Integrand size = 12, antiderivative size = 39 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {b c x}{2}+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )-\frac {1}{2} b c^2 \text {arctanh}\left (\frac {x}{c}\right ) \] Output:
1/2*b*c*x+1/2*x^2*(a+b*arctanh(c/x))-1/2*b*c^2*arctanh(x/c)
Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {b c x}{2}+\frac {a x^2}{2}+\frac {1}{2} b x^2 \text {arctanh}\left (\frac {c}{x}\right )+\frac {1}{4} b c^2 \log (-c+x)-\frac {1}{4} b c^2 \log (c+x) \] Input:
Integrate[x*(a + b*ArcTanh[c/x]),x]
Output:
(b*c*x)/2 + (a*x^2)/2 + (b*x^2*ArcTanh[c/x])/2 + (b*c^2*Log[-c + x])/4 - ( b*c^2*Log[c + x])/4
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6452, 772, 262, 220}
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.
\(\displaystyle \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} b c \int \frac {1}{1-\frac {c^2}{x^2}}dx+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 772 |
\(\displaystyle \frac {1}{2} b c \int \frac {x^2}{x^2-c^2}dx+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} b c \left (c^2 \int \frac {1}{x^2-c^2}dx+x\right )+\frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} x^2 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )+\frac {1}{2} b c \left (x-c \text {arctanh}\left (\frac {x}{c}\right )\right )\) |
Input:
Int[x*(a + b*ArcTanh[c/x]),x]
Output:
(x^2*(a + b*ArcTanh[c/x]))/2 + (b*c*(x - c*ArcTanh[x/c]))/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && ILtQ[n, 0] && IntegerQ[p]
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 = 0.47 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,x^{2}}{2}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,c^{2}}{2}+\frac {a \,x^{2}}{2}+\frac {b c x}{2}+\frac {a \,c^{2}}{2}\) | \(43\) |
parts | \(\frac {a \,x^{2}}{2}-b \,c^{2} \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\) | \(55\) |
derivativedivides | \(-c^{2} \left (-\frac {a \,x^{2}}{2 c^{2}}+b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\right )\) | \(59\) |
default | \(-c^{2} \left (-\frac {a \,x^{2}}{2 c^{2}}+b \left (-\frac {x^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 c^{2}}-\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {x}{2 c}+\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )\right )\) | \(59\) |
orering | \(-\left (c^{2}-x^{2}\right ) \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )\right )+\frac {\left (c -x \right ) \left (x +c \right ) \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )-\frac {b c}{x \left (1-\frac {c^{2}}{x^{2}}\right )}\right )}{2}\) | \(62\) |
risch | \(\frac {x^{2} b \ln \left (x +c \right )}{4}-\frac {x^{2} b \ln \left (c -x \right )}{4}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )}{8}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{3}}{8}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )}{8}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{8}-\frac {i \pi b \,x^{2}}{4}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{8}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{4}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{8}+\frac {i \pi b \,x^{2} \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{8}-\frac {i \pi b \,x^{2} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{3}}{8}+\frac {a \,x^{2}}{2}+\frac {b c x}{2}+\frac {b \,c^{2} \ln \left (x -c \right )}{4}-\frac {b \,c^{2} \ln \left (x +c \right )}{4}\) | \(311\) |
Input:
int(x*(a+b*arctanh(c/x)),x,method=_RETURNVERBOSE)
Output:
1/2*arctanh(c/x)*b*x^2-1/2*arctanh(c/x)*b*c^2+1/2*a*x^2+1/2*b*c*x+1/2*a*c^ 2
Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{2} \, b c x + \frac {1}{2} \, a x^{2} - \frac {1}{4} \, {\left (b c^{2} - b x^{2}\right )} \log \left (-\frac {c + x}{c - x}\right ) \] Input:
integrate(x*(a+b*arctanh(c/x)),x, algorithm="fricas")
Output:
1/2*b*c*x + 1/2*a*x^2 - 1/4*(b*c^2 - b*x^2)*log(-(c + x)/(c - x))
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {a x^{2}}{2} - \frac {b c^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2} + \frac {b c x}{2} + \frac {b x^{2} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2} \] Input:
integrate(x*(a+b*atanh(c/x)),x)
Output:
a*x**2/2 - b*c**2*atanh(c/x)/2 + b*c*x/2 + b*x**2*atanh(c/x)/2
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (\frac {c}{x}\right ) - {\left (c \log \left (c + x\right ) - c \log \left (-c + x\right ) - 2 \, x\right )} c\right )} b \] Input:
integrate(x*(a+b*arctanh(c/x)),x, algorithm="maxima")
Output:
1/2*a*x^2 + 1/4*(2*x^2*arctanh(c/x) - (c*log(c + x) - c*log(-c + x) - 2*x) *c)*b
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (33) = 66\).
Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 3.33 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=-\frac {\frac {b {\left (c + x\right )} c^{3} \log \left (-\frac {c + x}{c - x}\right )}{{\left (c - x\right )} {\left (\frac {{\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {2 \, {\left (c + x\right )}}{c - x} + 1\right )}} + \frac {b c^{3} + \frac {2 \, a {\left (c + x\right )} c^{3}}{c - x} + \frac {b {\left (c + x\right )} c^{3}}{c - x}}{\frac {{\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {2 \, {\left (c + x\right )}}{c - x} + 1}}{c} \] Input:
integrate(x*(a+b*arctanh(c/x)),x, algorithm="giac")
Output:
-(b*(c + x)*c^3*log(-(c + x)/(c - x))/((c - x)*((c + x)^2/(c - x)^2 + 2*(c + x)/(c - x) + 1)) + (b*c^3 + 2*a*(c + x)*c^3/(c - x) + b*(c + x)*c^3/(c - x))/((c + x)^2/(c - x)^2 + 2*(c + x)/(c - x) + 1))/c
Time = 3.58 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {a\,x^2}{2}-\frac {b\,c^2\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2}+\frac {b\,x^2\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2}+\frac {b\,c\,x}{2} \] Input:
int(x*(a + b*atanh(c/x)),x)
Output:
(a*x^2)/2 - (b*c^2*atanh(c/x))/2 + (b*x^2*atanh(c/x))/2 + (b*c*x)/2
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b \,c^{2}}{2}+\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b \,x^{2}}{2}+\frac {a \,x^{2}}{2}+\frac {b c x}{2} \] Input:
int(x*(a+b*atanh(c/x)),x)
Output:
( - atanh(c/x)*b*c**2 + atanh(c/x)*b*x**2 + a*x**2 + b*c*x)/2