Integrand size = 16, antiderivative size = 141 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\frac {3 b \left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 c^2}+\frac {3 b x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \log \left (\frac {2}{1-c x^2}\right )}{2 c^2}-\frac {3 b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right )}{4 c^2} \]
3/4*b*(a+b*arctanh(c*x^2))^2/c^2+3/4*b*x^2*(a+b*arctanh(c*x^2))^2/c-1/4*(a +b*arctanh(c*x^2))^3/c^2+1/4*x^4*(a+b*arctanh(c*x^2))^3-3/2*b^2*(a+b*arcta nh(c*x^2))*ln(2/(-c*x^2+1))/c^2-3/4*b^3*polylog(2,1-2/(-c*x^2+1))/c^2
Time = 0.36 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.31 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\frac {6 b^2 \left (-1+c x^2\right ) \left (a+b+a c x^2\right ) \text {arctanh}\left (c x^2\right )^2+2 b^3 \left (-1+c^2 x^4\right ) \text {arctanh}\left (c x^2\right )^3+6 b \text {arctanh}\left (c x^2\right ) \left (a c x^2 \left (2 b+a c x^2\right )-2 b^2 \log \left (1+e^{-2 \text {arctanh}\left (c x^2\right )}\right )\right )+a \left (6 a b c x^2+2 a^2 c^2 x^4+3 a b \log \left (1-c x^2\right )-3 a b \log \left (1+c x^2\right )+6 b^2 \log \left (1-c^2 x^4\right )\right )+6 b^3 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c x^2\right )}\right )}{8 c^2} \]
(6*b^2*(-1 + c*x^2)*(a + b + a*c*x^2)*ArcTanh[c*x^2]^2 + 2*b^3*(-1 + c^2*x ^4)*ArcTanh[c*x^2]^3 + 6*b*ArcTanh[c*x^2]*(a*c*x^2*(2*b + a*c*x^2) - 2*b^2 *Log[1 + E^(-2*ArcTanh[c*x^2])]) + a*(6*a*b*c*x^2 + 2*a^2*c^2*x^4 + 3*a*b* Log[1 - c*x^2] - 3*a*b*Log[1 + c*x^2] + 6*b^2*Log[1 - c^2*x^4]) + 6*b^3*Po lyLog[2, -E^(-2*ArcTanh[c*x^2])])/(8*c^2)
Time = 1.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6454, 6452, 6542, 6436, 6510, 6546, 6470, 2849, 2752}
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.
| \(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle \frac {1}{2} \int x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3dx^2\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{1-c^2 x^4}dx^2\right )\) |
| \(\Big \downarrow \) 6542 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{1-c^2 x^4}dx^2}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c x^2\right )\right )^2dx^2}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6436 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{1-c^2 x^4}dx^2}{c^2}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \int \frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b c^3}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \int \frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6546 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b c^3}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \left (\frac {\int \frac {a+b \text {arctanh}\left (c x^2\right )}{1-c x^2}dx^2}{c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 6470 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b c^3}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x^2}\right ) \left (a+b \text {arctanh}\left (c x^2\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{1-c x^2}\right )}{1-c^2 x^4}dx^2}{c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b c^3}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \left (\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c x^2}\right )}{1-\frac {2}{1-c x^2}}d\frac {1}{1-c x^2}}{c}+\frac {\log \left (\frac {2}{1-c x^2}\right ) \left (a+b \text {arctanh}\left (c x^2\right )\right )}{c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b c^3}-\frac {x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x^2}\right ) \left (a+b \text {arctanh}\left (c x^2\right )\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right )}{2 c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
((x^4*(a + b*ArcTanh[c*x^2])^3)/2 - (3*b*c*((a + b*ArcTanh[c*x^2])^3/(3*b* c^3) - (x^2*(a + b*ArcTanh[c*x^2])^2 - 2*b*c*(-1/2*(a + b*ArcTanh[c*x^2])^ 2/(b*c^2) + (((a + b*ArcTanh[c*x^2])*Log[2/(1 - c*x^2)])/c + (b*PolyLog[2, 1 - 2/(1 - c*x^2)])/(2*c))/c))/c^2))/2)/2
