Integrand size = 12, antiderivative size = 39 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{4} b c \log \left (c^2-x^4\right ) \] Output:
1/2*a*x^2+1/2*b*x^2*arctanh(c/x^2)+1/4*b*c*ln(-x^4+c^2)
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{4} b c \log \left (-c^2+x^4\right ) \] Input:
Integrate[x*(a + b*ArcTanh[c/x^2]),x]
Output:
(a*x^2)/2 + (b*x^2*ArcTanh[c/x^2])/2 + (b*c*Log[-c^2 + x^4])/4
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6452, 795, 792}
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.
\(\displaystyle \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle b c \int \frac {1}{\left (1-\frac {c^2}{x^4}\right ) x}dx+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle b c \int \frac {x^3}{x^4-c^2}dx+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 792 |
\(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} b c \log \left (c^2-x^4\right )\) |
Input:
Int[x*(a + b*ArcTanh[c/x^2]),x]
Output:
(x^2*(a + b*ArcTanh[c/x^2]))/2 + (b*c*Log[c^2 - x^4])/4
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveConten t[a + b*x^n, x]]/(b*n), x] /; FreeQ[{a, b, m, n}, x] && EqQ[m, n - 1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
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 = 0.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {b \,x^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{2}+\frac {\ln \left (x^{2}-c \right ) b c}{2}+\frac {a \,x^{2}}{2}+\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b c}{2}\) | \(42\) |
derivativedivides | \(\frac {a \,x^{2}}{2}-b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{2}+c \left (\ln \left (\frac {1}{x}\right )-\frac {\ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {\ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )\right )\) | \(50\) |
default | \(\frac {a \,x^{2}}{2}-b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{2}+c \left (\ln \left (\frac {1}{x}\right )-\frac {\ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {\ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )\right )\) | \(50\) |
parts | \(\frac {a \,x^{2}}{2}+b \left (\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{2}-c \left (\ln \left (\frac {1}{x}\right )-\frac {\ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {\ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )\right )\) | \(50\) |
risch | \(\frac {x^{2} b \ln \left (x^{2}+c \right )}{4}-\frac {b \,x^{2} \ln \left (-x^{2}+c \right )}{4}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{8}-\frac {i \pi b \,x^{2}}{4}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{8}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{8}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{8}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{8}-\frac {i \pi b \,x^{2} {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}}{8}-\frac {i \pi b \,x^{2} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}}{8}+\frac {i \pi b \,x^{2} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{4}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{8}+\frac {a \,x^{2}}{2}+\frac {b c \ln \left (x^{4}-c^{2}\right )}{4}\) | \(328\) |
Input:
int(x*(a+b*arctanh(c/x^2)),x,method=_RETURNVERBOSE)
Output:
1/2*b*x^2*arctanh(c/x^2)+1/2*ln(x^2-c)*b*c+1/2*a*x^2+1/2*arctanh(c/x^2)*b* c
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{4} \, b x^{2} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{2} \, a x^{2} + \frac {1}{4} \, b c \log \left (x^{4} - c^{2}\right ) \] Input:
integrate(x*(a+b*arctanh(c/x^2)),x, algorithm="fricas")
Output:
1/4*b*x^2*log((x^2 + c)/(x^2 - c)) + 1/2*a*x^2 + 1/4*b*c*log(x^4 - c^2)
Time = 1.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.56 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^{2}}{2} + \frac {b c \log {\left (x - \sqrt {- c} \right )}}{2} + \frac {b c \log {\left (x + \sqrt {- c} \right )}}{2} - \frac {b c \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} + \frac {b x^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2} \] Input:
integrate(x*(a+b*atanh(c/x**2)),x)
Output:
a*x**2/2 + b*c*log(x - sqrt(-c))/2 + b*c*log(x + sqrt(-c))/2 - b*c*atanh(c /x**2)/2 + b*x**2*atanh(c/x**2)/2
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + c \log \left (x^{4} - c^{2}\right )\right )} b \] Input:
integrate(x*(a+b*arctanh(c/x^2)),x, algorithm="maxima")
Output:
1/2*a*x^2 + 1/4*(2*x^2*arctanh(c/x^2) + c*log(x^4 - c^2))*b
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (33) = 66\).
Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 4.72 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {{\left (c^{2} {\left (\log \left (\frac {{\left | -x^{2} - c \right |}}{{\left | -x^{2} + c \right |}}\right ) - \log \left ({\left | \frac {x^{2} + c}{x^{2} - c} - 1 \right |}\right )\right )} + \frac {c^{2} \log \left (-\frac {\frac {c {\left (\frac {x^{2} + c}{{\left (x^{2} - c\right )} c} - \frac {1}{c}\right )}}{\frac {x^{2} + c}{x^{2} - c} + 1} + 1}{\frac {c {\left (\frac {x^{2} + c}{{\left (x^{2} - c\right )} c} - \frac {1}{c}\right )}}{\frac {x^{2} + c}{x^{2} - c} + 1} - 1}\right )}{\frac {x^{2} + c}{x^{2} - c} - 1}\right )} b}{2 \, c} \] Input:
integrate(x*(a+b*arctanh(c/x^2)),x, algorithm="giac")
Output:
1/2*a*x^2 + 1/2*(c^2*(log(abs(-x^2 - c)/abs(-x^2 + c)) - log(abs((x^2 + c) /(x^2 - c) - 1))) + c^2*log(-(c*((x^2 + c)/((x^2 - c)*c) - 1/c)/((x^2 + c) /(x^2 - c) + 1) + 1)/(c*((x^2 + c)/((x^2 - c)*c) - 1/c)/((x^2 + c)/(x^2 - c) + 1) - 1))/((x^2 + c)/(x^2 - c) - 1))*b/c
Time = 3.70 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (x^2+c\right )}{4}+\frac {b\,c\,\ln \left (x^4-c^2\right )}{4}-\frac {b\,x^2\,\ln \left (x^2-c\right )}{4} \] Input:
int(x*(a + b*atanh(c/x^2)),x)
Output:
(a*x^2)/2 + (b*x^2*log(c + x^2))/4 + (b*c*log(x^4 - c^2))/4 - (b*x^2*log(x ^2 - c))/4
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b c}{2}+\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,x^{2}}{2}+\frac {\mathrm {log}\left (x^{2}+c \right ) b c}{2}+\frac {a \,x^{2}}{2} \] Input:
int(x*(a+b*atanh(c/x^2)),x)
Output:
( - atanh(c/x**2)*b*c + atanh(c/x**2)*b*x**2 + log(c + x**2)*b*c + a*x**2) /2