Integrand size = 18, antiderivative size = 108 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {b d x}{4 c^3}+\frac {b d x^2}{10 c^2}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4+\frac {1}{4} d x^4 (a+b \text {arctanh}(c x))+\frac {1}{5} c d x^5 (a+b \text {arctanh}(c x))+\frac {9 b d \log (1-c x)}{40 c^4}-\frac {b d \log (1+c x)}{40 c^4} \]
1/4*b*d*x/c^3+1/10*b*d*x^2/c^2+1/12*b*d*x^3/c+1/20*b*d*x^4+1/4*d*x^4*(a+b* arctanh(c*x))+1/5*c*d*x^5*(a+b*arctanh(c*x))+9/40*b*d*ln(-c*x+1)/c^4-1/40* b*d*ln(c*x+1)/c^4
Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {d \left (30 b c x+12 b c^2 x^2+10 b c^3 x^3+30 a c^4 x^4+6 b c^4 x^4+24 a c^5 x^5+6 b c^4 x^4 (5+4 c x) \text {arctanh}(c x)+27 b \log (1-c x)-3 b \log (1+c x)\right )}{120 c^4} \]
(d*(30*b*c*x + 12*b*c^2*x^2 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 6*b*c^4*x^4 + 24*a*c^5*x^5 + 6*b*c^4*x^4*(5 + 4*c*x)*ArcTanh[c*x] + 27*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(120*c^4)
Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int x^3 (c d x+d) (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle -b c \int \frac {d x^4 (4 c x+5)}{20 \left (1-c^2 x^2\right )}dx+\frac {1}{5} c d x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d x^4 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{20} b c d \int \frac {x^4 (4 c x+5)}{1-c^2 x^2}dx+\frac {1}{5} c d x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d x^4 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 523 |
\(\displaystyle -\frac {1}{20} b c d \int \left (-\frac {4 x^3}{c}-\frac {5 x^2}{c^2}-\frac {4 x}{c^3}+\frac {4 c x+5}{c^4 \left (1-c^2 x^2\right )}-\frac {5}{c^4}\right )dx+\frac {1}{5} c d x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d x^4 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} c d x^5 (a+b \text {arctanh}(c x))+\frac {1}{4} d x^4 (a+b \text {arctanh}(c x))-\frac {1}{20} b c d \left (\frac {5 \text {arctanh}(c x)}{c^5}-\frac {5 x}{c^4}-\frac {2 x^2}{c^3}-\frac {5 x^3}{3 c^2}-\frac {2 \log \left (1-c^2 x^2\right )}{c^5}-\frac {x^4}{c}\right )\) |
(d*x^4*(a + b*ArcTanh[c*x]))/4 + (c*d*x^5*(a + b*ArcTanh[c*x]))/5 - (b*c*d *((-5*x)/c^4 - (2*x^2)/c^3 - (5*x^3)/(3*c^2) - x^4/c + (5*ArcTanh[c*x])/c^ 5 - (2*Log[1 - c^2*x^2])/c^5))/20
3.1.1.3.1 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 = 0.74 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85
method | result | size |
parts | \(a d \left (\frac {1}{5} c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {b d \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{10}+\frac {c x}{4}+\frac {9 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{4}}\) | \(92\) |
derivativedivides | \(\frac {a d \left (\frac {1}{5} c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{10}+\frac {c x}{4}+\frac {9 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{4}}\) | \(98\) |
default | \(\frac {a d \left (\frac {1}{5} c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {c^{5} x^{5} \operatorname {arctanh}\left (c x \right )}{5}+\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{12}+\frac {c^{2} x^{2}}{10}+\frac {c x}{4}+\frac {9 \ln \left (c x -1\right )}{40}-\frac {\ln \left (c x +1\right )}{40}\right )}{c^{4}}\) | \(98\) |
parallelrisch | \(\frac {12 b \,c^{5} d \,\operatorname {arctanh}\left (c x \right ) x^{5}+12 c^{5} d \,x^{5} a +15 x^{4} \operatorname {arctanh}\left (c x \right ) b \,c^{4} d +15 a \,c^{4} d \,x^{4}+3 d \,x^{4} c^{4} b +5 c^{3} x^{3} b d +6 b \,c^{2} d \,x^{2}+15 b c d x +12 \ln \left (c x -1\right ) b d -3 \,\operatorname {arctanh}\left (c x \right ) b d +6 b d}{60 c^{4}}\) | \(113\) |
risch | \(\frac {d b \,x^{4} \left (4 c x +5\right ) \ln \left (c x +1\right )}{40}-\frac {d c b \,x^{5} \ln \left (-c x +1\right )}{10}+\frac {d c \,x^{5} a}{5}-\frac {d b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {d \,x^{4} a}{4}+\frac {b d \,x^{4}}{20}+\frac {b d \,x^{3}}{12 c}+\frac {b d \,x^{2}}{10 c^{2}}+\frac {b d x}{4 c^{3}}-\frac {b d \ln \left (c x +1\right )}{40 c^{4}}+\frac {9 b d \ln \left (-c x +1\right )}{40 c^{4}}\) | \(127\) |
a*d*(1/5*c*x^5+1/4*x^4)+b*d/c^4*(1/5*c^5*x^5*arctanh(c*x)+1/4*c^4*x^4*arct anh(c*x)+1/20*c^4*x^4+1/12*c^3*x^3+1/10*c^2*x^2+1/4*c*x+9/40*ln(c*x-1)-1/4 0*ln(c*x+1))
