Integrand size = 18, antiderivative size = 98 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=-\frac {b c d}{6 x^2}-\frac {b c^2 d}{2 x}-\frac {d (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c d (a+b \text {arctanh}(c x))}{2 x^2}+\frac {1}{3} b c^3 d \log (x)-\frac {5}{12} b c^3 d \log (1-c x)+\frac {1}{12} b c^3 d \log (1+c x) \]
-1/6*b*c*d/x^2-1/2*b*c^2*d/x-1/3*d*(a+b*arctanh(c*x))/x^3-1/2*c*d*(a+b*arc tanh(c*x))/x^2+1/3*b*c^3*d*ln(x)-5/12*b*c^3*d*ln(-c*x+1)+1/12*b*c^3*d*ln(c *x+1)
Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=-\frac {d \left (4 a+6 a c x+2 b c x+6 b c^2 x^2+2 b (2+3 c x) \text {arctanh}(c x)-4 b c^3 x^3 \log (x)+5 b c^3 x^3 \log (1-c x)-b c^3 x^3 \log (1+c x)\right )}{12 x^3} \]
-1/12*(d*(4*a + 6*a*c*x + 2*b*c*x + 6*b*c^2*x^2 + 2*b*(2 + 3*c*x)*ArcTanh[ c*x] - 4*b*c^3*x^3*Log[x] + 5*b*c^3*x^3*Log[1 - c*x] - b*c^3*x^3*Log[1 + c *x]))/x^3
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 523, 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.
\(\displaystyle \int \frac {(c d x+d) (a+b \text {arctanh}(c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle -b c \int -\frac {d (3 c x+2)}{6 x^3 \left (1-c^2 x^2\right )}dx-\frac {d (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c d (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} b c d \int \frac {3 c x+2}{x^3 \left (1-c^2 x^2\right )}dx-\frac {d (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c d (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {1}{6} b c d \int \left (-\frac {5 c^3}{2 (c x-1)}+\frac {c^3}{2 (c x+1)}+\frac {2 c^2}{x}+\frac {3 c}{x^2}+\frac {2}{x^3}\right )dx-\frac {d (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c d (a+b \text {arctanh}(c x))}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c d (a+b \text {arctanh}(c x))}{2 x^2}+\frac {1}{6} b c d \left (2 c^2 \log (x)-\frac {5}{2} c^2 \log (1-c x)+\frac {1}{2} c^2 \log (c x+1)-\frac {3 c}{x}-\frac {1}{x^2}\right )\) |
-1/3*(d*(a + b*ArcTanh[c*x]))/x^3 - (c*d*(a + b*ArcTanh[c*x]))/(2*x^2) + ( b*c*d*(-x^(-2) - (3*c)/x + 2*c^2*Log[x] - (5*c^2*Log[1 - c*x])/2 + (c^2*Lo g[1 + c*x])/2))/6
3.1.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
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 = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88
method | result | size |
parts | \(a d \left (-\frac {1}{3 x^{3}}-\frac {c}{2 x^{2}}\right )+b d \,c^{3} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{2 c x}+\frac {\ln \left (c x \right )}{3}\right )\) | \(86\) |
derivativedivides | \(c^{3} \left (a d \left (-\frac {1}{2 c^{2} x^{2}}-\frac {1}{3 c^{3} x^{3}}\right )+b d \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{2 c x}+\frac {\ln \left (c x \right )}{3}\right )\right )\) | \(92\) |
default | \(c^{3} \left (a d \left (-\frac {1}{2 c^{2} x^{2}}-\frac {1}{3 c^{3} x^{3}}\right )+b d \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{2 c x}+\frac {\ln \left (c x \right )}{3}\right )\right )\) | \(92\) |
parallelrisch | \(-\frac {2 \ln \left (c x -1\right ) x^{3} b \,c^{3} d -2 b d \,c^{3} \ln \left (x \right ) x^{3}-x^{3} \operatorname {arctanh}\left (c x \right ) b d \,c^{3}+3 a \,c^{3} d \,x^{3}+c^{3} x^{3} b d +3 b \,c^{2} d \,x^{2}+3 b c d x \,\operatorname {arctanh}\left (c x \right )+3 a c d x +b c d x +2 \,\operatorname {arctanh}\left (c x \right ) b d +2 a d}{6 x^{3}}\) | \(111\) |
risch | \(-\frac {b d \left (3 c x +2\right ) \ln \left (c x +1\right )}{12 x^{3}}+\frac {d \left (b \,c^{3} \ln \left (c x +1\right ) x^{3}-5 b \,x^{3} \ln \left (-c x +1\right ) c^{3}+4 b \,c^{3} \ln \left (-x \right ) x^{3}-6 b \,c^{2} x^{2}+3 b c x \ln \left (-c x +1\right )-6 c x a -2 b c x +2 b \ln \left (-c x +1\right )-4 a \right )}{12 x^{3}}\) | \(115\) |
