a*x**5/5 , True))
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b \]
1/5*a*x^5 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^ 2*x^2 - 1)/c^6))*b
Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (49) = 98\).
Time = 0.29 (sec) , antiderivative size = 403, normalized size of antiderivative = 7.07 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, c {\left (\frac {{\left (\frac {5 \, {\left (c x + 1\right )}^{4} b}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{6}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{6}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{6}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{6}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{6}}{c x - 1} - c^{6}} + \frac {2 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} a}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} a}{{\left (c x - 1\right )}^{2}} + a + \frac {2 \, {\left (c x + 1\right )}^{4} b}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} b}{{\left (c x - 1\right )}^{3}} + \frac {4 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} b}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{5} c^{6}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{6}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{6}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{6}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{6}}{c x - 1} - c^{6}} - \frac {b \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{6}} + \frac {b \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{6}}\right )} \]
1/5*c*((5*(c*x + 1)^4*b/(c*x - 1)^4 + 10*(c*x + 1)^2*b/(c*x - 1)^2 + b)*lo g(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^6/(c*x - 1)^5 - 5*(c*x + 1)^4*c^6/( c*x - 1)^4 + 10*(c*x + 1)^3*c^6/(c*x - 1)^3 - 10*(c*x + 1)^2*c^6/(c*x - 1) ^2 + 5*(c*x + 1)*c^6/(c*x - 1) - c^6) + 2*(5*(c*x + 1)^4*a/(c*x - 1)^4 + 1 0*(c*x + 1)^2*a/(c*x - 1)^2 + a + 2*(c*x + 1)^4*b/(c*x - 1)^4 - 4*(c*x + 1 )^3*b/(c*x - 1)^3 + 4*(c*x + 1)^2*b/(c*x - 1)^2 - 2*(c*x + 1)*b/(c*x - 1)) /((c*x + 1)^5*c^6/(c*x - 1)^5 - 5*(c*x + 1)^4*c^6/(c*x - 1)^4 + 10*(c*x + 1)^3*c^6/(c*x - 1)^3 - 10*(c*x + 1)^2*c^6/(c*x - 1)^2 + 5*(c*x + 1)*c^6/(c *x - 1) - c^6) - b*log(-(c*x + 1)/(c*x - 1) + 1)/c^6 + b*log(-(c*x + 1)/(c *x - 1))/c^6)
Time = 3.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int x^4 (a+b \text {arctanh}(c x)) \, dx=\frac {a\,x^5}{5}+\frac {\frac {b\,\ln \left (c^2\,x^2-1\right )}{10}+\frac {b\,c^2\,x^2}{10}+\frac {b\,c^4\,x^4}{20}}{c^5}+\frac {b\,x^5\,\mathrm {atanh}\left (c\,x\right )}{5} \]