Integrand size = 14, antiderivative size = 113 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {a b x}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 x \text {arctanh}(c x)}{2 c^3}+\frac {b x^3 (a+b \text {arctanh}(c x))}{6 c}-\frac {(a+b \text {arctanh}(c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2+\frac {b^2 \log \left (1-c^2 x^2\right )}{3 c^4} \] Output:
1/2*a*b*x/c^3+1/12*b^2*x^2/c^2+1/2*b^2*x*arctanh(c*x)/c^3+1/6*b*x^3*(a+b*a rctanh(c*x))/c-1/4*(a+b*arctanh(c*x))^2/c^4+1/4*x^4*(a+b*arctanh(c*x))^2+1 /3*b^2*ln(-c^2*x^2+1)/c^4
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {6 a b c x+b^2 c^2 x^2+2 a b c^3 x^3+3 a^2 c^4 x^4+2 b c x \left (3 a c^3 x^3+b \left (3+c^2 x^2\right )\right ) \text {arctanh}(c x)+3 b^2 \left (-1+c^4 x^4\right ) \text {arctanh}(c x)^2+b (3 a+4 b) \log (1-c x)-3 a b \log (1+c x)+4 b^2 \log (1+c x)}{12 c^4} \] Input:
Integrate[x^3*(a + b*ArcTanh[c*x])^2,x]
Output:
(6*a*b*c*x + b^2*c^2*x^2 + 2*a*b*c^3*x^3 + 3*a^2*c^4*x^4 + 2*b*c*x*(3*a*c^ 3*x^3 + b*(3 + c^2*x^2))*ArcTanh[c*x] + 3*b^2*(-1 + c^4*x^4)*ArcTanh[c*x]^ 2 + b*(3*a + 4*b)*Log[1 - c*x] - 3*a*b*Log[1 + c*x] + 4*b^2*Log[1 + c*x])/ (12*c^4)
Time = 0.96 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6452, 6542, 6452, 243, 49, 2009, 6542, 2009, 6510}
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.
| \(\displaystyle \int x^3 (a+b \text {arctanh}(c x))^2 \, dx\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \int \frac {x^4 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx\) |
| \(\Big \downarrow \) 6542 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c^2}-\frac {\int x^2 (a+b \text {arctanh}(c x))dx}{c^2}\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{3} b c \int \frac {x^3}{1-c^2 x^2}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \int \frac {x^2}{1-c^2 x^2}dx^2}{c^2}\right )\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2-1\right )}\right )dx^2}{c^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\int \frac {x^2 (a+b \text {arctanh}(c x))}{1-c^2 x^2}dx}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 6542 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}(c x)}{1-c^2 x^2}dx}{c^2}-\frac {\int (a+b \text {arctanh}(c x))dx}{c^2}}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}(c x)}{1-c^2 x^2}dx}{c^2}-\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x))^2-\frac {1}{2} b c \left (\frac {\frac {(a+b \text {arctanh}(c x))^2}{2 b c^3}-\frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^3 (a+b \text {arctanh}(c x))-\frac {1}{6} b c \left (-\frac {x^2}{c^2}-\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{c^2}\right )\) |
Input:
Int[x^3*(a + b*ArcTanh[c*x])^2,x]
Output:
(x^4*(a + b*ArcTanh[c*x])^2)/4 - (b*c*(-(((x^3*(a + b*ArcTanh[c*x]))/3 - ( b*c*(-(x^2/c^2) - Log[1 - c^2*x^2]/c^4))/6)/c^2) + ((a + b*ArcTanh[c*x])^2 /(2*b*c^3) - (a*x + b*x*ArcTanh[c*x] + (b*Log[1 - c^2*x^2])/(2*c))/c^2)/c^ 2))/2
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTanh[c* x])^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.27
| method | result | size |
