Integrand size = 20, antiderivative size = 147 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {b c d^2}{12 x^3}-\frac {b c^2 d^2}{3 x^2}-\frac {3 b c^3 d^2}{4 x}-\frac {d^2 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {2 c d^2 (a+b \text {arctanh}(c x))}{3 x^3}-\frac {c^2 d^2 (a+b \text {arctanh}(c x))}{2 x^2}+\frac {2}{3} b c^4 d^2 \log (x)-\frac {17}{24} b c^4 d^2 \log (1-c x)+\frac {1}{24} b c^4 d^2 \log (1+c x) \]
-1/12*b*c*d^2/x^3-1/3*b*c^2*d^2/x^2-3/4*b*c^3*d^2/x-1/4*d^2*(a+b*arctanh(c *x))/x^4-2/3*c*d^2*(a+b*arctanh(c*x))/x^3-1/2*c^2*d^2*(a+b*arctanh(c*x))/x ^2+2/3*b*c^4*d^2*ln(x)-17/24*b*c^4*d^2*ln(-c*x+1)+1/24*b*c^4*d^2*ln(c*x+1)
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.78 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {d^2 \left (6 a+16 a c x+2 b c x+12 a c^2 x^2+8 b c^2 x^2+18 b c^3 x^3+2 b \left (3+8 c x+6 c^2 x^2\right ) \text {arctanh}(c x)-16 b c^4 x^4 \log (x)+17 b c^4 x^4 \log (1-c x)-b c^4 x^4 \log (1+c x)\right )}{24 x^4} \]
-1/24*(d^2*(6*a + 16*a*c*x + 2*b*c*x + 12*a*c^2*x^2 + 8*b*c^2*x^2 + 18*b*c ^3*x^3 + 2*b*(3 + 8*c*x + 6*c^2*x^2)*ArcTanh[c*x] - 16*b*c^4*x^4*Log[x] + 17*b*c^4*x^4*Log[1 - c*x] - b*c^4*x^4*Log[1 + c*x]))/x^4
Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6498, 27, 2333, 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.
| \(\displaystyle \int \frac {(c d x+d)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx\) |
| \(\Big \downarrow \) 6498 |
| \(\displaystyle -b c \int -\frac {d^2 \left (6 c^2 x^2+8 c x+3\right )}{12 x^4 \left (1-c^2 x^2\right )}dx-\frac {c^2 d^2 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {d^2 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {2 c d^2 (a+b \text {arctanh}(c x))}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{12} b c d^2 \int \frac {6 c^2 x^2+8 c x+3}{x^4 \left (1-c^2 x^2\right )}dx-\frac {c^2 d^2 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {d^2 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {2 c d^2 (a+b \text {arctanh}(c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {1}{12} b c d^2 \int \left (-\frac {17 c^4}{2 (c x-1)}+\frac {c^4}{2 (c x+1)}+\frac {8 c^3}{x}+\frac {9 c^2}{x^2}+\frac {8 c}{x^3}+\frac {3}{x^4}\right )dx-\frac {c^2 d^2 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {d^2 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {2 c d^2 (a+b \text {arctanh}(c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {c^2 d^2 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {d^2 (a+b \text {arctanh}(c x))}{4 x^4}-\frac {2 c d^2 (a+b \text {arctanh}(c x))}{3 x^3}+\frac {1}{12} b c d^2 \left (8 c^3 \log (x)-\frac {17}{2} c^3 \log (1-c x)+\frac {1}{2} c^3 \log (c x+1)-\frac {9 c^2}{x}-\frac {4 c}{x^2}-\frac {1}{x^3}\right )\) |
-1/4*(d^2*(a + b*ArcTanh[c*x]))/x^4 - (2*c*d^2*(a + b*ArcTanh[c*x]))/(3*x^ 3) - (c^2*d^2*(a + b*ArcTanh[c*x]))/(2*x^2) + (b*c*d^2*(-x^(-3) - (4*c)/x^ 2 - (9*c^2)/x + 8*c^3*Log[x] - (17*c^3*Log[1 - c*x])/2 + (c^3*Log[1 + c*x] )/2))/12
3.1.18.3.1 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[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 1.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.80
| method | result | size |
| parts | \(a \,d^{2} \left (-\frac {1}{4 x^{4}}-\frac {2 c}{3 x^{3}}-\frac {c^{2}}{2 x^{2}}\right )+d^{2} b \,c^{4} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{24}-\frac {17 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{3 c^{2} x^{2}}-\frac {3}{4 c x}+\frac {2 \ln \left (c x \right )}{3}\right )\) | \(118\) |
| derivativedivides | \(c^{4} \left (a \,d^{2} \left (-\frac {1}{2 c^{2} x^{2}}-\frac {2}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+d^{2} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{24}-\frac {17 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{3 c^{2} x^{2}}-\frac {3}{4 c x}+\frac {2 \ln \left (c x \right )}{3}\right )\right )\) | \(124\) |
