Integrand size = 12, antiderivative size = 65 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=-\frac {b c}{20 x^4}-\frac {b c^3}{10 x^2}-\frac {a+b \text {arctanh}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right ) \] Output:
-1/20*b*c/x^4-1/10*b*c^3/x^2-1/5*(a+b*arctanh(c*x))/x^5+1/5*b*c^5*ln(x)-1/ 10*b*c^5*ln(-c^2*x^2+1)
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {b c}{20 x^4}-\frac {b c^3}{10 x^2}-\frac {b \text {arctanh}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x])/x^6,x]
Output:
-1/5*a/x^5 - (b*c)/(20*x^4) - (b*c^3)/(10*x^2) - (b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 - c^2*x^2])/10
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6452, 243, 54, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{5} b c \int \frac {1}{x^5 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{5 x^5}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{10} b c \int \frac {1}{x^6 \left (1-c^2 x^2\right )}dx^2-\frac {a+b \text {arctanh}(c x)}{5 x^5}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{10} b c \int \left (-\frac {c^6}{c^2 x^2-1}+\frac {c^4}{x^2}+\frac {c^2}{x^4}+\frac {1}{x^6}\right )dx^2-\frac {a+b \text {arctanh}(c x)}{5 x^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} b c \left (c^4 \log \left (x^2\right )-\frac {c^2}{x^2}-c^4 \log \left (1-c^2 x^2\right )-\frac {1}{2 x^4}\right )-\frac {a+b \text {arctanh}(c x)}{5 x^5}\) |
Input:
Int[(a + b*ArcTanh[c*x])/x^6,x]
Output:
-1/5*(a + b*ArcTanh[c*x])/x^5 + (b*c*(-1/2*1/x^4 - c^2/x^2 + c^4*Log[x^2] - c^4*Log[1 - c^2*x^2]))/10
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98
method | result | size |
parts | \(-\frac {a}{5 x^{5}}+b \,c^{5} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{10 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}-\frac {\ln \left (c x -1\right )}{10}\right )\) | \(64\) |
derivativedivides | \(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{10 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}-\frac {\ln \left (c x -1\right )}{10}\right )\right )\) | \(68\) |
default | \(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{5 c^{5} x^{5}}-\frac {1}{20 c^{4} x^{4}}-\frac {1}{10 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{5}-\frac {\ln \left (c x +1\right )}{10}-\frac {\ln \left (c x -1\right )}{10}\right )\right )\) | \(68\) |
risch | \(-\frac {b \ln \left (c x +1\right )}{10 x^{5}}+\frac {4 b \,c^{5} \ln \left (x \right ) x^{5}-2 b \,c^{5} \ln \left (c^{2} x^{2}-1\right ) x^{5}-2 b \,c^{3} x^{3}-b c x +2 b \ln \left (-c x +1\right )-4 a}{20 x^{5}}\) | \(77\) |
parallelrisch | \(\frac {4 b \,c^{5} \ln \left (x \right ) x^{5}-4 \ln \left (c x -1\right ) x^{5} b \,c^{5}-4 b \,\operatorname {arctanh}\left (c x \right ) x^{5} c^{5}-2 b \,c^{5} x^{5}-2 b \,c^{3} x^{3}-b c x -4 b \,\operatorname {arctanh}\left (c x \right )-4 a}{20 x^{5}}\) | \(79\) |
Input:
int((a+b*arctanh(c*x))/x^6,x,method=_RETURNVERBOSE)
Output:
-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+1/5* ln(c*x)-1/10*ln(c*x+1)-1/10*ln(c*x-1))
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=-\frac {2 \, b c^{5} x^{5} \log \left (c^{2} x^{2} - 1\right ) - 4 \, b c^{5} x^{5} \log \left (x\right ) + 2 \, b c^{3} x^{3} + b c x + 2 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 4 \, a}{20 \, x^{5}} \] Input:
integrate((a+b*arctanh(c*x))/x^6,x, algorithm="fricas")
Output:
-1/20*(2*b*c^5*x^5*log(c^2*x^2 - 1) - 4*b*c^5*x^5*log(x) + 2*b*c^3*x^3 + b *c*x + 2*b*log(-(c*x + 1)/(c*x - 1)) + 4*a)/x^5