| parallelrisch | \(\frac {3 b^{2} \operatorname {arctanh}\left (c x \right )^{2} x^{4} c^{4}+6 a b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+3 a^{2} c^{4} x^{4}+2 b^{2} \operatorname {arctanh}\left (c x \right ) x^{3} c^{3}+2 a b \,c^{3} x^{3}+b^{2} c^{2} x^{2}+6 b^{2} \operatorname {arctanh}\left (c x \right ) x c +6 a b c x -3 b^{2} \operatorname {arctanh}\left (c x \right )^{2}+8 \ln \left (c x -1\right ) b^{2}-6 \,\operatorname {arctanh}\left (c x \right ) a b +8 \,\operatorname {arctanh}\left (c x \right ) b^{2}+b^{2}}{12 c^{4}}\) | \(143\) |
| parts | \(\frac {a^{2} x^{4}}{4}+\frac {b^{2} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )^{2}}{4}+\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{6}+\frac {c x \,\operatorname {arctanh}\left (c x \right )}{2}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (c x -1\right )^{2}}{16}+\frac {\ln \left (c x +1\right )^{2}}{16}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {c^{2} x^{2}}{12}+\frac {\ln \left (c x -1\right )}{3}+\frac {\ln \left (c x +1\right )}{3}\right )}{c^{4}}+\frac {2 a b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(208\) |
| derivativedivides | \(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )^{2}}{4}+\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{6}+\frac {c x \,\operatorname {arctanh}\left (c x \right )}{2}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (c x -1\right )^{2}}{16}+\frac {\ln \left (c x +1\right )^{2}}{16}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {c^{2} x^{2}}{12}+\frac {\ln \left (c x -1\right )}{3}+\frac {\ln \left (c x +1\right )}{3}\right )+2 a b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(209\) |
| default | \(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )^{2}}{4}+\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{6}+\frac {c x \,\operatorname {arctanh}\left (c x \right )}{2}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (c x -1\right )^{2}}{16}+\frac {\ln \left (c x +1\right )^{2}}{16}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{8}+\frac {c^{2} x^{2}}{12}+\frac {\ln \left (c x -1\right )}{3}+\frac {\ln \left (c x +1\right )}{3}\right )+2 a b \left (\frac {c^{4} x^{4} \operatorname {arctanh}\left (c x \right )}{4}+\frac {c^{3} x^{3}}{12}+\frac {c x}{4}+\frac {\ln \left (c x -1\right )}{8}-\frac {\ln \left (c x +1\right )}{8}\right )}{c^{4}}\) | \(209\) |
| risch | \(\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (c x +1\right )^{2}}{16 c^{4}}+\frac {b \left (-3 b \,x^{4} \ln \left (-c x +1\right ) c^{4}+6 a \,c^{4} x^{4}+2 b \,c^{3} x^{3}+6 b c x +3 b \ln \left (-c x +1\right )\right ) \ln \left (c x +1\right )}{24 c^{4}}+\frac {\ln \left (-c x +1\right )^{2} b^{2} x^{4}}{16}-\frac {\ln \left (-c x +1\right ) a b \,x^{4}}{4}+\frac {a^{2} x^{4}}{4}-\frac {b^{2} x^{3} \ln \left (-c x +1\right )}{12 c}+\frac {a b \,x^{3}}{6 c}+\frac {b^{2} x^{2}}{12 c^{2}}-\frac {b^{2} x \ln \left (-c x +1\right )}{4 c^{3}}-\frac {b^{2} \ln \left (-c x +1\right )^{2}}{16 c^{4}}+\frac {a b x}{2 c^{3}}-\frac {b \ln \left (c x +1\right ) a}{4 c^{4}}+\frac {b^{2} \ln \left (c x +1\right )}{3 c^{4}}+\frac {b \ln \left (-c x +1\right ) a}{4 c^{4}}+\frac {b^{2} \ln \left (-c x +1\right )}{3 c^{4}}\) | \(264\) |