a*d*(-1/3/x^3-1/2*c/x^2)+b*d*c^3*(-1/2/c^2/x^2*arctanh(c*x)-1/3/c^3/x^3*ar ctanh(c*x)+1/12*ln(c*x+1)-5/12*ln(c*x-1)-1/6/c^2/x^2-1/2/c/x+1/3*ln(c*x))
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {b c^{3} d x^{3} \log \left (c x + 1\right ) - 5 \, b c^{3} d x^{3} \log \left (c x - 1\right ) + 4 \, b c^{3} d x^{3} \log \left (x\right ) - 6 \, b c^{2} d x^{2} - 2 \, {\left (3 \, a + b\right )} c d x - 4 \, a d - {\left (3 \, b c d x + 2 \, b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{12 \, x^{3}} \]
1/12*(b*c^3*d*x^3*log(c*x + 1) - 5*b*c^3*d*x^3*log(c*x - 1) + 4*b*c^3*d*x^ 3*log(x) - 6*b*c^2*d*x^2 - 2*(3*a + b)*c*d*x - 4*a*d - (3*b*c*d*x + 2*b*d) *log(-(c*x + 1)/(c*x - 1)))/x^3
Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.19 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=\begin {cases} - \frac {a c d}{2 x^{2}} - \frac {a d}{3 x^{3}} + \frac {b c^{3} d \log {\left (x \right )}}{3} - \frac {b c^{3} d \log {\left (x - \frac {1}{c} \right )}}{3} + \frac {b c^{3} d \operatorname {atanh}{\left (c x \right )}}{6} - \frac {b c^{2} d}{2 x} - \frac {b c d \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} - \frac {b c d}{6 x^{2}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a d}{3 x^{3}} & \text {otherwise} \end {cases} \]
Piecewise((-a*c*d/(2*x**2) - a*d/(3*x**3) + b*c**3*d*log(x)/3 - b*c**3*d*l og(x - 1/c)/3 + b*c**3*d*atanh(c*x)/6 - b*c**2*d/(2*x) - b*c*d*atanh(c*x)/ (2*x**2) - b*c*d/(6*x**2) - b*d*atanh(c*x)/(3*x**3), Ne(c, 0)), (-a*d/(3*x **3), True))
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c d - \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b d - \frac {a c d}{2 \, x^{2}} - \frac {a d}{3 \, x^{3}} \]
1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*c*d - 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*d - 1/2*a*c*d/x^2 - 1/3*a*d/x^3
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (84) = 168\).
Time = 0.29 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.12 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {1}{3} \, {\left (b c^{2} d \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{2} d \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {6 \, {\left (c x + 1\right )}^{2} b c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} b c^{2} d}{c x - 1} + b c^{2} d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {12 \, {\left (c x + 1\right )}^{2} a c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {6 \, {\left (c x + 1\right )} a c^{2} d}{c x - 1} + 2 \, a c^{2} d + \frac {5 \, {\left (c x + 1\right )}^{2} b c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {8 \, {\left (c x + 1\right )} b c^{2} d}{c x - 1} + 3 \, b c^{2} d}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \]
1/3*(b*c^2*d*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^2*d*log(-(c*x + 1)/(c*x - 1)) + (6*(c*x + 1)^2*b*c^2*d/(c*x - 1)^2 + 3*(c*x + 1)*b*c^2*d/(c*x - 1) + b*c^2*d)*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) + (12*(c*x + 1)^2*a*c^2*d/(c* x - 1)^2 + 6*(c*x + 1)*a*c^2*d/(c*x - 1) + 2*a*c^2*d + 5*(c*x + 1)^2*b*c^2 *d/(c*x - 1)^2 + 8*(c*x + 1)*b*c^2*d/(c*x - 1) + 3*b*c^2*d)/((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 = 3.36 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^4} \, dx=\frac {b\,c^3\,d\,\ln \left (x\right )}{3}-\frac {a\,c\,d}{2\,x^2}-\frac {b\,c\,d}{6\,x^2}-\frac {b\,d\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3}-\frac {b\,c^3\,d\,\ln \left (c^2\,x^2-1\right )}{6}-\frac {b\,c^2\,d}{2\,x}-\frac {a\,d}{3\,x^3}-\frac {b\,c^4\,d\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{2\,\sqrt {-c^2}}-\frac {b\,c\,d\,\mathrm {atanh}\left (c\,x\right )}{2\,x^2} \]