Integrand size = 18, antiderivative size = 96 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {b d x}{4 c^2}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3+\frac {1}{3} d x^3 (a+b \text {arctanh}(c x))+\frac {1}{4} c d x^4 (a+b \text {arctanh}(c x))+\frac {7 b d \log (1-c x)}{24 c^3}+\frac {b d \log (1+c x)}{24 c^3} \] Output:
1/4*b*d*x/c^2+1/6*b*d*x^2/c+1/12*b*d*x^3+1/3*d*x^3*(a+b*arctanh(c*x))+1/4* c*d*x^4*(a+b*arctanh(c*x))+7/24*b*d*ln(-c*x+1)/c^3+1/24*b*d*ln(c*x+1)/c^3
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {d \left (6 b c x+4 b c^2 x^2+8 a c^3 x^3+2 b c^3 x^3+6 a c^4 x^4+2 b c^3 x^3 (4+3 c x) \text {arctanh}(c x)+7 b \log (1-c x)+b \log (1+c x)\right )}{24 c^3} \] Input:
Integrate[x^2*(d + c*d*x)*(a + b*ArcTanh[c*x]),x]
Output:
(d*(6*b*c*x + 4*b*c^2*x^2 + 8*a*c^3*x^3 + 2*b*c^3*x^3 + 6*a*c^4*x^4 + 2*b* c^3*x^3*(4 + 3*c*x)*ArcTanh[c*x] + 7*b*Log[1 - c*x] + b*Log[1 + c*x]))/(24 *c^3)
Time = 0.32 (sec) , antiderivative size = 89, 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, 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.
\(\displaystyle \int x^2 (c d x+d) (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle -b c \int \frac {d x^3 (3 c x+4)}{12 \left (1-c^2 x^2\right )}dx+\frac {1}{4} c d x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{12} b c d \int \frac {x^3 (3 c x+4)}{1-c^2 x^2}dx+\frac {1}{4} c d x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 523 |
\(\displaystyle -\frac {1}{12} b c d \int \left (-\frac {3 x^2}{c}-\frac {4 x}{c^2}+\frac {4 c x+3}{c^3 \left (1-c^2 x^2\right )}-\frac {3}{c^3}\right )dx+\frac {1}{4} c d x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d x^3 (a+b \text {arctanh}(c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} c d x^4 (a+b \text {arctanh}(c x))+\frac {1}{3} d x^3 (a+b \text {arctanh}(c x))-\frac {1}{12} b c d \left (\frac {3 \text {arctanh}(c x)}{c^4}-\frac {3 x}{c^3}-\frac {2 x^2}{c^2}-\frac {2 \log \left (1-c^2 x^2\right )}{c^4}-\frac {x^3}{c}\right )\) |
Input:
Int[x^2*(d + c*d*x)*(a + b*ArcTanh[c*x]),x]
Output:
(d*x^3*(a + b*ArcTanh[c*x]))/3 + (c*d*x^4*(a + b*ArcTanh[c*x]))/4 - (b*c*d *((-3*x)/c^3 - (2*x^2)/c^2 - x^3/c + (3*ArcTanh[c*x])/c^4 - (2*Log[1 - c^2 *x^2])/c^4))/12
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 = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88
method | result | size |
parts | \(a d \left (\frac {1}{4} c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {d b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{6}+\frac {c x}{4}+\frac {7 \ln \left (c x -1\right )}{24}+\frac {\ln \left (c x +1\right )}{24}\right )}{c^{3}}\) | \(84\) |
derivativedivides | \(\frac {a d \left (\frac {1}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{6}+\frac {c x}{4}+\frac {7 \ln \left (c x -1\right )}{24}+\frac {\ln \left (c x +1\right )}{24}\right )}{c^{3}}\) | \(90\) |