Time = 0.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=\begin {cases} - \frac {a}{5 x^{5}} + \frac {b c^{5} \log {\left (x \right )}}{5} - \frac {b c^{5} \log {\left (x - \frac {1}{c} \right )}}{5} - \frac {b c^{5} \operatorname {atanh}{\left (c x \right )}}{5} - \frac {b c^{3}}{10 x^{2}} - \frac {b c}{20 x^{4}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a}{5 x^{5}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c*x))/x**6,x)
Output:
Piecewise((-a/(5*x**5) + b*c**5*log(x)/5 - b*c**5*log(x - 1/c)/5 - b*c**5* atanh(c*x)/5 - b*c**3/(10*x**2) - b*c/(20*x**4) - b*atanh(c*x)/(5*x**5), N e(c, 0)), (-a/(5*x**5), True))
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \] Input:
integrate((a+b*arctanh(c*x))/x^6,x, algorithm="maxima")
Output:
-1/20*((2*c^4*log(c^2*x^2 - 1) - 2*c^4*log(x^2) + (2*c^2*x^2 + 1)/x^4)*c + 4*arctanh(c*x)/x^5)*b - 1/5*a/x^5
Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 397, normalized size of antiderivative = 6.11 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=\frac {1}{5} \, {\left (b c^{4} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{4} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {5 \, {\left (c x + 1\right )}^{4} b c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} b c^{4}}{{\left (c x - 1\right )}^{2}} + b c^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {2 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} a c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} a c^{4}}{{\left (c x - 1\right )}^{2}} + a c^{4} + \frac {2 \, {\left (c x + 1\right )}^{4} b c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3} b c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {4 \, {\left (c x + 1\right )}^{2} b c^{4}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )} b c^{4}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \] Input:
integrate((a+b*arctanh(c*x))/x^6,x, algorithm="giac")
Output:
1/5*(b*c^4*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^4*log(-(c*x + 1)/(c*x - 1)) + (5*(c*x + 1)^4*b*c^4/(c*x - 1)^4 + 10*(c*x + 1)^2*b*c^4/(c*x - 1)^2 + b *c^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4/( c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*( c*x + 1)/(c*x - 1) + 1) + 2*(5*(c*x + 1)^4*a*c^4/(c*x - 1)^4 + 10*(c*x + 1 )^2*a*c^4/(c*x - 1)^2 + a*c^4 + 2*(c*x + 1)^4*b*c^4/(c*x - 1)^4 + 4*(c*x + 1)^3*b*c^4/(c*x - 1)^3 + 4*(c*x + 1)^2*b*c^4/(c*x - 1)^2 + 2*(c*x + 1)*b* c^4/(c*x - 1))/((c*x + 1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4/(c*x - 1)^4 + 10*( c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c*x - 1 ) + 1))*c
Time = 3.82 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=\frac {b\,c^5\,\ln \left (x\right )}{5}-\frac {b\,c^5\,\ln \left (c^2\,x^2-1\right )}{10}-\frac {\frac {b\,c^3\,x^3}{2}+\frac {b\,c\,x}{4}+a}{5\,x^5}-\frac {b\,\ln \left (c\,x+1\right )}{10\,x^5}+\frac {b\,\ln \left (1-c\,x\right )}{10\,x^5} \] Input:
int((a + b*atanh(c*x))/x^6,x)
Output:
(b*c^5*log(x))/5 - (b*c^5*log(c^2*x^2 - 1))/10 - (a + (b*c^3*x^3)/2 + (b*c *x)/4)/(5*x^5) - (b*log(c*x + 1))/(10*x^5) + (b*log(1 - c*x))/(10*x^5)
Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \text {arctanh}(c x)}{x^6} \, dx=\frac {-4 \mathit {atanh} \left (c x \right ) b \,c^{5} x^{5}-4 \mathit {atanh} \left (c x \right ) b -4 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{5} x^{5}+4 \,\mathrm {log}\left (x \right ) b \,c^{5} x^{5}-4 a -2 b \,c^{3} x^{3}-b c x}{20 x^{5}} \] Input:
int((a+b*atanh(c*x))/x^6,x)
Output:
( - 4*atanh(c*x)*b*c**5*x**5 - 4*atanh(c*x)*b - 4*log(c**2*x - c)*b*c**5*x **5 + 4*log(x)*b*c**5*x**5 - 4*a - 2*b*c**3*x**3 - b*c*x)/(20*x**5)