Integrand size = 18, antiderivative size = 173 \[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {2 a b \sqrt {x}}{3 c^5}+\frac {8 b^2 x}{45 c^4}+\frac {b^2 x^2}{30 c^2}+\frac {2 b^2 \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )}{3 c^5}+\frac {2 b x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{9 c^3}+\frac {2 b x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{15 c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{3 c^6}+\frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2+\frac {23 b^2 \log \left (1-c^2 x\right )}{45 c^6} \]
8/45*b^2*x/c^4+1/30*b^2*x^2/c^2+2/9*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))/c^3 +2/15*b*x^(5/2)*(a+b*arctanh(c*x^(1/2)))/c-1/3*(a+b*arctanh(c*x^(1/2)))^2/ c^6+1/3*x^3*(a+b*arctanh(c*x^(1/2)))^2+23/45*b^2*ln(-c^2*x+1)/c^6+2/3*a*b* x^(1/2)/c^5+2/3*b^2*arctanh(c*x^(1/2))*x^(1/2)/c^5
Time = 0.08 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.12 \[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {60 a b c \sqrt {x}+16 b^2 c^2 x+20 a b c^3 x^{3/2}+3 b^2 c^4 x^2+12 a b c^5 x^{5/2}+30 a^2 c^6 x^3+4 b c \sqrt {x} \left (15 a c^5 x^{5/2}+b \left (15+5 c^2 x+3 c^4 x^2\right )\right ) \text {arctanh}\left (c \sqrt {x}\right )+30 b^2 \left (-1+c^6 x^3\right ) \text {arctanh}\left (c \sqrt {x}\right )^2+2 b (15 a+23 b) \log \left (1-c \sqrt {x}\right )-30 a b \log \left (1+c \sqrt {x}\right )+46 b^2 \log \left (1+c \sqrt {x}\right )}{90 c^6} \]
(60*a*b*c*Sqrt[x] + 16*b^2*c^2*x + 20*a*b*c^3*x^(3/2) + 3*b^2*c^4*x^2 + 12 *a*b*c^5*x^(5/2) + 30*a^2*c^6*x^3 + 4*b*c*Sqrt[x]*(15*a*c^5*x^(5/2) + b*(1 5 + 5*c^2*x + 3*c^4*x^2))*ArcTanh[c*Sqrt[x]] + 30*b^2*(-1 + c^6*x^3)*ArcTa nh[c*Sqrt[x]]^2 + 2*b*(15*a + 23*b)*Log[1 - c*Sqrt[x]] - 30*a*b*Log[1 + c* Sqrt[x]] + 46*b^2*Log[1 + c*Sqrt[x]])/(90*c^6)
Time = 1.43 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.31, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6454, 6452, 6542, 6452, 243, 49, 2009, 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.
\(\displaystyle \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle 2 \int x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{5} b c \int \frac {x^{5/2}}{1-c^2 x}d\sqrt {x}}{c^2}\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \int \frac {x}{1-c^2 x}dx}{c^2}\right )\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \int \left (-\frac {x}{c^2}-\frac {1}{c^4 \left (c^2 x-1\right )}-\frac {1}{c^4}\right )dx}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{3} b c \int \frac {x^{3/2}}{1-c^2 x}d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \int \frac {x}{1-c^2 x}dx}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x-1\right )}\right )dx}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\frac {\int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\frac {\int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {a \sqrt {x}+b \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )+\frac {b \log \left (1-c^2 x\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle 2 \left (\frac {1}{6} x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{3} b c \left (\frac {\frac {\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^3}-\frac {a \sqrt {x}+b \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )+\frac {b \log \left (1-c^2 x\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}\right )\right )\) |
2*((x^3*(a + b*ArcTanh[c*Sqrt[x]])^2)/6 - (b*c*(-(((x^(5/2)*(a + b*ArcTanh [c*Sqrt[x]]))/5 - (b*c*(-(x/c^4) - x/(2*c^2) - Log[1 - c^2*x]/c^6))/10)/c^ 2) + (-(((x^(3/2)*(a + b*ArcTanh[c*Sqrt[x]]))/3 - (b*c*(-(x/c^2) - Log[1 - c^2*x]/c^4))/6)/c^2) + ((a + b*ArcTanh[c*Sqrt[x]])^2/(2*b*c^3) - (a*Sqrt[ x] + b*Sqrt[x]*ArcTanh[c*Sqrt[x]] + (b*Log[1 - c^2*x])/(2*c))/c^2)/c^2)/c^ 2))/3)
3.2.96.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(278\) vs. \(2(137)=274\).