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {24 \, a c^{5} d x^{5} + 6 \, {\left (5 \, a + b\right )} c^{4} d x^{4} + 10 \, b c^{3} d x^{3} + 12 \, b c^{2} d x^{2} + 30 \, b c d x - 3 \, b d \log \left (c x + 1\right ) + 27 \, b d \log \left (c x - 1\right ) + 3 \, {\left (4 \, b c^{5} d x^{5} + 5 \, b c^{4} d x^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{120 \, c^{4}} \]
1/120*(24*a*c^5*d*x^5 + 6*(5*a + b)*c^4*d*x^4 + 10*b*c^3*d*x^3 + 12*b*c^2* d*x^2 + 30*b*c*d*x - 3*b*d*log(c*x + 1) + 27*b*d*log(c*x - 1) + 3*(4*b*c^5 *d*x^5 + 5*b*c^4*d*x^4)*log(-(c*x + 1)/(c*x - 1)))/c^4
Time = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.15 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c d x^{5}}{5} + \frac {a d x^{4}}{4} + \frac {b c d x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b d x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d x^{4}}{20} + \frac {b d x^{3}}{12 c} + \frac {b d x^{2}}{10 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{5 c^{4}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{20 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a*c*d*x**5/5 + a*d*x**4/4 + b*c*d*x**5*atanh(c*x)/5 + b*d*x**4* atanh(c*x)/4 + b*d*x**4/20 + b*d*x**3/(12*c) + b*d*x**2/(10*c**2) + b*d*x/ (4*c**3) + b*d*log(x - 1/c)/(5*c**4) - b*d*atanh(c*x)/(20*c**4), Ne(c, 0)) , (a*d*x**4/4, True))
Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {1}{5} \, a c d x^{5} + \frac {1}{4} \, a d x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c d + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d \]
1/5*a*c*d*x^5 + 1/4*a*d*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x ^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*b*c*d + 1/24*(6*x^4*arctanh(c*x) + c*(2 *(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*d
Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (92) = 184\).
Time = 0.29 (sec) , antiderivative size = 491, normalized size of antiderivative = 4.55 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {1}{15} \, c {\left (\frac {3 \, {\left (\frac {10 \, {\left (c x + 1\right )}^{4} b d}{{\left (c x - 1\right )}^{4}} - \frac {5 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} + \frac {15 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} - \frac {5 \, {\left (c x + 1\right )} b d}{c x - 1} + b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{5}}{c x - 1} - c^{5}} + \frac {\frac {60 \, {\left (c x + 1\right )}^{4} a d}{{\left (c x - 1\right )}^{4}} - \frac {30 \, {\left (c x + 1\right )}^{3} a d}{{\left (c x - 1\right )}^{3}} + \frac {90 \, {\left (c x + 1\right )}^{2} a d}{{\left (c x - 1\right )}^{2}} - \frac {30 \, {\left (c x + 1\right )} a d}{c x - 1} + 6 \, a d + \frac {27 \, {\left (c x + 1\right )}^{4} b d}{{\left (c x - 1\right )}^{4}} - \frac {69 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} + \frac {79 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} - \frac {47 \, {\left (c x + 1\right )} b d}{c x - 1} + 10 \, b d}{\frac {{\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{5}}{c x - 1} - c^{5}} - \frac {3 \, b d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{5}} + \frac {3 \, b d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{5}}\right )} \]
1/15*c*(3*(10*(c*x + 1)^4*b*d/(c*x - 1)^4 - 5*(c*x + 1)^3*b*d/(c*x - 1)^3 + 15*(c*x + 1)^2*b*d/(c*x - 1)^2 - 5*(c*x + 1)*b*d/(c*x - 1) + b*d)*log(-( c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^5/(c*x - 1)^5 - 5*(c*x + 1)^4*c^5/(c*x - 1)^4 + 10*(c*x + 1)^3*c^5/(c*x - 1)^3 - 10*(c*x + 1)^2*c^5/(c*x - 1)^2 + 5*(c*x + 1)*c^5/(c*x - 1) - c^5) + (60*(c*x + 1)^4*a*d/(c*x - 1)^4 - 30*( c*x + 1)^3*a*d/(c*x - 1)^3 + 90*(c*x + 1)^2*a*d/(c*x - 1)^2 - 30*(c*x + 1) *a*d/(c*x - 1) + 6*a*d + 27*(c*x + 1)^4*b*d/(c*x - 1)^4 - 69*(c*x + 1)^3*b *d/(c*x - 1)^3 + 79*(c*x + 1)^2*b*d/(c*x - 1)^2 - 47*(c*x + 1)*b*d/(c*x - 1) + 10*b*d)/((c*x + 1)^5*c^5/(c*x - 1)^5 - 5*(c*x + 1)^4*c^5/(c*x - 1)^4 + 10*(c*x + 1)^3*c^5/(c*x - 1)^3 - 10*(c*x + 1)^2*c^5/(c*x - 1)^2 + 5*(c*x + 1)*c^5/(c*x - 1) - c^5) - 3*b*d*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 3*b *d*log(-(c*x + 1)/(c*x - 1))/c^5)
Time = 3.66 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95 \[ \int x^3 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {\frac {b\,c\,d\,x}{4}-\frac {d\,\left (15\,b\,\mathrm {atanh}\left (c\,x\right )-6\,b\,\ln \left (c^2\,x^2-1\right )\right )}{60}+\frac {b\,c^2\,d\,x^2}{10}+\frac {b\,c^3\,d\,x^3}{12}}{c^4}+\frac {d\,\left (15\,a\,x^4+3\,b\,x^4+15\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c\,d\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{60} \]