Integrand size = 14, antiderivative size = 48 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {b}{6 c x^2}-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}-\frac {b \log \left (1-\frac {c^2}{x^2}\right )}{6 c^3} \] Output:
-1/6*b/c/x^2-1/3*(a+b*arctanh(c/x))/x^3-1/6*b*ln(1-c^2/x^2)/c^3
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b}{6 c x^2}-\frac {b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (-c^2+x^2\right )}{6 c^3} \] Input:
Integrate[(a + b*ArcTanh[c/x])/x^4,x]
Output:
-1/3*a/x^3 - b/(6*c*x^2) - (b*ArcTanh[c/x])/(3*x^3) + (b*Log[x])/(3*c^3) - (b*Log[-c^2 + x^2])/(6*c^3)
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6452, 795, 243, 25, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -\frac {1}{3} b c \int \frac {1}{\left (1-\frac {c^2}{x^2}\right ) x^5}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle -\frac {1}{3} b c \int \frac {1}{x^3 \left (x^2-c^2\right )}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {1}{6} b c \int -\frac {1}{x^4 \left (c^2-x^2\right )}dx^2-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{6} b c \int \frac {1}{x^4 \left (c^2-x^2\right )}dx^2-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{6} b c \int \left (\frac {1}{c^4 x^2}+\frac {1}{c^2 x^4}+\frac {1}{c^4 \left (c^2-x^2\right )}\right )dx^2-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{3 x^3}-\frac {1}{6} b c \left (-\frac {\log \left (x^2\right )}{c^4}+\frac {1}{c^2 x^2}+\frac {\log \left (c^2-x^2\right )}{c^4}\right )\) |
Input:
Int[(a + b*ArcTanh[c/x])/x^4,x]
Output:
-1/3*(a + b*ArcTanh[c/x])/x^3 - (b*c*(1/(c^2*x^2) - Log[x^2]/c^4 + Log[c^2 - x^2]/c^4))/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[(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 = 0.42 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19
method | result | size |
parts | \(-\frac {a}{3 x^{3}}-\frac {b \left (\frac {c^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2}}{6 x^{2}}+\frac {\ln \left (\frac {c}{x}-1\right )}{6}+\frac {\ln \left (1+\frac {c}{x}\right )}{6}\right )}{c^{3}}\) | \(57\) |
derivativedivides | \(-\frac {\frac {a \,c^{3}}{3 x^{3}}+b \left (\frac {c^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2}}{6 x^{2}}+\frac {\ln \left (\frac {c}{x}-1\right )}{6}+\frac {\ln \left (1+\frac {c}{x}\right )}{6}\right )}{c^{3}}\) | \(61\) |
default | \(-\frac {\frac {a \,c^{3}}{3 x^{3}}+b \left (\frac {c^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2}}{6 x^{2}}+\frac {\ln \left (\frac {c}{x}-1\right )}{6}+\frac {\ln \left (1+\frac {c}{x}\right )}{6}\right )}{c^{3}}\) | \(61\) |
parallelrisch | \(\frac {2 b \ln \left (x \right ) x^{3}-2 \ln \left (x -c \right ) x^{3} b -2 b \,x^{3} \operatorname {arctanh}\left (\frac {c}{x}\right )-2 \,\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,c^{3}-b \,c^{2} x -2 a \,c^{3}}{6 x^{3} c^{3}}\) | \(67\) |
risch | \(-\frac {b \ln \left (x +c \right )}{6 x^{3}}-\frac {2 i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}+i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )+i \pi b \,c^{3} \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}-i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )-i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}-2 i \pi b \,c^{3}-i \pi b \,c^{3} \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}-i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{3}+i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}-i \pi b \,c^{3} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{3}-4 b \ln \left (x \right ) x^{3}+2 b \ln \left (c^{2}-x^{2}\right ) x^{3}-2 b \ln \left (c -x \right ) c^{3}+4 a \,c^{3}+2 b \,c^{2} x}{12 c^{3} x^{3}}\) | \(324\) |
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*c^2/x^2+1/6*ln(c/x-1)+1/6*l n(1+c/x))
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {b x^{3} \log \left (-c^{2} + x^{2}\right ) - 2 \, b x^{3} \log \left (x\right ) + b c^{3} \log \left (-\frac {c + x}{c - x}\right ) + 2 \, a c^{3} + b c^{2} x}{6 \, c^{3} x^{3}} \] Input:
integrate((a+b*arctanh(c/x))/x^4,x, algorithm="fricas")
Output:
-1/6*(b*x^3*log(-c^2 + x^2) - 2*b*x^3*log(x) + b*c^3*log(-(c + x)/(c - x)) + 2*a*c^3 + b*c^2*x)/(c^3*x^3)
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=\begin {cases} - \frac {a}{3 x^{3}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3 x^{3}} - \frac {b}{6 c x^{2}} + \frac {b \log {\left (x \right )}}{3 c^{3}} - \frac {b \log {\left (- c + x \right )}}{3 c^{3}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c/x))/x**4,x)
Output:
Piecewise((-a/(3*x**3) - b*atanh(c/x)/(3*x**3) - b/(6*c*x**2) + b*log(x)/( 3*c**3) - b*log(-c + x)/(3*c**3) - b*atanh(c/x)/(3*c**3), Ne(c, 0)), (-a/( 3*x**3), True))
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {1}{6} \, {\left (c {\left (\frac {\log \left (-c^{2} + x^{2}\right )}{c^{4}} - \frac {\log \left (x^{2}\right )}{c^{4}} + \frac {1}{c^{2} x^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (\frac {c}{x}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*arctanh(c/x))/x^4,x, algorithm="maxima")
Output:
-1/6*(c*(log(-c^2 + x^2)/c^4 - log(x^2)/c^4 + 1/(c^2*x^2)) + 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. 234 vs. \(2 (42) = 84\).
