Integrand size = 14, antiderivative size = 45 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{12} b c x^4+\frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} b c^3 \log \left (c^2-x^4\right ) \] Output:
1/12*b*c*x^4+1/6*x^6*(a+b*arctanh(c/x^2))+1/12*b*c^3*ln(-x^4+c^2)
Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{12} b c x^4+\frac {a x^6}{6}+\frac {1}{6} b x^6 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{12} b c^3 \log \left (-c^2+x^4\right ) \] Input:
Integrate[x^5*(a + b*ArcTanh[c/x^2]),x]
Output:
(b*c*x^4)/12 + (a*x^6)/6 + (b*x^6*ArcTanh[c/x^2])/6 + (b*c^3*Log[-c^2 + x^ 4])/12
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6452, 795, 798, 25, 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.
\(\displaystyle \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} b c \int \frac {x^3}{1-\frac {c^2}{x^4}}dx+\frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{3} b c \int \frac {x^7}{x^4-c^2}dx+\frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{12} b c \int -\frac {x^4}{c^2-x^4}dx^4+\frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{12} b c \int \frac {x^4}{c^2-x^4}dx^4\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{12} b c \int \left (\frac {c^2}{c^2-x^4}-1\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} x^6 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{12} b c \left (c^2 \log \left (c^2-x^4\right )+x^4\right )\) |
Input:
Int[x^5*(a + b*ArcTanh[c/x^2]),x]
Output:
(x^6*(a + b*ArcTanh[c/x^2]))/6 + (b*c*(x^4 + c^2*Log[c^2 - x^4]))/12
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_)^(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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/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 = 0.73 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.31
method | result | size |
parallelrisch | \(\frac {x^{6} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b}{6}+\frac {a \,x^{6}}{6}+\frac {b c \,x^{4}}{12}+\frac {\ln \left (x^{2}-c \right ) b \,c^{3}}{6}+\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,c^{3}}{6}+\frac {b \,c^{3}}{12}\) | \(59\) |
parts | \(\frac {a \,x^{6}}{6}+b \left (\frac {x^{6} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{6}-\frac {c \left (-\frac {c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{4}-\frac {c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {x^{4}}{4}+c^{2} \ln \left (\frac {1}{x}\right )\right )}{3}\right )\) | \(65\) |
derivativedivides | \(\frac {a \,x^{6}}{6}-b \left (-\frac {x^{6} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{6}+\frac {c \left (-\frac {c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{4}-\frac {c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {x^{4}}{4}+c^{2} \ln \left (\frac {1}{x}\right )\right )}{3}\right )\) | \(66\) |
default | \(\frac {a \,x^{6}}{6}-b \left (-\frac {x^{6} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{6}+\frac {c \left (-\frac {c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{4}-\frac {c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {x^{4}}{4}+c^{2} \ln \left (\frac {1}{x}\right )\right )}{3}\right )\) | \(66\) |
risch | \(\frac {b \,x^{6} \ln \left (x^{2}+c \right )}{12}-\frac {b \,x^{6} \ln \left (-x^{2}+c \right )}{12}-\frac {i \pi b \,x^{6} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}}{24}-\frac {i \pi b \,x^{6}}{12}+\frac {i \pi b \,x^{6} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{24}-\frac {i \pi b \,x^{6} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{24}-\frac {i \pi b \,x^{6} \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{24}+\frac {i \pi b \,x^{6} \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{24}+\frac {i \pi b \,x^{6} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{24}+\frac {i \pi b \,x^{6} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{12}-\frac {i \pi b \,x^{6} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{24}-\frac {i \pi b \,x^{6} {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}}{24}+\frac {a \,x^{6}}{6}+\frac {b c \,x^{4}}{12}+\frac {b \,c^{3} \ln \left (x^{4}-c^{2}\right )}{12}\) | \(337\) |
Input:
int(x^5*(a+b*arctanh(c/x^2)),x,method=_RETURNVERBOSE)
Output:
1/6*x^6*arctanh(c/x^2)*b+1/6*a*x^6+1/12*b*c*x^4+1/6*ln(x^2-c)*b*c^3+1/6*ar ctanh(c/x^2)*b*c^3+1/12*b*c^3
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{12} \, b x^{6} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{6} \, a x^{6} + \frac {1}{12} \, b c x^{4} + \frac {1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \] Input:
integrate(x^5*(a+b*arctanh(c/x^2)),x, algorithm="fricas")
Output:
1/12*b*x^6*log((x^2 + c)/(x^2 - c)) + 1/6*a*x^6 + 1/12*b*c*x^4 + 1/12*b*c^ 3*log(x^4 - c^2)
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (37) = 74\).
Time = 2.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.67 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^{6}}{6} + \frac {b c^{3} \log {\left (x - \sqrt {- c} \right )}}{6} + \frac {b c^{3} \log {\left (x + \sqrt {- c} \right )}}{6} - \frac {b c^{3} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6} + \frac {b c x^{4}}{12} + \frac {b x^{6} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{6} \] Input:
integrate(x**5*(a+b*atanh(c/x**2)),x)
Output:
a*x**6/6 + b*c**3*log(x - sqrt(-c))/6 + b*c**3*log(x + sqrt(-c))/6 - b*c** 3*atanh(c/x**2)/6 + b*c*x**4/12 + b*x**6*atanh(c/x**2)/6
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{6} \, a x^{6} + \frac {1}{12} \, {\left (2 \, x^{6} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (x^{4} + c^{2} \log \left (x^{4} - c^{2}\right )\right )} c\right )} b \] Input:
integrate(x^5*(a+b*arctanh(c/x^2)),x, algorithm="maxima")
Output:
1/6*a*x^6 + 1/12*(2*x^6*arctanh(c/x^2) + (x^4 + c^2*log(x^4 - c^2))*c)*b
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{12} \, b x^{6} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{6} \, a x^{6} + \frac {1}{12} \, b c x^{4} + \frac {1}{12} \, b c^{3} \log \left (x^{4} - c^{2}\right ) \] Input:
integrate(x^5*(a+b*arctanh(c/x^2)),x, algorithm="giac")
Output:
1/12*b*x^6*log((x^2 + c)/(x^2 - c)) + 1/6*a*x^6 + 1/12*b*c*x^4 + 1/12*b*c^ 3*log(x^4 - c^2)
Time = 4.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a\,x^6}{6}+\frac {b\,c^3\,\ln \left (x^4-c^2\right )}{12}+\frac {b\,x^6\,\ln \left (x^2+c\right )}{12}+\frac {b\,c\,x^4}{12}-\frac {b\,x^6\,\ln \left (x^2-c\right )}{12} \] Input:
int(x^5*(a + b*atanh(c/x^2)),x)
Output:
(a*x^6)/6 + (b*c^3*log(x^4 - c^2))/12 + (b*x^6*log(c + x^2))/12 + (b*c*x^4 )/12 - (b*x^6*log(x^2 - c))/12
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,c^{3}}{6}+\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,x^{6}}{6}+\frac {\mathrm {log}\left (x^{2}+c \right ) b \,c^{3}}{6}+\frac {a \,x^{6}}{6}+\frac {b c \,x^{4}}{12} \] Input:
int(x^5*(a+b*atanh(c/x^2)),x)
Output:
( - 2*atanh(c/x**2)*b*c**3 + 2*atanh(c/x**2)*b*x**6 + 2*log(c + x**2)*b*c* *3 + 2*a*x**6 + b*c*x**4)/12