Integrand size = 14, antiderivative size = 43 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=-\frac {b}{2 c x}-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}+\frac {b \text {arctanh}\left (\frac {x}{c}\right )}{2 c^2} \] Output:
-1/2*b/c/x-1/2*(a+b*arctanh(c/x))/x^2+1/2*b*arctanh(x/c)/c^2
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.40 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b}{2 c x}-\frac {b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}-\frac {b \log (-c+x)}{4 c^2}+\frac {b \log (c+x)}{4 c^2} \] Input:
Integrate[(a + b*ArcTanh[c/x])/x^3,x]
Output:
-1/2*a/x^2 - b/(2*c*x) - (b*ArcTanh[c/x])/(2*x^2) - (b*Log[-c + x])/(4*c^2 ) + (b*Log[c + x])/(4*c^2)
Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6452, 795, 264, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -\frac {1}{2} b c \int \frac {1}{\left (1-\frac {c^2}{x^2}\right ) x^4}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle -\frac {1}{2} b c \int \frac {1}{x^2 \left (x^2-c^2\right )}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {1}{2} b c \left (\frac {\int \frac {1}{x^2-c^2}dx}{c^2}+\frac {1}{c^2 x}\right )-\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{2 x^2}-\frac {1}{2} b c \left (\frac {1}{c^2 x}-\frac {\text {arctanh}\left (\frac {x}{c}\right )}{c^3}\right )\) |
Input:
Int[(a + b*ArcTanh[c/x])/x^3,x]
Output:
-1/2*(a + b*ArcTanh[c/x])/x^2 - (b*c*(1/(c^2*x) - ArcTanh[x/c]/c^3))/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*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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 = 0.43 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(-\frac {-\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,x^{2}+\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,c^{2}+b c x +a \,c^{2}}{2 x^{2} c^{2}}\) | \(42\) |
parts | \(-\frac {a}{2 x^{2}}-\frac {b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\) | \(55\) |
derivativedivides | \(-\frac {\frac {a \,c^{2}}{2 x^{2}}+b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\) | \(59\) |
default | \(-\frac {\frac {a \,c^{2}}{2 x^{2}}+b \left (\frac {c^{2} \operatorname {arctanh}\left (\frac {c}{x}\right )}{2 x^{2}}+\frac {c}{2 x}+\frac {\ln \left (\frac {c}{x}-1\right )}{4}-\frac {\ln \left (1+\frac {c}{x}\right )}{4}\right )}{c^{2}}\) | \(59\) |
orering | \(-\frac {2 \left (c^{2}-x^{2}\right ) \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )\right )}{x^{2} c^{2}}-\frac {\left (c -x \right ) \left (x +c \right ) x^{2} \left (-\frac {b c}{x^{5} \left (1-\frac {c^{2}}{x^{2}}\right )}-\frac {3 \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )\right )}{x^{4}}\right )}{2 c^{2}}\) | \(80\) |
risch | \(-\frac {b \ln \left (x +c \right )}{4 x^{2}}-\frac {-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )+i \pi b \,c^{2} \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}+i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )+2 i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{3}+i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}-2 i \pi b \,c^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{3}+2 b \ln \left (x -c \right ) x^{2}-2 x^{2} b \ln \left (x +c \right )-2 b \ln \left (c -x \right ) c^{2}+4 a \,c^{2}+4 b c x}{8 c^{2} x^{2}}\) | \(320\) |
Input:
int((a+b*arctanh(c/x))/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(-arctanh(c/x)*b*x^2+arctanh(c/x)*b*c^2+b*c*x+a*c^2)/x^2/c^2
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=-\frac {2 \, a c^{2} + 2 \, b c x + {\left (b c^{2} - b x^{2}\right )} \log \left (-\frac {c + x}{c - x}\right )}{4 \, c^{2} x^{2}} \] Input:
integrate((a+b*arctanh(c/x))/x^3,x, algorithm="fricas")
Output:
-1/4*(2*a*c^2 + 2*b*c*x + (b*c^2 - b*x^2)*log(-(c + x)/(c - x)))/(c^2*x^2)
Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=\begin {cases} - \frac {a}{2 x^{2}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2 x^{2}} - \frac {b}{2 c x} + \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\- \frac {a}{2 x^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c/x))/x**3,x)
Output:
Piecewise((-a/(2*x**2) - b*atanh(c/x)/(2*x**2) - b/(2*c*x) + b*atanh(c/x)/ (2*c**2), Ne(c, 0)), (-a/(2*x**2), True))
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=\frac {1}{4} \, {\left (c {\left (\frac {\log \left (c + x\right )}{c^{3}} - \frac {\log \left (-c + x\right )}{c^{3}} - \frac {2}{c^{2} x}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*arctanh(c/x))/x^3,x, algorithm="maxima")
Output:
1/4*(c*(log(c + x)/c^3 - log(-c + x)/c^3 - 2/(c^2*x)) - 2*arctanh(c/x)/x^2 )*b - 1/2*a/x^2
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (37) = 74\).
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.86 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=-\frac {\frac {b {\left (c + x\right )} \log \left (-\frac {c + x}{c - x}\right )}{{\left (\frac {{\left (c + x\right )}^{2} c}{{\left (c - x\right )}^{2}} - \frac {2 \, {\left (c + x\right )} c}{c - x} + c\right )} {\left (c - x\right )}} - \frac {b - \frac {2 \, a {\left (c + x\right )}}{c - x} - \frac {b {\left (c + x\right )}}{c - x}}{\frac {{\left (c + x\right )}^{2} c}{{\left (c - x\right )}^{2}} - \frac {2 \, {\left (c + x\right )} c}{c - x} + c}}{c} \] Input:
integrate((a+b*arctanh(c/x))/x^3,x, algorithm="giac")
Output:
-(b*(c + x)*log(-(c + x)/(c - x))/(((c + x)^2*c/(c - x)^2 - 2*(c + x)*c/(c - x) + c)*(c - x)) - (b - 2*a*(c + x)/(c - x) - b*(c + x)/(c - x))/((c + x)^2*c/(c - x)^2 - 2*(c + x)*c/(c - x) + c))/c
Time = 3.52 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=\frac {b\,c\,\mathrm {atan}\left (\frac {x}{\sqrt {-c^2}}\right )}{2\,{\left (-c^2\right )}^{3/2}}-\frac {b}{2\,c\,x}-\frac {b\,\mathrm {atanh}\left (\frac {c}{x}\right )}{2\,x^2}-\frac {a}{2\,x^2} \] Input:
int((a + b*atanh(c/x))/x^3,x)
Output:
(b*c*atan(x/(-c^2)^(1/2)))/(2*(-c^2)^(3/2)) - b/(2*c*x) - (b*atanh(c/x))/( 2*x^2) - a/(2*x^2)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x}\right )}{x^3} \, dx=\frac {-\mathit {atanh} \left (\frac {c}{x}\right ) b \,c^{2}+\mathit {atanh} \left (\frac {c}{x}\right ) b \,x^{2}-a \,c^{2}-b c x}{2 c^{2} x^{2}} \] Input:
int((a+b*atanh(c/x))/x^3,x)
Output:
( - atanh(c/x)*b*c**2 + atanh(c/x)*b*x**2 - a*c**2 - b*c*x)/(2*c**2*x**2)