Input:
int(x^3*(a+b*arctanh(c*x))^2,x,method=_RETURNVERBOSE)
Output:
1/12*(3*b^2*arctanh(c*x)^2*x^4*c^4+6*a*b*arctanh(c*x)*x^4*c^4+3*a^2*c^4*x^ 4+2*b^2*arctanh(c*x)*x^3*c^3+2*a*b*c^3*x^3+b^2*c^2*x^2+6*b^2*arctanh(c*x)* x*c+6*a*b*c*x-3*b^2*arctanh(c*x)^2+8*ln(c*x-1)*b^2-6*arctanh(c*x)*a*b+8*ar ctanh(c*x)*b^2+b^2)/c^4
Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.42 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {12 \, a^{2} c^{4} x^{4} + 8 \, a b c^{3} x^{3} + 4 \, b^{2} c^{2} x^{2} + 24 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x + 1\right ) + 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{4} + b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{48 \, c^{4}} \] Input:
integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="fricas")
Output:
1/48*(12*a^2*c^4*x^4 + 8*a*b*c^3*x^3 + 4*b^2*c^2*x^2 + 24*a*b*c*x + 3*(b^2 *c^4*x^4 - b^2)*log(-(c*x + 1)/(c*x - 1))^2 - 4*(3*a*b - 4*b^2)*log(c*x + 1) + 4*(3*a*b + 4*b^2)*log(c*x - 1) + 4*(3*a*b*c^4*x^4 + b^2*c^3*x^3 + 3*b ^2*c*x)*log(-(c*x + 1)/(c*x - 1)))/c^4
Time = 0.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.49 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {a b x^{3}}{6 c} + \frac {a b x}{2 c^{3}} - \frac {a b \operatorname {atanh}{\left (c x \right )}}{2 c^{4}} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x \right )}}{4} + \frac {b^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{6 c} + \frac {b^{2} x^{2}}{12 c^{2}} + \frac {b^{2} x \operatorname {atanh}{\left (c x \right )}}{2 c^{3}} + \frac {2 b^{2} \log {\left (x - \frac {1}{c} \right )}}{3 c^{4}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{4 c^{4}} + \frac {2 b^{2} \operatorname {atanh}{\left (c x \right )}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(a+b*atanh(c*x))**2,x)
Output:
Piecewise((a**2*x**4/4 + a*b*x**4*atanh(c*x)/2 + a*b*x**3/(6*c) + a*b*x/(2 *c**3) - a*b*atanh(c*x)/(2*c**4) + b**2*x**4*atanh(c*x)**2/4 + b**2*x**3*a tanh(c*x)/(6*c) + b**2*x**2/(12*c**2) + b**2*x*atanh(c*x)/(2*c**3) + 2*b** 2*log(x - 1/c)/(3*c**4) - b**2*atanh(c*x)**2/(4*c**4) + 2*b**2*atanh(c*x)/ (3*c**4), Ne(c, 0)), (a**2*x**4/4, True))
Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.67 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c x\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{48} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {4 \, c^{2} x^{2} - 2 \, {\left (3 \, \log \left (c x - 1\right ) - 8\right )} \log \left (c x + 1\right ) + 3 \, \log \left (c x + 1\right )^{2} + 3 \, \log \left (c x - 1\right )^{2} + 16 \, \log \left (c x - 1\right )}{c^{4}}\right )} b^{2} \] Input:
integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="maxima")
Output:
1/4*b^2*x^4*arctanh(c*x)^2 + 1/4*a^2*x^4 + 1/12*(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))*a*b + 1/4 8*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)*a rctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*log(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16*log(c*x - 1))/c^4)*b^2
Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (99) = 198\).