| default | \(c^{4} \left (a \,d^{2} \left (-\frac {1}{2 c^{2} x^{2}}-\frac {2}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+d^{2} b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {2 \,\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{24}-\frac {17 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{3 c^{2} x^{2}}-\frac {3}{4 c x}+\frac {2 \ln \left (c x \right )}{3}\right )\right )\) | \(124\) |
| risch | \(-\frac {d^{2} b \left (6 c^{2} x^{2}+8 c x +3\right ) \ln \left (c x +1\right )}{24 x^{4}}+\frac {d^{2} \left (b \,c^{4} \ln \left (c x +1\right ) x^{4}-17 b \,x^{4} \ln \left (-c x +1\right ) c^{4}+16 b \,c^{4} \ln \left (-x \right ) x^{4}-18 b \,c^{3} x^{3}+6 b \,x^{2} \ln \left (-c x +1\right ) c^{2}-12 a \,c^{2} x^{2}-8 b \,c^{2} x^{2}+8 b c x \ln \left (-c x +1\right )-16 c x a -2 b c x +3 b \ln \left (-c x +1\right )-6 a \right )}{24 x^{4}}\) | \(161\) |
| parallelrisch | \(-\frac {8 \ln \left (c x -1\right ) x^{4} b \,c^{4} d^{2}-8 b \,c^{4} d^{2} \ln \left (x \right ) x^{4}-x^{4} \operatorname {arctanh}\left (c x \right ) b \,c^{4} d^{2}+6 a \,c^{4} d^{2} x^{4}+4 b \,c^{4} d^{2} x^{4}+9 b \,c^{3} d^{2} x^{3}+6 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{2}+6 a \,c^{2} d^{2} x^{2}+4 b \,c^{2} d^{2} x^{2}+8 b c \,d^{2} x \,\operatorname {arctanh}\left (c x \right )+8 a \,d^{2} c x +b c \,d^{2} x +3 b \,d^{2} \operatorname {arctanh}\left (c x \right )+3 a \,d^{2}}{12 x^{4}}\) | \(174\) |
a*d^2*(-1/4/x^4-2/3*c/x^3-1/2*c^2/x^2)+d^2*b*c^4*(-1/2/c^2/x^2*arctanh(c*x )-2/3/c^3/x^3*arctanh(c*x)-1/4/c^4/x^4*arctanh(c*x)+1/24*ln(c*x+1)-17/24*l n(c*x-1)-1/12/c^3/x^3-1/3/c^2/x^2-3/4/c/x+2/3*ln(c*x))
Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {b c^{4} d^{2} x^{4} \log \left (c x + 1\right ) - 17 \, b c^{4} d^{2} x^{4} \log \left (c x - 1\right ) + 16 \, b c^{4} d^{2} x^{4} \log \left (x\right ) - 18 \, b c^{3} d^{2} x^{3} - 4 \, {\left (3 \, a + 2 \, b\right )} c^{2} d^{2} x^{2} - 2 \, {\left (8 \, a + b\right )} c d^{2} x - 6 \, a d^{2} - {\left (6 \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + 3 \, b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, x^{4}} \]
1/24*(b*c^4*d^2*x^4*log(c*x + 1) - 17*b*c^4*d^2*x^4*log(c*x - 1) + 16*b*c^ 4*d^2*x^4*log(x) - 18*b*c^3*d^2*x^3 - 4*(3*a + 2*b)*c^2*d^2*x^2 - 2*(8*a + b)*c*d^2*x - 6*a*d^2 - (6*b*c^2*d^2*x^2 + 8*b*c*d^2*x + 3*b*d^2)*log(-(c* x + 1)/(c*x - 1)))/x^4
Time = 0.46 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.29 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=\begin {cases} - \frac {a c^{2} d^{2}}{2 x^{2}} - \frac {2 a c d^{2}}{3 x^{3}} - \frac {a d^{2}}{4 x^{4}} + \frac {2 b c^{4} d^{2} \log {\left (x \right )}}{3} - \frac {2 b c^{4} d^{2} \log {\left (x - \frac {1}{c} \right )}}{3} + \frac {b c^{4} d^{2} \operatorname {atanh}{\left (c x \right )}}{12} - \frac {3 b c^{3} d^{2}}{4 x} - \frac {b c^{2} d^{2} \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} - \frac {b c^{2} d^{2}}{3 x^{2}} - \frac {2 b c d^{2} \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} - \frac {b c d^{2}}{12 x^{3}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a d^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \]
Piecewise((-a*c**2*d**2/(2*x**2) - 2*a*c*d**2/(3*x**3) - a*d**2/(4*x**4) + 2*b*c**4*d**2*log(x)/3 - 2*b*c**4*d**2*log(x - 1/c)/3 + b*c**4*d**2*atanh (c*x)/12 - 3*b*c**3*d**2/(4*x) - b*c**2*d**2*atanh(c*x)/(2*x**2) - b*c**2* d**2/(3*x**2) - 2*b*c*d**2*atanh(c*x)/(3*x**3) - b*c*d**2/(12*x**3) - b*d* *2*atanh(c*x)/(4*x**4), Ne(c, 0)), (-a*d**2/(4*x**4), True))
Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.21 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{2} d^{2} - \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c d^{2} + \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b d^{2} - \frac {a c^{2} d^{2}}{2 \, x^{2}} - \frac {2 \, a c d^{2}}{3 \, x^{3}} - \frac {a d^{2}}{4 \, x^{4}} \]
1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*c^2 *d^2 - 1/3*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c* x)/x^3)*b*c*d^2 + 1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^ 2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b*d^2 - 1/2*a*c^2*d^2/x^2 - 2/3*a* c*d^2/x^3 - 1/4*a*d^2/x^4
Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (129) = 258\).