3.1.77.3.1 Defintions of rubi rules used
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Simp[b*c*n*p Int[x^n*((a + b*ArcTanh[c*x^n]) ^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && (EqQ[n, 1] || EqQ[p, 1])
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_)), x_Symbol ] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c *(p/e) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^2*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2 , 0]
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]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*e*(p + 1)), x] + Simp[1/ (c*d) Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.35 (sec) , antiderivative size = 798, normalized size of antiderivative = 5.66
1/32*b^3*(c^2*x^4-1)/c^2*ln(c*x^2+1)^3+3/32*b^2*(-b*c^2*ln(-c*x^2+1)*x^4+2 *a*c^2*x^4+2*b*c*x^2+b*ln(-c*x^2+1)-2*a+2*b)/c^2*ln(c*x^2+1)^2+(3/32*b^3*( c^2*x^4-1)/c^2*ln(-c*x^2+1)^2-3/32*b^2*(2*a*c*x^2+b)^2/c^2/a*ln(-c*x^2+1)- 3/32*b*(-4*a^3*c^2*x^4-8*a^2*b*c*x^2-4*ln(-c*x^2+1)*a^2*b-4*ln(-c*x^2+1)*a *b^2-ln(-c*x^2+1)*b^3-4*a*b^2)/a/c^2)*ln(c*x^2+1)-3/8*b/c^2*ln(c*x^2+1)*a^ 2+3/4*b^2/c^2*ln(c*x^2+1)*a+3/16/c^2*b^2*a*ln(c*x^2-1)+3/8/c^2*b^3*ln(-c*x ^2+1)-3/4/c*a*b^2*x^2*ln(-c*x^2+1)+1/4*a^3*x^4-3/8/c^2*b^3*ln(c*x^2-1)-3/8 /c^2*b^3*ln(c*x^2+1)-1/32*b^3*x^4*ln(-c*x^2+1)^3-3/16*b^3/c^2*ln(-c*x^2+1) ^2+1/32*b^3/c^2*ln(-c*x^2+1)^3-3/16*b^3/c^2+3/16/c*b^3*x^2*ln(-c*x^2+1)^2- 3/8*a^2*b*x^4*ln(-c*x^2+1)+3/8*a^2*b/c^2*ln(c*x^2-1)+3/16*a*b^2*x^4*ln(-c* x^2+1)^2+9/16/c^2*a*b^2*ln(-c*x^2+1)-3/16/c^2*a*b^2*ln(-c*x^2+1)^2+3/4/c*a ^2*b*x^2+3/4/c*b^2*Sum(-(ln(x-_alpha)*ln(-c*x^2+1)+2*c*(-1/2*ln(x-_alpha)* (ln((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_ alpha*c-2,index=1))+ln((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)-x+_alpha)/R ootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)))/c-1/2*(dilog((RootOf(_Z^2*c+2*_Z*_ alpha*c-2,index=1)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=1))+dilog ((RootOf(_Z^2*c+2*_Z*_alpha*c-2,index=2)-x+_alpha)/RootOf(_Z^2*c+2*_Z*_alp ha*c-2,index=2)))/c))*b/c,_alpha=RootOf(_Z^2*c+1))
\[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]
integral(b^3*x^3*arctanh(c*x^2)^3 + 3*a*b^2*x^3*arctanh(c*x^2)^2 + 3*a^2*b *x^3*arctanh(c*x^2) + a^3*x^3, x)
\[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\int x^{3} \left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}\, dx \]
\[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]
3/4*a*b^2*x^4*arctanh(c*x^2)^2 + 1/4*a^3*x^4 + 3/8*(2*x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3))*a^2*b + 3/16*(4* c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3)*arctanh(c*x^2) - ( 2*(log(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^ 2 - 4*log(c*x^2 - 1))/c^2)*a*b^2 - 1/128*(4*x^4*log(-c*x^2 + 1)^3 + 3*c^3* (x^4/c^3 + log(c^2*x^4 - 1)/c^5) - 6*c*((c*x^4 + 2*x^2)/c^2 + 2*log(c*x^2 - 1)/c^3)*log(-c*x^2 + 1)^2 + 21*c^2*(2*x^2/c^3 - log(c*x^2 + 1)/c^4 + log (c*x^2 - 1)/c^4) + c*(6*(c^2*x^4 + 6*c*x^2 + 2*log(c*x^2 - 1)^2 + 6*log(c* x^2 - 1))*log(-c*x^2 + 1)/c^3 - (3*c^2*x^4 + 42*c*x^2 + 4*log(c*x^2 - 1)^3 + 18*log(c*x^2 - 1)^2 + 42*log(c*x^2 - 1))/c^3) - 1152*c*integrate(1/4*x^ 3*log(c*x^2 + 1)/(c^3*x^4 - c), x) - 2*(12*c*x^2*log(c*x^2 + 1)^2 + 2*(c^2 *x^4 - 1)*log(c*x^2 + 1)^3 - 3*(c^2*x^4 - 2*c*x^2 - 2*(c^2*x^4 - 1)*log(c* x^2 + 1) + 1)*log(-c*x^2 + 1)^2 + 3*(c^2*x^4 + 6*c*x^2 - 2*(c^2*x^4 - 1)*l og(c*x^2 + 1)^2 - 8*(c*x^2 + 1)*log(c*x^2 + 1))*log(-c*x^2 + 1))/c^2 + 18* log(4*c^3*x^4 - 4*c)/c^2 - 384*integrate(1/4*x*log(c*x^2 + 1)/(c^3*x^4 - c ), x))*b^3
\[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]
Timed out. \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3 \, dx=\int x^3\,{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3 \,d x \]