Integrand size = 18, antiderivative size = 56 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=-\frac {b c d}{2 x}-\frac {d (1+c x)^2 (a+b \text {arctanh}(c x))}{2 x^2}+b c^2 d \log (x)-b c^2 d \log (1-c x) \] Output:
-1/2*b*c*d/x-1/2*d*(c*x+1)^2*(a+b*arctanh(c*x))/x^2+b*c^2*d*ln(x)-b*c^2*d* ln(-c*x+1)
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.36 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=-\frac {d \left (2 a+4 a c x+2 b c x+2 (b+2 b c x) \text {arctanh}(c x)-4 b c^2 x^2 \log (x)+3 b c^2 x^2 \log (1-c x)+b c^2 x^2 \log (1+c x)\right )}{4 x^2} \] Input:
Integrate[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^3,x]
Output:
-1/4*(d*(2*a + 4*a*c*x + 2*b*c*x + 2*(b + 2*b*c*x)*ArcTanh[c*x] - 4*b*c^2* x^2*Log[x] + 3*b*c^2*x^2*Log[1 - c*x] + b*c^2*x^2*Log[1 + c*x]))/x^2
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 86, 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.
\(\displaystyle \int \frac {(c d x+d) (a+b \text {arctanh}(c x))}{x^3} \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle -b c \int -\frac {d (c x+1)}{2 x^2 (1-c x)}dx-\frac {d (c x+1)^2 (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} b c d \int \frac {c x+1}{x^2 (1-c x)}dx-\frac {d (c x+1)^2 (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} b c d \int \left (-\frac {2 c^2}{c x-1}+\frac {2 c}{x}+\frac {1}{x^2}\right )dx-\frac {d (c x+1)^2 (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} b c d \left (2 c \log (x)-2 c \log (1-c x)-\frac {1}{x}\right )-\frac {d (c x+1)^2 (a+b \text {arctanh}(c x))}{2 x^2}\) |
Input:
Int[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^3,x]
Output:
-1/2*(d*(1 + c*x)^2*(a + b*ArcTanh[c*x]))/x^2 + (b*c*d*(-x^(-1) + 2*c*Log[ x] - 2*c*Log[1 - c*x]))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( x_))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[( a + b*ArcTanh[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 - c^2*x ^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && Intege rQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0 ]))
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.36
method | result | size |
parts | \(a d \left (-\frac {c}{x}-\frac {1}{2 x^{2}}\right )+d b \,c^{2} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+\ln \left (c x \right )-\frac {\ln \left (c x +1\right )}{4}\right )\) | \(76\) |
derivativedivides | \(c^{2} \left (a d \left (-\frac {1}{2 c^{2} x^{2}}-\frac {1}{c x}\right )+d b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+\ln \left (c x \right )-\frac {\ln \left (c x +1\right )}{4}\right )\right )\) | \(82\) |
default | \(c^{2} \left (a d \left (-\frac {1}{2 c^{2} x^{2}}-\frac {1}{c x}\right )+d b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}-\frac {3 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+\ln \left (c x \right )-\frac {\ln \left (c x +1\right )}{4}\right )\right )\) | \(82\) |
parallelrisch | \(\frac {2 b \,c^{2} d \ln \left (x \right ) x^{2}-2 \ln \left (c x -1\right ) x^{2} b \,c^{2} d -x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d -a \,c^{2} d \,x^{2}-2 b c d x \,\operatorname {arctanh}\left (c x \right )-2 a d x c -b c d x -\operatorname {arctanh}\left (c x \right ) b d -a d}{2 x^{2}}\) | \(93\) |
risch | \(-\frac {d b \left (2 c x +1\right ) \ln \left (c x +1\right )}{4 x^{2}}+\frac {d \left (4 b \,c^{2} \ln \left (-x \right ) x^{2}-3 b \,c^{2} x^{2} \ln \left (-c x +1\right )-b \,c^{2} \ln \left (c x +1\right ) x^{2}+2 b c x \ln \left (-c x +1\right )-4 a c x -2 b c x +b \ln \left (-c x +1\right )-2 a \right )}{4 x^{2}}\) | \(106\) |
Input:
int((c*d*x+d)*(a+b*arctanh(c*x))/x^3,x,method=_RETURNVERBOSE)
Output:
a*d*(-c/x-1/2/x^2)+d*b*c^2*(-1/2*arctanh(c*x)/c^2/x^2-arctanh(c*x)/c/x-3/4 *ln(c*x-1)-1/2/c/x+ln(c*x)-1/4*ln(c*x+1))
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.59 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=-\frac {b c^{2} d x^{2} \log \left (c x + 1\right ) + 3 \, b c^{2} d x^{2} \log \left (c x - 1\right ) - 4 \, b c^{2} d x^{2} \log \left (x\right ) + 2 \, {\left (2 \, a + b\right )} c d x + 2 \, a d + {\left (2 \, b c d x + b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, x^{2}} \] Input:
integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^3,x, algorithm="fricas")
Output:
-1/4*(b*c^2*d*x^2*log(c*x + 1) + 3*b*c^2*d*x^2*log(c*x - 1) - 4*b*c^2*d*x^ 2*log(x) + 2*(2*a + b)*c*d*x + 2*a*d + (2*b*c*d*x + b*d)*log(-(c*x + 1)/(c *x - 1)))/x^2
Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.70 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=\begin {cases} - \frac {a c d}{x} - \frac {a d}{2 x^{2}} + b c^{2} d \log {\left (x \right )} - b c^{2} d \log {\left (x - \frac {1}{c} \right )} - \frac {b c^{2} d \operatorname {atanh}{\left (c x \right )}}{2} - \frac {b c d \operatorname {atanh}{\left (c x \right )}}{x} - \frac {b c d}{2 x} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} & \text {for}\: c \neq 0 \\- \frac {a d}{2 x^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((c*d*x+d)*(a+b*atanh(c*x))/x**3,x)
Output:
Piecewise((-a*c*d/x - a*d/(2*x**2) + b*c**2*d*log(x) - b*c**2*d*log(x - 1/ c) - b*c**2*d*atanh(c*x)/2 - b*c*d*atanh(c*x)/x - b*c*d/(2*x) - b*d*atanh( c*x)/(2*x**2), Ne(c, 0)), (-a*d/(2*x**2), True))
Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.59 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=-\frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c d + \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b d - \frac {a c d}{x} - \frac {a d}{2 \, x^{2}} \] Input:
integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^3,x, algorithm="maxima")
Output:
-1/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c*d + 1/4*((c* log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d - a*c*d/x - 1/2*a*d/x^2
Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (52) = 104\).
Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.43 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx={\left (b c d \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c d \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {2 \, {\left (c x + 1\right )} b c d}{c x - 1} + b c d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {4 \, {\left (c x + 1\right )} a c d}{c x - 1} + 2 \, a c d + \frac {{\left (c x + 1\right )} b c d}{c x - 1} + b c d}{\frac {{\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \] Input:
integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^3,x, algorithm="giac")
Output:
(b*c*d*log(-(c*x + 1)/(c*x - 1) - 1) - b*c*d*log(-(c*x + 1)/(c*x - 1)) + ( 2*(c*x + 1)*b*c*d/(c*x - 1) + b*c*d)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^ 2/(c*x - 1)^2 + 2*(c*x + 1)/(c*x - 1) + 1) + (4*(c*x + 1)*a*c*d/(c*x - 1) + 2*a*c*d + (c*x + 1)*b*c*d/(c*x - 1) + b*c*d)/((c*x + 1)^2/(c*x - 1)^2 + 2*(c*x + 1)/(c*x - 1) + 1))*c
Time = 3.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=\frac {d\,\left (b\,c^2\,\mathrm {atanh}\left (c\,x\right )-b\,c^2\,\ln \left (c^2\,x^2-1\right )+2\,b\,c^2\,\ln \left (x\right )\right )}{2}-\frac {\frac {d\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{2}+\frac {d\,x\,\left (2\,a\,c+b\,c+2\,b\,c\,\mathrm {atanh}\left (c\,x\right )\right )}{2}}{x^2} \] Input:
int(((a + b*atanh(c*x))*(d + c*d*x))/x^3,x)
Output:
(d*(b*c^2*atanh(c*x) - b*c^2*log(c^2*x^2 - 1) + 2*b*c^2*log(x)))/2 - ((d*( a + b*atanh(c*x)))/2 + (d*x*(2*a*c + b*c + 2*b*c*atanh(c*x)))/2)/x^2
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.41 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^3} \, dx=\frac {d \left (-\mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-2 \mathit {atanh} \left (c x \right ) b c x -\mathit {atanh} \left (c x \right ) b -2 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{2} x^{2}+2 \,\mathrm {log}\left (x \right ) b \,c^{2} x^{2}-2 a c x -a -b c x \right )}{2 x^{2}} \] Input:
int((c*d*x+d)*(a+b*atanh(c*x))/x^3,x)
Output:
(d*( - atanh(c*x)*b*c**2*x**2 - 2*atanh(c*x)*b*c*x - atanh(c*x)*b - 2*log( c**2*x - c)*b*c**2*x**2 + 2*log(x)*b*c**2*x**2 - 2*a*c*x - a - b*c*x))/(2* x**2)