default | \(\frac {a d \left (\frac {1}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+d b \left (\frac {\operatorname {arctanh}\left (c x \right ) c^{4} x^{4}}{4}+\frac {\operatorname {arctanh}\left (c x \right ) c^{3} x^{3}}{3}+\frac {x^{3} c^{3}}{12}+\frac {c^{2} x^{2}}{6}+\frac {c x}{4}+\frac {7 \ln \left (c x -1\right )}{24}+\frac {\ln \left (c x +1\right )}{24}\right )}{c^{3}}\) | \(90\) |
parallelrisch | \(\frac {3 d b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+3 a \,c^{4} d \,x^{4}+4 d b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+4 a \,c^{3} d \,x^{3}+b \,c^{3} d \,x^{3}+2 b \,c^{2} d \,x^{2}+3 b c d x +4 \ln \left (c x -1\right ) b d +\operatorname {arctanh}\left (c x \right ) b d}{12 c^{3}}\) | \(97\) |
risch | \(\frac {d b \,x^{3} \left (3 c x +4\right ) \ln \left (c x +1\right )}{24}-\frac {d c b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {d c a \,x^{4}}{4}-\frac {d b \,x^{3} \ln \left (-c x +1\right )}{6}+\frac {d a \,x^{3}}{3}+\frac {b d \,x^{3}}{12}+\frac {b d \,x^{2}}{6 c}+\frac {b d x}{4 c^{2}}+\frac {7 b d \ln \left (-c x +1\right )}{24 c^{3}}+\frac {b d \ln \left (c x +1\right )}{24 c^{3}}\) | \(117\) |
Input:
int(x^2*(c*d*x+d)*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
a*d*(1/4*c*x^4+1/3*x^3)+d*b/c^3*(1/4*arctanh(c*x)*c^4*x^4+1/3*arctanh(c*x) *c^3*x^3+1/12*x^3*c^3+1/6*c^2*x^2+1/4*c*x+7/24*ln(c*x-1)+1/24*ln(c*x+1))
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, a c^{4} d x^{4} + 2 \, {\left (4 \, a + b\right )} c^{3} d x^{3} + 4 \, b c^{2} d x^{2} + 6 \, b c d x + b d \log \left (c x + 1\right ) + 7 \, b d \log \left (c x - 1\right ) + {\left (3 \, b c^{4} d x^{4} + 4 \, b c^{3} d x^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{3}} \] Input:
integrate(x^2*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/24*(6*a*c^4*d*x^4 + 2*(4*a + b)*c^3*d*x^3 + 4*b*c^2*d*x^2 + 6*b*c*d*x + b*d*log(c*x + 1) + 7*b*d*log(c*x - 1) + (3*b*c^4*d*x^4 + 4*b*c^3*d*x^3)*lo g(-(c*x + 1)/(c*x - 1)))/c^3
Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c d x^{4}}{4} + \frac {a d x^{3}}{3} + \frac {b c d x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b d x^{3}}{12} + \frac {b d x^{2}}{6 c} + \frac {b d x}{4 c^{2}} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{3 c^{3}} + \frac {b d \operatorname {atanh}{\left (c x \right )}}{12 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(c*d*x+d)*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*c*d*x**4/4 + a*d*x**3/3 + b*c*d*x**4*atanh(c*x)/4 + b*d*x**3* atanh(c*x)/3 + b*d*x**3/12 + b*d*x**2/(6*c) + b*d*x/(4*c**2) + b*d*log(x - 1/c)/(3*c**3) + b*d*atanh(c*x)/(12*c**3), Ne(c, 0)), (a*d*x**3/3, True))
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.15 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {1}{4} \, a c d x^{4} + \frac {1}{3} \, a d x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c d + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d \] Input:
integrate(x^2*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/4*a*c*d*x^4 + 1/3*a*d*x^3 + 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*c*d + 1/6*(2*x^3*arc tanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d
Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (82) = 164\).
