Integrand size = 16, antiderivative size = 139 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\frac {3}{4} b c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {3 b c \left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 x^2}+\frac {1}{4} c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^3-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{4 x^4}+\frac {3}{2} b^2 c^2 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \log \left (2-\frac {2}{1+c x^2}\right )-\frac {3}{4} b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right ) \]
3/4*b*c^2*(a+b*arctanh(c*x^2))^2-3/4*b*c*(a+b*arctanh(c*x^2))^2/x^2+1/4*c^ 2*(a+b*arctanh(c*x^2))^3-1/4*(a+b*arctanh(c*x^2))^3/x^4+3/2*b^2*c^2*(a+b*a rctanh(c*x^2))*ln(2-2/(c*x^2+1))-3/4*b^3*c^2*polylog(2,-1+2/(c*x^2+1))
Time = 0.23 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\frac {6 b^2 \left (-1+c x^2\right ) \left (a+a c x^2+b c x^2\right ) \text {arctanh}\left (c x^2\right )^2+2 b^3 \left (-1+c^2 x^4\right ) \text {arctanh}\left (c x^2\right )^3-6 b \text {arctanh}\left (c x^2\right ) \left (a^2+2 a b c x^2-2 b^2 c^2 x^4 \log \left (1-e^{-2 \text {arctanh}\left (c x^2\right )}\right )\right )+a \left (-2 a^2-6 a b c x^2-3 a b c^2 x^4 \log \left (1-c x^2\right )+3 a b c^2 x^4 \log \left (1+c x^2\right )+12 b^2 c^2 x^4 \log \left (\frac {c x^2}{\sqrt {1-c^2 x^4}}\right )\right )-6 b^3 c^2 x^4 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^2\right )}\right )}{8 x^4} \]
(6*b^2*(-1 + c*x^2)*(a + a*c*x^2 + b*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^2 + 2*a*b*c*x^2 - 2*b^2 *c^2*x^4*Log[1 - E^(-2*ArcTanh[c*x^2])]) + a*(-2*a^2 - 6*a*b*c*x^2 - 3*a*b *c^2*x^4*Log[1 - c*x^2] + 3*a*b*c^2*x^4*Log[1 + c*x^2] + 12*b^2*c^2*x^4*Lo g[(c*x^2)/Sqrt[1 - c^2*x^4]]) - 6*b^3*c^2*x^4*PolyLog[2, E^(-2*ArcTanh[c*x ^2])])/(8*x^4)
Time = 1.06 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6454, 6452, 6544, 6452, 6510, 6550, 6494, 2897}
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.
| \(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^6}dx^2\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^4 \left (1-c^2 x^4\right )}dx^2-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 6544 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{1-c^2 x^4}dx^2+\int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^4}dx^2\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{1-c^2 x^4}dx^2+2 b c \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^2 \left (1-c^2 x^4\right )}dx^2-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (2 b c \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^2 \left (1-c^2 x^4\right )}dx^2+\frac {c \left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 6550 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (2 b c \left (\int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^2 \left (c x^2+1\right )}dx^2+\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b}\right )+\frac {c \left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 6494 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{c x^2+1}\right )}{1-c^2 x^4}dx^2+\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \text {arctanh}\left (c x^2\right )\right )\right )+\frac {c \left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (2 b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x^2+1}-1\right )\right )+\frac {c \left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{2 x^4}\right )\) |
(-1/2*(a + b*ArcTanh[c*x^2])^3/x^4 + (3*b*c*(-((a + b*ArcTanh[c*x^2])^2/x^ 2) + (c*(a + b*ArcTanh[c*x^2])^3)/(3*b) + 2*b*c*((a + b*ArcTanh[c*x^2])^2/ (2*b) + (a + b*ArcTanh[c*x^2])*Log[2 - 2/(1 + c*x^2)] - (b*PolyLog[2, -1 + 2/(1 + c*x^2)])/2)))/2)/2
3.1.81.3.1 Defintions of rubi rules used
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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_.)/((x_)*((d_) + (e_.)*(x_))), x _Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Simp[b*c*(p/d) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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[1/d Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x ], x] - Simp[e/(d*f^2) 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] && LtQ[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*d*(p + 1)), x] + Simp[1/ d Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{2}\right )\right )}^{3}}{x^{5}}d x\]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
integral((b^3*arctanh(c*x^2)^3 + 3*a*b^2*arctanh(c*x^2)^2 + 3*a^2*b*arctan h(c*x^2) + a^3)/x^5, x)
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{5}}\, dx \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
3/8*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^ 4)*a^2*b + 3/16*((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) + 16*log(x))*c^2 + 4*(c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c*arctanh(c*x^2))*a*b^2 - 1/32*b^3*(((c^2* x^4 - 1)*log(-c*x^2 + 1)^3 + 3*(2*c*x^2 - (c^2*x^4 - 1)*log(c*x^2 + 1))*lo g(-c*x^2 + 1)^2)/x^4 + 4*integrate(-((c*x^2 - 1)*log(c*x^2 + 1)^3 + 3*(2*c ^2*x^4 - (c*x^2 - 1)*log(c*x^2 + 1)^2 - (c^3*x^6 - c*x^2)*log(c*x^2 + 1))* log(-c*x^2 + 1))/(c*x^7 - x^5), x)) - 3/4*a*b^2*arctanh(c*x^2)^2/x^4 - 1/4 *a^3/x^4
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^5} \,d x \]