Time = 0.14 (sec) , antiderivative size = 603, normalized size of antiderivative = 5.34 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx =\text {Too large to display} \] Input:
integrate(x^3*(a+b*arctanh(c*x))^2,x, algorithm="giac")
Output:
1/6*(3*((c*x + 1)^3*b^2/(c*x - 1)^3 + (c*x + 1)*b^2/(c*x - 1))*log(-(c*x + 1)/(c*x - 1))^2/((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) + 2 *(6*(c*x + 1)^3*a*b/(c*x - 1)^3 + 6*(c*x + 1)*a*b/(c*x - 1) + 3*(c*x + 1)^ 3*b^2/(c*x - 1)^3 - 6*(c*x + 1)^2*b^2/(c*x - 1)^2 + 5*(c*x + 1)*b^2/(c*x - 1) - 2*b^2)*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) + 2*(6*(c*x + 1)^3*a^2/(c*x - 1)^3 + 6*(c*x + 1)*a^2/(c *x - 1) + 6*(c*x + 1)^3*a*b/(c*x - 1)^3 - 12*(c*x + 1)^2*a*b/(c*x - 1)^2 + 10*(c*x + 1)*a*b/(c*x - 1) - 4*a*b + (c*x + 1)^3*b^2/(c*x - 1)^3 - 2*(c*x + 1)^2*b^2/(c*x - 1)^2 + (c*x + 1)*b^2/(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) - 4*b^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 4*b^2*log(-(c*x + 1)/(c*x - 1))/c^5)*c
Time = 3.94 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.19 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {4\,b^2\,\ln \left (c^2\,x^2-1\right )-3\,b^2\,{\mathrm {atanh}\left (c\,x\right )}^2+3\,a^2\,c^4\,x^4+b^2\,c^2\,x^2-6\,a\,b\,\mathrm {atanh}\left (c\,x\right )+2\,b^2\,c^3\,x^3\,\mathrm {atanh}\left (c\,x\right )+6\,b^2\,c\,x\,\mathrm {atanh}\left (c\,x\right )+3\,b^2\,c^4\,x^4\,{\mathrm {atanh}\left (c\,x\right )}^2+2\,a\,b\,c^3\,x^3+6\,a\,b\,c\,x+6\,a\,b\,c^4\,x^4\,\mathrm {atanh}\left (c\,x\right )}{12\,c^4} \] Input:
int(x^3*(a + b*atanh(c*x))^2,x)
Output:
(4*b^2*log(c^2*x^2 - 1) - 3*b^2*atanh(c*x)^2 + 3*a^2*c^4*x^4 + b^2*c^2*x^2 - 6*a*b*atanh(c*x) + 2*b^2*c^3*x^3*atanh(c*x) + 6*b^2*c*x*atanh(c*x) + 3* b^2*c^4*x^4*atanh(c*x)^2 + 2*a*b*c^3*x^3 + 6*a*b*c*x + 6*a*b*c^4*x^4*atanh (c*x))/(12*c^4)
Time = 0.17 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.27 \[ \int x^3 (a+b \text {arctanh}(c x))^2 \, dx=\frac {3 \mathit {atanh} \left (c x \right )^{2} b^{2} c^{4} x^{4}-3 \mathit {atanh} \left (c x \right )^{2} b^{2}+6 \mathit {atanh} \left (c x \right ) a b \,c^{4} x^{4}-6 \mathit {atanh} \left (c x \right ) a b +2 \mathit {atanh} \left (c x \right ) b^{2} c^{3} x^{3}+6 \mathit {atanh} \left (c x \right ) b^{2} c x +8 \mathit {atanh} \left (c x \right ) b^{2}+8 \,\mathrm {log}\left (c^{2} x -c \right ) b^{2}+3 a^{2} c^{4} x^{4}+2 a b \,c^{3} x^{3}+6 a b c x +b^{2} c^{2} x^{2}}{12 c^{4}} \] Input:
int(x^3*(a+b*atanh(c*x))^2,x)
Output:
(3*atanh(c*x)**2*b**2*c**4*x**4 - 3*atanh(c*x)**2*b**2 + 6*atanh(c*x)*a*b* c**4*x**4 - 6*atanh(c*x)*a*b + 2*atanh(c*x)*b**2*c**3*x**3 + 6*atanh(c*x)* b**2*c*x + 8*atanh(c*x)*b**2 + 8*log(c**2*x - c)*b**2 + 3*a**2*c**4*x**4 + 2*a*b*c**3*x**3 + 6*a*b*c*x + b**2*c**2*x**2)/(12*c**4)