Integrand size = 16, antiderivative size = 125 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {a b x^3}{6 c^3}+\frac {b^2 x^6}{36 c^2}+\frac {b^2 x^3 \text {arctanh}\left (c x^3\right )}{6 c^3}+\frac {b x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{18 c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^6\right )}{9 c^4} \] Output:
1/6*a*b*x^3/c^3+1/36*b^2*x^6/c^2+1/6*b^2*x^3*arctanh(c*x^3)/c^3+1/18*b*x^9 *(a+b*arctanh(c*x^3))/c-1/12*(a+b*arctanh(c*x^3))^2/c^4+1/12*x^12*(a+b*arc tanh(c*x^3))^2+1/9*b^2*ln(-c^2*x^6+1)/c^4
Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.17 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {6 a b c x^3+b^2 c^2 x^6+2 a b c^3 x^9+3 a^2 c^4 x^{12}+2 b c x^3 \left (3 a c^3 x^9+b \left (3+c^2 x^6\right )\right ) \text {arctanh}\left (c x^3\right )+3 b^2 \left (-1+c^4 x^{12}\right ) \text {arctanh}\left (c x^3\right )^2+b (3 a+4 b) \log \left (1-c x^3\right )-3 a b \log \left (1+c x^3\right )+4 b^2 \log \left (1+c x^3\right )}{36 c^4} \] Input:
Integrate[x^11*(a + b*ArcTanh[c*x^3])^2,x]
Output:
(6*a*b*c*x^3 + b^2*c^2*x^6 + 2*a*b*c^3*x^9 + 3*a^2*c^4*x^12 + 2*b*c*x^3*(3 *a*c^3*x^9 + b*(3 + c^2*x^6))*ArcTanh[c*x^3] + 3*b^2*(-1 + c^4*x^12)*ArcTa nh[c*x^3]^2 + b*(3*a + 4*b)*Log[1 - c*x^3] - 3*a*b*Log[1 + c*x^3] + 4*b^2* Log[1 + c*x^3])/(36*c^4)
Time = 0.91 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6454, 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.
| \(\displaystyle \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle \frac {1}{3} \int x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2dx^3\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \int \frac {x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3\right )\) |
| \(\Big \downarrow \) 6542 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\int x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )dx^3}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{3} b c \int \frac {x^9}{1-c^2 x^6}dx^3}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \int \frac {x^6}{1-c^2 x^6}dx^6}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^6-1\right )}\right )dx^6}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \left (-\frac {x^6}{c^2}-\frac {\log \left (1-c^2 x^6\right )}{c^4}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6542 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c x^3\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c x^3\right )\right )dx^3}{c^2}}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \left (-\frac {x^6}{c^2}-\frac {\log \left (1-c^2 x^6\right )}{c^4}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c x^3\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {a x^3+b x^3 \text {arctanh}\left (c x^3\right )+\frac {b \log \left (1-c^2 x^6\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \left (-\frac {x^6}{c^2}-\frac {\log \left (1-c^2 x^6\right )}{c^4}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{4} x^{12} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^3}-\frac {a x^3+b x^3 \text {arctanh}\left (c x^3\right )+\frac {b \log \left (1-c^2 x^6\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{6} b c \left (-\frac {x^6}{c^2}-\frac {\log \left (1-c^2 x^6\right )}{c^4}\right )}{c^2}\right )\right )\) |