Time = 7.55 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.61
method | result | size |
parts | \(\frac {a^{2} x^{3}}{3}+\frac {2 b^{2} \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{6}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{15}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{9}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{3}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{6}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{6}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{24}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{24}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {c^{4} x^{2}}{60}+\frac {4 c^{2} x}{45}+\frac {23 \ln \left (c \sqrt {x}-1\right )}{90}+\frac {23 \ln \left (1+c \sqrt {x}\right )}{90}\right )}{c^{6}}+\frac {4 a b \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )}{6}+\frac {c^{5} x^{\frac {5}{2}}}{30}+\frac {c^{3} x^{\frac {3}{2}}}{18}+\frac {c \sqrt {x}}{6}+\frac {\ln \left (c \sqrt {x}-1\right )}{12}-\frac {\ln \left (1+c \sqrt {x}\right )}{12}\right )}{c^{6}}\) | \(279\) |
derivativedivides | \(\frac {\frac {a^{2} c^{6} x^{3}}{3}+2 b^{2} \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{6}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{15}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{9}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{3}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{6}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{6}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{24}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{24}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {c^{4} x^{2}}{60}+\frac {4 c^{2} x}{45}+\frac {23 \ln \left (c \sqrt {x}-1\right )}{90}+\frac {23 \ln \left (1+c \sqrt {x}\right )}{90}\right )+4 a b \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )}{6}+\frac {c^{5} x^{\frac {5}{2}}}{30}+\frac {c^{3} x^{\frac {3}{2}}}{18}+\frac {c \sqrt {x}}{6}+\frac {\ln \left (c \sqrt {x}-1\right )}{12}-\frac {\ln \left (1+c \sqrt {x}\right )}{12}\right )}{c^{6}}\) | \(280\) |
default | \(\frac {\frac {a^{2} c^{6} x^{3}}{3}+2 b^{2} \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{6}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{15}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{9}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{3}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{6}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{6}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{24}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{24}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{12}+\frac {c^{4} x^{2}}{60}+\frac {4 c^{2} x}{45}+\frac {23 \ln \left (c \sqrt {x}-1\right )}{90}+\frac {23 \ln \left (1+c \sqrt {x}\right )}{90}\right )+4 a b \left (\frac {c^{6} x^{3} \operatorname {arctanh}\left (c \sqrt {x}\right )}{6}+\frac {c^{5} x^{\frac {5}{2}}}{30}+\frac {c^{3} x^{\frac {3}{2}}}{18}+\frac {c \sqrt {x}}{6}+\frac {\ln \left (c \sqrt {x}-1\right )}{12}-\frac {\ln \left (1+c \sqrt {x}\right )}{12}\right )}{c^{6}}\) | \(280\) |
1/3*a^2*x^3+2*b^2/c^6*(1/6*c^6*x^3*arctanh(c*x^(1/2))^2+1/15*arctanh(c*x^( 1/2))*c^5*x^(5/2)+1/9*arctanh(c*x^(1/2))*c^3*x^(3/2)+1/3*arctanh(c*x^(1/2) )*c*x^(1/2)+1/6*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)-1/6*arctanh(c*x^(1/2))* ln(1+c*x^(1/2))-1/12*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)+1/24*ln(c*x^(1/ 2)-1)^2+1/24*ln(1+c*x^(1/2))^2-1/12*(ln(1+c*x^(1/2))-ln(1/2*c*x^(1/2)+1/2) )*ln(-1/2*c*x^(1/2)+1/2)+1/60*c^4*x^2+4/45*c^2*x+23/90*ln(c*x^(1/2)-1)+23/ 90*ln(1+c*x^(1/2)))+4*a*b/c^6*(1/6*c^6*x^3*arctanh(c*x^(1/2))+1/30*c^5*x^( 5/2)+1/18*c^3*x^(3/2)+1/6*c*x^(1/2)+1/12*ln(c*x^(1/2)-1)-1/12*ln(1+c*x^(1/ 2)))
Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.39 \[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {60 \, a^{2} c^{6} x^{3} + 6 \, b^{2} c^{4} x^{2} + 32 \, b^{2} c^{2} x + 15 \, {\left (b^{2} c^{6} x^{3} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (15 \, a b c^{6} - 15 \, a b + 23 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (15 \, a b c^{6} - 15 \, a b - 23 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (15 \, a b c^{6} x^{3} - 15 \, a b c^{6} + {\left (3 \, b^{2} c^{5} x^{2} + 5 \, b^{2} c^{3} x + 15 \, b^{2} c\right )} \sqrt {x}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 8 \, {\left (3 \, a b c^{5} x^{2} + 5 \, a b c^{3} x + 15 \, a b c\right )} \sqrt {x}}{180 \, c^{6}} \]
1/180*(60*a^2*c^6*x^3 + 6*b^2*c^4*x^2 + 32*b^2*c^2*x + 15*(b^2*c^6*x^3 - b ^2)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1))^2 + 4*(15*a*b*c^6 - 15*a*b + 23*b^2)*log(c*sqrt(x) + 1) - 4*(15*a*b*c^6 - 15*a*b - 23*b^2)*log(c*sqr t(x) - 1) + 4*(15*a*b*c^6*x^3 - 15*a*b*c^6 + (3*b^2*c^5*x^2 + 5*b^2*c^3*x + 15*b^2*c)*sqrt(x))*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(3*a* b*c^5*x^2 + 5*a*b*c^3*x + 15*a*b*c)*sqrt(x))/c^6
\[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\int x^{2} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{2}\, dx \]
Time = 0.20 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.39 \[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {1}{3} \, b^{2} x^{3} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{3} \, a^{2} x^{3} + \frac {1}{45} \, {\left (30 \, x^{3} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )}\right )} a b + \frac {1}{180} \, {\left (4 \, c {\left (\frac {2 \, {\left (3 \, c^{4} x^{\frac {5}{2}} + 5 \, c^{2} x^{\frac {3}{2}} + 15 \, \sqrt {x}\right )}}{c^{6}} - \frac {15 \, \log \left (c \sqrt {x} + 1\right )}{c^{7}} + \frac {15 \, \log \left (c \sqrt {x} - 1\right )}{c^{7}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {6 \, c^{4} x^{2} + 32 \, c^{2} x - 2 \, {\left (15 \, \log \left (c \sqrt {x} - 1\right ) - 46\right )} \log \left (c \sqrt {x} + 1\right ) + 15 \, \log \left (c \sqrt {x} + 1\right )^{2} + 15 \, \log \left (c \sqrt {x} - 1\right )^{2} + 92 \, \log \left (c \sqrt {x} - 1\right )}{c^{6}}\right )} b^{2} \]
1/3*b^2*x^3*arctanh(c*sqrt(x))^2 + 1/3*a^2*x^3 + 1/45*(30*x^3*arctanh(c*sq rt(x)) + c*(2*(3*c^4*x^(5/2) + 5*c^2*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c* sqrt(x) + 1)/c^7 + 15*log(c*sqrt(x) - 1)/c^7))*a*b + 1/180*(4*c*(2*(3*c^4* x^(5/2) + 5*c^2*x^(3/2) + 15*sqrt(x))/c^6 - 15*log(c*sqrt(x) + 1)/c^7 + 15 *log(c*sqrt(x) - 1)/c^7)*arctanh(c*sqrt(x)) + (6*c^4*x^2 + 32*c^2*x - 2*(1 5*log(c*sqrt(x) - 1) - 46)*log(c*sqrt(x) + 1) + 15*log(c*sqrt(x) + 1)^2 + 15*log(c*sqrt(x) - 1)^2 + 92*log(c*sqrt(x) - 1))/c^6)*b^2
\[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )}^{2} x^{2} \,d x } \]
Time = 3.91 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.07 \[ \int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {46\,b^2\,\ln \left (c^2\,x-1\right )-30\,b^2\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2-60\,a\,b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+16\,b^2\,c^2\,x+30\,a^2\,c^6\,x^3+3\,b^2\,c^4\,x^2+30\,b^2\,c^6\,x^3\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2+60\,b^2\,c\,\sqrt {x}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+60\,a\,b\,c\,\sqrt {x}+20\,b^2\,c^3\,x^{3/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+12\,b^2\,c^5\,x^{5/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+20\,a\,b\,c^3\,x^{3/2}+12\,a\,b\,c^5\,x^{5/2}+60\,a\,b\,c^6\,x^3\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{90\,c^6} \]
(46*b^2*log(c^2*x - 1) - 30*b^2*atanh(c*x^(1/2))^2 - 60*a*b*atanh(c*x^(1/2 )) + 16*b^2*c^2*x + 30*a^2*c^6*x^3 + 3*b^2*c^4*x^2 + 30*b^2*c^6*x^3*atanh( c*x^(1/2))^2 + 60*b^2*c*x^(1/2)*atanh(c*x^(1/2)) + 60*a*b*c*x^(1/2) + 20*b ^2*c^3*x^(3/2)*atanh(c*x^(1/2)) + 12*b^2*c^5*x^(5/2)*atanh(c*x^(1/2)) + 20 *a*b*c^3*x^(3/2) + 12*a*b*c^5*x^(5/2) + 60*a*b*c^6*x^3*atanh(c*x^(1/2)))/( 90*c^6)