Time = 0.28 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.93 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {1}{3} \, {\left (2 \, b c^{3} d^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 2 \, b c^{3} d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} b c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} b c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b c^{3} d^{2}}{c x - 1} + b c^{3} d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {24 \, {\left (c x + 1\right )}^{3} a c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {24 \, {\left (c x + 1\right )}^{2} a c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {16 \, {\left (c x + 1\right )} a c^{3} d^{2}}{c x - 1} + 4 \, a c^{3} d^{2} + \frac {10 \, {\left (c x + 1\right )}^{3} b c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {23 \, {\left (c x + 1\right )}^{2} b c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {18 \, {\left (c x + 1\right )} b c^{3} d^{2}}{c x - 1} + 5 \, b c^{3} d^{2}}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \]
1/3*(2*b*c^3*d^2*log(-(c*x + 1)/(c*x - 1) - 1) - 2*b*c^3*d^2*log(-(c*x + 1 )/(c*x - 1)) + 2*(6*(c*x + 1)^3*b*c^3*d^2/(c*x - 1)^3 + 6*(c*x + 1)^2*b*c^ 3*d^2/(c*x - 1)^2 + 4*(c*x + 1)*b*c^3*d^2/(c*x - 1) + b*c^3*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6* (c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1) + (24*(c*x + 1)^3*a*c ^3*d^2/(c*x - 1)^3 + 24*(c*x + 1)^2*a*c^3*d^2/(c*x - 1)^2 + 16*(c*x + 1)*a *c^3*d^2/(c*x - 1) + 4*a*c^3*d^2 + 10*(c*x + 1)^3*b*c^3*d^2/(c*x - 1)^3 + 23*(c*x + 1)^2*b*c^3*d^2/(c*x - 1)^2 + 18*(c*x + 1)*b*c^3*d^2/(c*x - 1) + 5*b*c^3*d^2)/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1))*c
Time = 3.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14 \[ \int \frac {(d+c d x)^2 (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {2\,b\,c^4\,d^2\,\ln \left (x\right )}{3}-\frac {b\,c^4\,d^2\,\ln \left (c^2\,x^2-1\right )}{3}-\frac {a\,c^2\,d^2}{2\,x^2}-\frac {b\,c^2\,d^2}{3\,x^2}-\frac {3\,b\,c^3\,d^2}{4\,x}-\frac {a\,d^2}{4\,x^4}-\frac {2\,a\,c\,d^2}{3\,x^3}-\frac {b\,c\,d^2}{12\,x^3}-\frac {b\,d^2\,\mathrm {atanh}\left (c\,x\right )}{4\,x^4}-\frac {3\,b\,c^5\,d^2\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{4\,\sqrt {-c^2}}-\frac {2\,b\,c\,d^2\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3}-\frac {b\,c^2\,d^2\,\mathrm {atanh}\left (c\,x\right )}{2\,x^2} \]
(2*b*c^4*d^2*log(x))/3 - (b*c^4*d^2*log(c^2*x^2 - 1))/3 - (a*c^2*d^2)/(2*x ^2) - (b*c^2*d^2)/(3*x^2) - (3*b*c^3*d^2)/(4*x) - (a*d^2)/(4*x^4) - (2*a*c *d^2)/(3*x^3) - (b*c*d^2)/(12*x^3) - (b*d^2*atanh(c*x))/(4*x^4) - (3*b*c^5 *d^2*atan((c^2*x)/(-c^2)^(1/2)))/(4*(-c^2)^(1/2)) - (2*b*c*d^2*atanh(c*x)) /(3*x^3) - (b*c^2*d^2*atanh(c*x))/(2*x^2)