Integrand size = 12, antiderivative size = 54 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=-\frac {b c}{6 x^2}-\frac {a+b \text {arctanh}(c x)}{3 x^3}+\frac {1}{3} b c^3 \log (x)-\frac {1}{6} b c^3 \log \left (1-c^2 x^2\right ) \] Output:
-1/6*b*c/x^2-1/3*(a+b*arctanh(c*x))/x^3+1/3*b*c^3*ln(x)-1/6*b*c^3*ln(-c^2* x^2+1)
Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b c}{6 x^2}-\frac {b \text {arctanh}(c x)}{3 x^3}+\frac {1}{3} b c^3 \log (x)-\frac {1}{6} b c^3 \log \left (1-c^2 x^2\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x])/x^4,x]
Output:
-1/3*a/x^3 - (b*c)/(6*x^2) - (b*ArcTanh[c*x])/(3*x^3) + (b*c^3*Log[x])/3 - (b*c^3*Log[1 - c^2*x^2])/6
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} b c \int \frac {1}{x^3 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{6} b c \int \frac {1}{x^4 \left (1-c^2 x^2\right )}dx^2-\frac {a+b \text {arctanh}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{6} b c \int \left (-\frac {c^4}{c^2 x^2-1}+\frac {c^2}{x^2}+\frac {1}{x^4}\right )dx^2-\frac {a+b \text {arctanh}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}\) |
Input:
Int[(a + b*ArcTanh[c*x])/x^4,x]
Output:
-1/3*(a + b*ArcTanh[c*x])/x^3 + (b*c*(-x^(-2) + c^2*Log[x^2] - c^2*Log[1 - c^2*x^2]))/6
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 = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04
method | result | size |
parts | \(-\frac {a}{3 x^{3}}+b \,c^{3} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\ln \left (c x +1\right )}{6}-\frac {1}{6 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{6}\right )\) | \(56\) |
derivativedivides | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\ln \left (c x +1\right )}{6}-\frac {1}{6 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{6}\right )\right )\) | \(60\) |
default | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\ln \left (c x +1\right )}{6}-\frac {1}{6 c^{2} x^{2}}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x -1\right )}{6}\right )\right )\) | \(60\) |
risch | \(-\frac {b \ln \left (c x +1\right )}{6 x^{3}}+\frac {2 b \,c^{3} \ln \left (x \right ) x^{3}-b \,c^{3} \ln \left (c^{2} x^{2}-1\right ) x^{3}-b c x +b \ln \left (-c x +1\right )-2 a}{6 x^{3}}\) | \(67\) |
parallelrisch | \(\frac {2 b \,c^{3} \ln \left (x \right ) x^{3}-2 \ln \left (c x -1\right ) x^{3} b \,c^{3}-2 b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-b \,c^{3} x^{3}-b c x -2 b \,\operatorname {arctanh}\left (c x \right )-2 a}{6 x^{3}}\) | \(70\) |
Input:
int((a+b*arctanh(c*x))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*a/x^3+b*c^3*(-1/3/c^3/x^3*arctanh(c*x)-1/6*ln(c*x+1)-1/6/c^2/x^2+1/3* ln(c*x)-1/6*ln(c*x-1))
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=-\frac {b c^{3} x^{3} \log \left (c^{2} x^{2} - 1\right ) - 2 \, b c^{3} x^{3} \log \left (x\right ) + b c x + b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a}{6 \, x^{3}} \] Input:
integrate((a+b*arctanh(c*x))/x^4,x, algorithm="fricas")
Output:
-1/6*(b*c^3*x^3*log(c^2*x^2 - 1) - 2*b*c^3*x^3*log(x) + b*c*x + b*log(-(c* x + 1)/(c*x - 1)) + 2*a)/x^3
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=\begin {cases} - \frac {a}{3 x^{3}} + \frac {b c^{3} \log {\left (x \right )}}{3} - \frac {b c^{3} \log {\left (x - \frac {1}{c} \right )}}{3} - \frac {b c^{3} \operatorname {atanh}{\left (c x \right )}}{3} - \frac {b c}{6 x^{2}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c*x))/x**4,x)
Output:
Piecewise((-a/(3*x**3) + b*c**3*log(x)/3 - b*c**3*log(x - 1/c)/3 - b*c**3* atanh(c*x)/3 - b*c/(6*x**2) - b*atanh(c*x)/(3*x**3), Ne(c, 0)), (-a/(3*x** 3), True))
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*arctanh(c*x))/x^4,x, algorithm="maxima")
Output:
-1/6*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3 )*b - 1/3*a/x^3
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (46) = 92\).
Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 4.65 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=\frac {1}{3} \, {\left (b c^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {3 \, {\left (c x + 1\right )}^{2} b c^{2}}{{\left (c x - 1\right )}^{2}} + b c^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {2 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} a c^{2}}{{\left (c x - 1\right )}^{2}} + a c^{2} + \frac {{\left (c x + 1\right )}^{2} b c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {{\left (c x + 1\right )} b c^{2}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \] Input:
integrate((a+b*arctanh(c*x))/x^4,x, algorithm="giac")
Output:
1/3*(b*c^2*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^2*log(-(c*x + 1)/(c*x - 1)) + (3*(c*x + 1)^2*b*c^2/(c*x - 1)^2 + b*c^2)*log(-(c*x + 1)/(c*x - 1))/((c *x + 1)^3/(c*x - 1)^3 + 3*(c*x + 1)^2/(c*x - 1)^2 + 3*(c*x + 1)/(c*x - 1) + 1) + 2*(3*(c*x + 1)^2*a*c^2/(c*x - 1)^2 + a*c^2 + (c*x + 1)^2*b*c^2/(c*x - 1)^2 + (c*x + 1)*b*c^2/(c*x - 1))/((c*x + 1)^3/(c*x - 1)^3 + 3*(c*x + 1 )^2/(c*x - 1)^2 + 3*(c*x + 1)/(c*x - 1) + 1))*c
Time = 3.73 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=\frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^2-1\right )}{6}-\frac {\frac {a}{3}+\frac {b\,\mathrm {atanh}\left (c\,x\right )}{3}+\frac {b\,c\,x}{6}}{x^3} \] Input:
int((a + b*atanh(c*x))/x^4,x)
Output:
(b*c^3*log(x))/3 - (b*c^3*log(c^2*x^2 - 1))/6 - (a/3 + (b*atanh(c*x))/3 + (b*c*x)/6)/x^3
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4} \, dx=\frac {-2 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}-2 \mathit {atanh} \left (c x \right ) b -2 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{3} x^{3}+2 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{3}-2 a -b c x}{6 x^{3}} \] Input:
int((a+b*atanh(c*x))/x^4,x)
Output:
( - 2*atanh(c*x)*b*c**3*x**3 - 2*atanh(c*x)*b - 2*log(c**2*x - c)*b*c**3*x **3 + 2*log(x)*b*c**3*x**3 - 2*a - b*c*x)/(6*x**3)