Time = 0.13 (sec) , antiderivative size = 234, normalized size of antiderivative = 4.88 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {\frac {{\left (b + \frac {3 \, b {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{3} c^{2}}{{\left (c - x\right )}^{3}} - \frac {3 \, {\left (c + x\right )}^{2} c^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )} c^{2}}{c - x} - c^{2}} + \frac {2 \, {\left (a + \frac {3 \, a {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {b {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} - \frac {b {\left (c + x\right )}}{c - x}\right )}}{\frac {{\left (c + x\right )}^{3} c^{2}}{{\left (c - x\right )}^{3}} - \frac {3 \, {\left (c + x\right )}^{2} c^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )} c^{2}}{c - x} - c^{2}} - \frac {b \log \left (-\frac {c + x}{c - x} + 1\right )}{c^{2}} + \frac {b \log \left (-\frac {c + x}{c - x}\right )}{c^{2}}}{3 \, c} \] Input:
integrate((a+b*arctanh(c/x))/x^4,x, algorithm="giac")
Output:
-1/3*((b + 3*b*(c + x)^2/(c - x)^2)*log(-(c + x)/(c - x))/((c + x)^3*c^2/( c - x)^3 - 3*(c + x)^2*c^2/(c - x)^2 + 3*(c + x)*c^2/(c - x) - c^2) + 2*(a + 3*a*(c + x)^2/(c - x)^2 + b*(c + x)^2/(c - x)^2 - b*(c + x)/(c - x))/(( c + x)^3*c^2/(c - x)^3 - 3*(c + x)^2*c^2/(c - x)^2 + 3*(c + x)*c^2/(c - x) - c^2) - b*log(-(c + x)/(c - x) + 1)/c^2 + b*log(-(c + x)/(c - x))/c^2)/c
Time = 3.49 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {\frac {a}{3}+\frac {b\,\mathrm {atanh}\left (\frac {c}{x}\right )}{3}}{x^3}-\frac {\frac {b\,x^3\,\ln \left (x^2-c^2\right )}{6}-\frac {b\,x^3\,\ln \left (x\right )}{3}+\frac {b\,c^2\,x}{6}}{c^3\,x^3} \] Input:
int((a + b*atanh(c/x))/x^4,x)
Output:
- (a/3 + (b*atanh(c/x))/3)/x^3 - ((b*x^3*log(x^2 - c^2))/6 - (b*x^3*log(x) )/3 + (b*c^2*x)/6)/(c^3*x^3)
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^4} \, dx=\frac {-2 \mathit {atanh} \left (\frac {c}{x}\right ) b \,c^{3}+2 \mathit {atanh} \left (\frac {c}{x}\right ) b \,x^{3}-2 \,\mathrm {log}\left (-c -x \right ) b \,x^{3}+2 \,\mathrm {log}\left (x \right ) b \,x^{3}-2 a \,c^{3}-b \,c^{2} x}{6 c^{3} x^{3}} \] Input:
int((a+b*atanh(c/x))/x^4,x)
Output:
( - 2*atanh(c/x)*b*c**3 + 2*atanh(c/x)*b*x**3 - 2*log( - c - x)*b*x**3 + 2 *log(x)*b*x**3 - 2*a*c**3 - b*c**2*x)/(6*c**3*x**3)