Integrand size = 12, antiderivative size = 48 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {b x}{4 c^3}+\frac {b x^3}{12 c}-\frac {b \text {arctanh}(c x)}{4 c^4}+\frac {1}{4} x^4 (a+b \text {arctanh}(c x)) \] Output:
1/4*b*x/c^3+1/12*b*x^3/c-1/4*b*arctanh(c*x)/c^4+1/4*x^4*(a+b*arctanh(c*x))
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.46 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {b x}{4 c^3}+\frac {b x^3}{12 c}+\frac {a x^4}{4}+\frac {1}{4} b x^4 \text {arctanh}(c x)+\frac {b \log (1-c x)}{8 c^4}-\frac {b \log (1+c x)}{8 c^4} \] Input:
Integrate[x^3*(a + b*ArcTanh[c*x]),x]
Output:
(b*x)/(4*c^3) + (b*x^3)/(12*c) + (a*x^4)/4 + (b*x^4*ArcTanh[c*x])/4 + (b*L og[1 - c*x])/(8*c^4) - (b*Log[1 + c*x])/(8*c^4)
Time = 0.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6452, 254, 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.
\(\displaystyle \int x^3 (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))-\frac {1}{4} b c \int \frac {x^4}{1-c^2 x^2}dx\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))-\frac {1}{4} b c \int \left (-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}-\frac {1}{c^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))-\frac {1}{4} b c \left (\frac {\text {arctanh}(c x)}{c^5}-\frac {x}{c^4}-\frac {x^3}{3 c^2}\right )\) |
Input:
Int[x^3*(a + b*ArcTanh[c*x]),x]
Output:
(x^4*(a + b*ArcTanh[c*x]))/4 - (b*c*(-(x/c^4) - x^3/(3*c^2) + ArcTanh[c*x] /c^5))/4
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {-3 b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}-3 a \,c^{4} x^{4}-b \,c^{3} x^{3}-3 b c x +3 b \,\operatorname {arctanh}\left (c x \right )}{12 c^{4}}\) | \(50\) |
parts | \(\frac {x^{4} a}{4}+\frac {b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(54\) |
derivativedivides | \(\frac {\frac {a \,c^{4} x^{4}}{4}+b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(58\) |
default | \(\frac {\frac {a \,c^{4} x^{4}}{4}+b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(58\) |
risch | \(\frac {b \,x^{4} \ln \left (c x +1\right )}{8}-\frac {b \,x^{4} \ln \left (-c x +1\right )}{8}+\frac {x^{4} a}{4}+\frac {b \,x^{3}}{12 c}+\frac {b x}{4 c^{3}}+\frac {b \ln \left (-c x +1\right )}{8 c^{4}}-\frac {b \ln \left (c x +1\right )}{8 c^{4}}\) | \(74\) |
orering | \(\frac {\left (c^{4} x^{4}+c^{2} x^{2}-2\right ) \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )}{2 c^{4}}-\frac {\left (c^{2} x^{2}+3\right ) \left (c x -1\right ) \left (c x +1\right ) \left (3 x^{2} \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )+\frac {x^{3} b c}{-c^{2} x^{2}+1}\right )}{12 x^{2} c^{4}}\) | \(90\) |
Input:
int(x^3*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
Output:
-1/12*(-3*b*arctanh(c*x)*x^4*c^4-3*a*c^4*x^4-b*c^3*x^3-3*b*c*x+3*b*arctanh (c*x))/c^4
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, a c^{4} x^{4} + 2 \, b c^{3} x^{3} + 6 \, b c x + 3 \, {\left (b c^{4} x^{4} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{4}} \] Input:
integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="fricas")
Output:
1/24*(6*a*c^4*x^4 + 2*b*c^3*x^3 + 6*b*c*x + 3*(b*c^4*x^4 - b)*log(-(c*x + 1)/(c*x - 1)))/c^4
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b x^{3}}{12 c} + \frac {b x}{4 c^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{4 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(a+b*atanh(c*x)),x)
Output:
Piecewise((a*x**4/4 + b*x**4*atanh(c*x)/4 + b*x**3/(12*c) + b*x/(4*c**3) - b*atanh(c*x)/(4*c**4), Ne(c, 0)), (a*x**4/4, True))
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b \] Input:
integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="maxima")
Output:
1/4*a*x^4 + 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
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (40) = 80\).
Time = 0.12 (sec) , antiderivative size = 296, normalized size of antiderivative = 6.17 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c x + 1\right )}^{3} b}{{\left (c x - 1\right )}^{3}} + \frac {{\left (c x + 1\right )} b}{c x - 1}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} + \frac {\frac {6 \, {\left (c x + 1\right )}^{3} a}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a}{c x - 1} + \frac {3 \, {\left (c x + 1\right )}^{3} b}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b}{c x - 1} - 2 \, b}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}}\right )} \] Input:
integrate(x^3*(a+b*arctanh(c*x)),x, algorithm="giac")
Output:
1/3*c*(3*((c*x + 1)^3*b/(c*x - 1)^3 + (c*x + 1)*b/(c*x - 1))*log(-(c*x + 1 )/(c*x - 1))/((c*x + 1)^4*c^5/(c*x - 1)^4 - 4*(c*x + 1)^3*c^5/(c*x - 1)^3 + 6*(c*x + 1)^2*c^5/(c*x - 1)^2 - 4*(c*x + 1)*c^5/(c*x - 1) + c^5) + (6*(c *x + 1)^3*a/(c*x - 1)^3 + 6*(c*x + 1)*a/(c*x - 1) + 3*(c*x + 1)^3*b/(c*x - 1)^3 - 6*(c*x + 1)^2*b/(c*x - 1)^2 + 5*(c*x + 1)*b/(c*x - 1) - 2*b)/((c*x + 1)^4*c^5/(c*x - 1)^4 - 4*(c*x + 1)^3*c^5/(c*x - 1)^3 + 6*(c*x + 1)^2*c^ 5/(c*x - 1)^2 - 4*(c*x + 1)*c^5/(c*x - 1) + c^5))
Time = 3.75 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {a\,x^4}{4}+\frac {\frac {b\,c^3\,x^3}{12}-\frac {b\,\mathrm {atanh}\left (c\,x\right )}{4}+\frac {b\,c\,x}{4}}{c^4}+\frac {b\,x^4\,\mathrm {atanh}\left (c\,x\right )}{4} \] Input:
int(x^3*(a + b*atanh(c*x)),x)
Output:
(a*x^4)/4 + ((b*c^3*x^3)/12 - (b*atanh(c*x))/4 + (b*c*x)/4)/c^4 + (b*x^4*a tanh(c*x))/4
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x^3 (a+b \text {arctanh}(c x)) \, dx=\frac {3 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-3 \mathit {atanh} \left (c x \right ) b +3 a \,c^{4} x^{4}+b \,c^{3} x^{3}+3 b c x}{12 c^{4}} \] Input:
int(x^3*(a+b*atanh(c*x)),x)
Output:
(3*atanh(c*x)*b*c**4*x**4 - 3*atanh(c*x)*b + 3*a*c**4*x**4 + b*c**3*x**3 + 3*b*c*x)/(12*c**4)