Time = 0.12 (sec) , antiderivative size = 394, normalized size of antiderivative = 4.10 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {1}{3} \, c {\left (\frac {{\left (\frac {6 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d}{c x - 1} - b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{4}}{c x - 1} + c^{4}} + \frac {\frac {12 \, {\left (c x + 1\right )}^{3} a d}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} a d}{{\left (c x - 1\right )}^{2}} + \frac {8 \, {\left (c x + 1\right )} a d}{c x - 1} - 2 \, a d + \frac {5 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} + \frac {7 \, {\left (c x + 1\right )} b d}{c x - 1} - 2 \, b d}{\frac {{\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{4}}{c x - 1} + c^{4}} - \frac {b d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{4}} + \frac {b d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{4}}\right )} \] Input:
integrate(x^2*(c*d*x+d)*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
1/3*c*((6*(c*x + 1)^3*b*d/(c*x - 1)^3 - 3*(c*x + 1)^2*b*d/(c*x - 1)^2 + 4* (c*x + 1)*b*d/(c*x - 1) - b*d)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4*c^4/ (c*x - 1)^4 - 4*(c*x + 1)^3*c^4/(c*x - 1)^3 + 6*(c*x + 1)^2*c^4/(c*x - 1)^ 2 - 4*(c*x + 1)*c^4/(c*x - 1) + c^4) + (12*(c*x + 1)^3*a*d/(c*x - 1)^3 - 6 *(c*x + 1)^2*a*d/(c*x - 1)^2 + 8*(c*x + 1)*a*d/(c*x - 1) - 2*a*d + 5*(c*x + 1)^3*b*d/(c*x - 1)^3 - 10*(c*x + 1)^2*b*d/(c*x - 1)^2 + 7*(c*x + 1)*b*d/ (c*x - 1) - 2*b*d)/((c*x + 1)^4*c^4/(c*x - 1)^4 - 4*(c*x + 1)^3*c^4/(c*x - 1)^3 + 6*(c*x + 1)^2*c^4/(c*x - 1)^2 - 4*(c*x + 1)*c^4/(c*x - 1) + c^4) - b*d*log(-(c*x + 1)/(c*x - 1) + 1)/c^4 + b*d*log(-(c*x + 1)/(c*x - 1))/c^4 )
Time = 3.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {\frac {b\,c\,d\,x}{4}-\frac {d\,\left (3\,b\,\mathrm {atanh}\left (c\,x\right )-2\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12}+\frac {b\,c^2\,d\,x^2}{6}}{c^3}+\frac {d\,\left (4\,a\,x^3+b\,x^3+4\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c\,d\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \] Input:
int(x^2*(a + b*atanh(c*x))*(d + c*d*x),x)
Output:
((b*c*d*x)/4 - (d*(3*b*atanh(c*x) - 2*b*log(c^2*x^2 - 1)))/12 + (b*c^2*d*x ^2)/6)/c^3 + (d*(4*a*x^3 + b*x^3 + 4*b*x^3*atanh(c*x)))/12 + (c*d*(3*a*x^4 + 3*b*x^4*atanh(c*x)))/12
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int x^2 (d+c d x) (a+b \text {arctanh}(c x)) \, dx=\frac {d \left (3 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}+4 \mathit {atanh} \left (c x \right ) b \,c^{3} x^{3}+\mathit {atanh} \left (c x \right ) b +4 \,\mathrm {log}\left (c^{2} x -c \right ) b +3 a \,c^{4} x^{4}+4 a \,c^{3} x^{3}+b \,c^{3} x^{3}+2 b \,c^{2} x^{2}+3 b c x \right )}{12 c^{3}} \] Input:
int(x^2*(c*d*x+d)*(a+b*atanh(c*x)),x)
Output:
(d*(3*atanh(c*x)*b*c**4*x**4 + 4*atanh(c*x)*b*c**3*x**3 + atanh(c*x)*b + 4 *log(c**2*x - c)*b + 3*a*c**4*x**4 + 4*a*c**3*x**3 + b*c**3*x**3 + 2*b*c** 2*x**2 + 3*b*c*x))/(12*c**3)