Input:
Int[x^11*(a + b*ArcTanh[c*x^3])^2,x]
Output:
((x^12*(a + b*ArcTanh[c*x^3])^2)/4 - (b*c*(-(((x^9*(a + b*ArcTanh[c*x^3])) /3 - (b*c*(-(x^6/c^2) - Log[1 - c^2*x^6]/c^4))/6)/c^2) + ((a + b*ArcTanh[c *x^3])^2/(2*b*c^3) - (a*x^3 + b*x^3*ArcTanh[c*x^3] + (b*Log[1 - c^2*x^6])/ (2*c))/c^2)/c^2))/2)/3
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_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x ], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl ify[(m + 1)/n]]
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 = 1.53 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.30
| method | result | size |
| parallelrisch | \(\frac {3 b^{2} \operatorname {arctanh}\left (c \,x^{3}\right )^{2} x^{12} c^{4}+6 a b \,\operatorname {arctanh}\left (c \,x^{3}\right ) x^{12} c^{4}+3 a^{2} c^{4} x^{12}+2 b^{2} \operatorname {arctanh}\left (c \,x^{3}\right ) x^{9} c^{3}+2 a b \,c^{3} x^{9}+b^{2} c^{2} x^{6}+6 b^{2} \operatorname {arctanh}\left (c \,x^{3}\right ) x^{3} c +6 a b c \,x^{3}-3 b^{2} \operatorname {arctanh}\left (c \,x^{3}\right )^{2}+8 \ln \left (c \,x^{3}-1\right ) b^{2}-6 \,\operatorname {arctanh}\left (c \,x^{3}\right ) a b +8 \,\operatorname {arctanh}\left (c \,x^{3}\right ) b^{2}+b^{2}}{36 c^{4}}\) | \(163\) |
| risch | \(\frac {b^{2} \left (x^{12} c^{4}-1\right ) \ln \left (c \,x^{3}+1\right )^{2}}{48 c^{4}}+\frac {b \left (-3 x^{12} b \ln \left (-c \,x^{3}+1\right ) c^{4}+6 a \,c^{4} x^{12}+2 b \,c^{3} x^{9}+6 b c \,x^{3}+3 b \ln \left (-c \,x^{3}+1\right )\right ) \ln \left (c \,x^{3}+1\right )}{72 c^{4}}+\frac {b^{2} x^{12} \ln \left (-c \,x^{3}+1\right )^{2}}{48}-\frac {a b \,x^{12} \ln \left (-c \,x^{3}+1\right )}{12}+\frac {a^{2} x^{12}}{12}-\frac {b^{2} x^{9} \ln \left (-c \,x^{3}+1\right )}{36 c}+\frac {a b \,x^{9}}{18 c}+\frac {b^{2} x^{6}}{36 c^{2}}-\frac {b^{2} x^{3} \ln \left (-c \,x^{3}+1\right )}{12 c^{3}}+\frac {a b \,x^{3}}{6 c^{3}}-\frac {b^{2} \ln \left (-c \,x^{3}+1\right )^{2}}{48 c^{4}}+\frac {b \ln \left (-c \,x^{3}+1\right ) a}{12 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{3}+1\right )}{9 c^{4}}-\frac {b \ln \left (-c \,x^{3}-1\right ) a}{12 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{3}-1\right )}{9 c^{4}}\) | \(298\) |
| default | \(\text {Expression too large to display}\) | \(2995\) |
| parts | \(\text {Expression too large to display}\) | \(2995\) |
Input:
int(x^11*(a+b*arctanh(c*x^3))^2,x,method=_RETURNVERBOSE)
Output:
1/36*(3*b^2*arctanh(c*x^3)^2*x^12*c^4+6*a*b*arctanh(c*x^3)*x^12*c^4+3*a^2* c^4*x^12+2*b^2*arctanh(c*x^3)*x^9*c^3+2*a*b*c^3*x^9+b^2*c^2*x^6+6*b^2*arct anh(c*x^3)*x^3*c+6*a*b*c*x^3-3*b^2*arctanh(c*x^3)^2+8*ln(c*x^3-1)*b^2-6*ar ctanh(c*x^3)*a*b+8*arctanh(c*x^3)*b^2+b^2)/c^4
Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.41 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {12 \, a^{2} c^{4} x^{12} + 8 \, a b c^{3} x^{9} + 4 \, b^{2} c^{2} x^{6} + 24 \, a b c x^{3} + 3 \, {\left (b^{2} c^{4} x^{12} - b^{2}\right )} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )^{2} - 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{3} + 1\right ) + 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{3} - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{12} + b^{2} c^{3} x^{9} + 3 \, b^{2} c x^{3}\right )} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{144 \, c^{4}} \] Input:
integrate(x^11*(a+b*arctanh(c*x^3))^2,x, algorithm="fricas")
Output:
1/144*(12*a^2*c^4*x^12 + 8*a*b*c^3*x^9 + 4*b^2*c^2*x^6 + 24*a*b*c*x^3 + 3* (b^2*c^4*x^12 - b^2)*log(-(c*x^3 + 1)/(c*x^3 - 1))^2 - 4*(3*a*b - 4*b^2)*l og(c*x^3 + 1) + 4*(3*a*b + 4*b^2)*log(c*x^3 - 1) + 4*(3*a*b*c^4*x^12 + b^2 *c^3*x^9 + 3*b^2*c*x^3)*log(-(c*x^3 + 1)/(c*x^3 - 1)))/c^4
Timed out. \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\text {Timed out} \] Input:
integrate(x**11*(a+b*atanh(c*x**3))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.74 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {1}{12} \, b^{2} x^{12} \operatorname {artanh}\left (c x^{3}\right )^{2} + \frac {1}{12} \, a^{2} x^{12} + \frac {1}{36} \, {\left (6 \, x^{12} \operatorname {artanh}\left (c x^{3}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{9} + 3 \, x^{3}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{3} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{3} - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{144} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{9} + 3 \, x^{3}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{3} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{3} - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c x^{3}\right ) + \frac {4 \, c^{2} x^{6} - 2 \, {\left (3 \, \log \left (c x^{3} - 1\right ) - 8\right )} \log \left (c x^{3} + 1\right ) + 3 \, \log \left (c x^{3} + 1\right )^{2} + 3 \, \log \left (c x^{3} - 1\right )^{2} + 16 \, \log \left (c x^{3} - 1\right )}{c^{4}}\right )} b^{2} \] Input:
integrate(x^11*(a+b*arctanh(c*x^3))^2,x, algorithm="maxima")
Output:
1/12*b^2*x^12*arctanh(c*x^3)^2 + 1/12*a^2*x^12 + 1/36*(6*x^12*arctanh(c*x^ 3) + c*(2*(c^2*x^9 + 3*x^3)/c^4 - 3*log(c*x^3 + 1)/c^5 + 3*log(c*x^3 - 1)/ c^5))*a*b + 1/144*(4*c*(2*(c^2*x^9 + 3*x^3)/c^4 - 3*log(c*x^3 + 1)/c^5 + 3 *log(c*x^3 - 1)/c^5)*arctanh(c*x^3) + (4*c^2*x^6 - 2*(3*log(c*x^3 - 1) - 8 )*log(c*x^3 + 1) + 3*log(c*x^3 + 1)^2 + 3*log(c*x^3 - 1)^2 + 16*log(c*x^3 - 1))/c^4)*b^2
Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {1}{12} \, a^{2} x^{12} + \frac {a b x^{9}}{18 \, c} + \frac {b^{2} x^{6}}{36 \, c^{2}} + \frac {1}{48} \, {\left (b^{2} x^{12} - \frac {b^{2}}{c^{4}}\right )} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )^{2} + \frac {a b x^{3}}{6 \, c^{3}} + \frac {1}{36} \, {\left (3 \, a b x^{12} + \frac {b^{2} x^{9}}{c} + \frac {3 \, b^{2} x^{3}}{c^{3}}\right )} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) - \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{3} + 1\right )}{36 \, c^{4}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{3} - 1\right )}{36 \, c^{4}} \] Input:
integrate(x^11*(a+b*arctanh(c*x^3))^2,x, algorithm="giac")
Output:
1/12*a^2*x^12 + 1/18*a*b*x^9/c + 1/36*b^2*x^6/c^2 + 1/48*(b^2*x^12 - b^2/c ^4)*log(-(c*x^3 + 1)/(c*x^3 - 1))^2 + 1/6*a*b*x^3/c^3 + 1/36*(3*a*b*x^12 + b^2*x^9/c + 3*b^2*x^3/c^3)*log(-(c*x^3 + 1)/(c*x^3 - 1)) - 1/36*(3*a*b - 4*b^2)*log(c*x^3 + 1)/c^4 + 1/36*(3*a*b + 4*b^2)*log(c*x^3 - 1)/c^4
Time = 4.34 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.68 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {a^2\,x^{12}}{12}+\frac {b^2\,\ln \left (c\,x^3-1\right )}{9\,c^4}+\frac {b^2\,\ln \left (c\,x^3+1\right )}{9\,c^4}-\frac {b^2\,{\ln \left (c\,x^3+1\right )}^2}{48\,c^4}-\frac {b^2\,{\ln \left (1-c\,x^3\right )}^2}{48\,c^4}+\frac {b^2\,x^6}{36\,c^2}+\frac {b^2\,x^{12}\,{\ln \left (c\,x^3+1\right )}^2}{48}+\frac {b^2\,x^{12}\,{\ln \left (1-c\,x^3\right )}^2}{48}+\frac {b^2\,x^3\,\ln \left (c\,x^3+1\right )}{12\,c^3}-\frac {b^2\,x^3\,\ln \left (1-c\,x^3\right )}{12\,c^3}+\frac {b^2\,x^9\,\ln \left (c\,x^3+1\right )}{36\,c}-\frac {b^2\,x^9\,\ln \left (1-c\,x^3\right )}{36\,c}+\frac {a\,b\,\ln \left (c\,x^3-1\right )}{12\,c^4}-\frac {a\,b\,\ln \left (c\,x^3+1\right )}{12\,c^4}+\frac {a\,b\,x^{12}\,\ln \left (c\,x^3+1\right )}{12}-\frac {a\,b\,x^{12}\,\ln \left (1-c\,x^3\right )}{12}+\frac {b^2\,\ln \left (c\,x^3+1\right )\,\ln \left (1-c\,x^3\right )}{24\,c^4}+\frac {a\,b\,x^3}{6\,c^3}+\frac {a\,b\,x^9}{18\,c}-\frac {b^2\,x^{12}\,\ln \left (c\,x^3+1\right )\,\ln \left (1-c\,x^3\right )}{24} \] Input:
int(x^11*(a + b*atanh(c*x^3))^2,x)
Output:
(a^2*x^12)/12 + (b^2*log(c*x^3 - 1))/(9*c^4) + (b^2*log(c*x^3 + 1))/(9*c^4 ) - (b^2*log(c*x^3 + 1)^2)/(48*c^4) - (b^2*log(1 - c*x^3)^2)/(48*c^4) + (b ^2*x^6)/(36*c^2) + (b^2*x^12*log(c*x^3 + 1)^2)/48 + (b^2*x^12*log(1 - c*x^ 3)^2)/48 + (b^2*x^3*log(c*x^3 + 1))/(12*c^3) - (b^2*x^3*log(1 - c*x^3))/(1 2*c^3) + (b^2*x^9*log(c*x^3 + 1))/(36*c) - (b^2*x^9*log(1 - c*x^3))/(36*c) + (a*b*log(c*x^3 - 1))/(12*c^4) - (a*b*log(c*x^3 + 1))/(12*c^4) + (a*b*x^ 12*log(c*x^3 + 1))/12 - (a*b*x^12*log(1 - c*x^3))/12 + (b^2*log(c*x^3 + 1) *log(1 - c*x^3))/(24*c^4) + (a*b*x^3)/(6*c^3) + (a*b*x^9)/(18*c) - (b^2*x^ 12*log(c*x^3 + 1)*log(1 - c*x^3))/24
Time = 0.20 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.46 \[ \int x^{11} \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {3 \mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2} c^{4} x^{12}-3 \mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2}+6 \mathit {atanh} \left (c \,x^{3}\right ) a b \,c^{4} x^{12}-6 \mathit {atanh} \left (c \,x^{3}\right ) a b +2 \mathit {atanh} \left (c \,x^{3}\right ) b^{2} c^{3} x^{9}+6 \mathit {atanh} \left (c \,x^{3}\right ) b^{2} c \,x^{3}-8 \mathit {atanh} \left (c \,x^{3}\right ) b^{2}+8 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) b^{2}+8 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) b^{2}+3 a^{2} c^{4} x^{12}+2 a b \,c^{3} x^{9}+6 a b c \,x^{3}+b^{2} c^{2} x^{6}}{36 c^{4}} \] Input:
int(x^11*(a+b*atanh(c*x^3))^2,x)
Output:
(3*atanh(c*x**3)**2*b**2*c**4*x**12 - 3*atanh(c*x**3)**2*b**2 + 6*atanh(c* x**3)*a*b*c**4*x**12 - 6*atanh(c*x**3)*a*b + 2*atanh(c*x**3)*b**2*c**3*x** 9 + 6*atanh(c*x**3)*b**2*c*x**3 - 8*atanh(c*x**3)*b**2 + 8*log(c**(2/3)*x* *2 - c**(1/3)*x + 1)*b**2 + 8*log(c**(2/3)*x + c**(1/3))*b**2 + 3*a**2*c** 4*x**12 + 2*a*b*c**3*x**9 + 6*a*b*c*x**3 + b**2*c**2*x**6)/(36*c**4)