Integrand size = 16, antiderivative size = 136 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\frac {1}{2} b c^2 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {b c \left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 x^3}+\frac {1}{6} c^2 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{6 x^6}+b^2 c^2 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{2} b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \]
1/2*b*c^2*(a+b*arctanh(c*x^3))^2-1/2*b*c*(a+b*arctanh(c*x^3))^2/x^3+1/6*c^ 2*(a+b*arctanh(c*x^3))^3-1/6*(a+b*arctanh(c*x^3))^3/x^6+b^2*c^2*(a+b*arcta nh(c*x^3))*ln(2-2/(c*x^3+1))-1/2*b^3*c^2*polylog(2,-1+2/(c*x^3+1))
Time = 0.23 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\frac {6 b^2 \left (-1+c x^3\right ) \left (a+a c x^3+b c x^3\right ) \text {arctanh}\left (c x^3\right )^2+2 b^3 \left (-1+c^2 x^6\right ) \text {arctanh}\left (c x^3\right )^3-6 b \text {arctanh}\left (c x^3\right ) \left (a^2+2 a b c x^3-2 b^2 c^2 x^6 \log \left (1-e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )+a \left (-2 a^2-6 a b c x^3-3 a b c^2 x^6 \log \left (1-c x^3\right )+3 a b c^2 x^6 \log \left (1+c x^3\right )+12 b^2 c^2 x^6 \log \left (\frac {c x^3}{\sqrt {1-c^2 x^6}}\right )\right )-6 b^3 c^2 x^6 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{12 x^6} \]
(6*b^2*(-1 + c*x^3)*(a + a*c*x^3 + b*c*x^3)*ArcTanh[c*x^3]^2 + 2*b^3*(-1 + c^2*x^6)*ArcTanh[c*x^3]^3 - 6*b*ArcTanh[c*x^3]*(a^2 + 2*a*b*c*x^3 - 2*b^2 *c^2*x^6*Log[1 - E^(-2*ArcTanh[c*x^3])]) + a*(-2*a^2 - 6*a*b*c*x^3 - 3*a*b *c^2*x^6*Log[1 - c*x^3] + 3*a*b*c^2*x^6*Log[1 + c*x^3] + 12*b^2*c^2*x^6*Lo g[(c*x^3)/Sqrt[1 - c^2*x^6]]) - 6*b^3*c^2*x^6*PolyLog[2, E^(-2*ArcTanh[c*x ^3])])/(12*x^6)
Time = 1.03 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx\) |
| \(\Big \downarrow \) 6454 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^9}dx^3\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^6 \left (1-c^2 x^6\right )}dx^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 6544 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{1-c^2 x^6}dx^3+\int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^6}dx^3\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 6452 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (c^2 \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{1-c^2 x^6}dx^3+2 b c \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 6510 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (2 b c \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3+\frac {c \left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 6550 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (2 b c \left (\int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (c x^3+1\right )}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}\right )+\frac {c \left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 6494 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{c x^3+1}\right )}{1-c^2 x^6}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )\right )+\frac {c \left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} b c \left (2 b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right )\right )+\frac {c \left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{2 x^6}\right )\) |
(-1/2*(a + b*ArcTanh[c*x^3])^3/x^6 + (3*b*c*(-((a + b*ArcTanh[c*x^3])^2/x^ 3) + (c*(a + b*ArcTanh[c*x^3])^3)/(3*b) + 2*b*c*((a + b*ArcTanh[c*x^3])^2/ (2*b) + (a + b*ArcTanh[c*x^3])*Log[2 - 2/(1 + c*x^3)] - (b*PolyLog[2, -1 + 2/(1 + c*x^3)])/2)))/2)/3
3.2.29.3.1 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]
\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{3}\right )\right )}^{3}}{x^{7}}d x\]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{7}} \,d x } \]
integral((b^3*arctanh(c*x^3)^3 + 3*a*b^2*arctanh(c*x^3)^2 + 3*a^2*b*arctan h(c*x^3) + a^3)/x^7, x)
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{7}} \,d x } \]
1/4*((c*log(c*x^3 + 1) - c*log(c*x^3 - 1) - 2/x^3)*c - 2*arctanh(c*x^3)/x^ 6)*a^2*b + 1/8*((2*(log(c*x^3 - 1) - 2)*log(c*x^3 + 1) - log(c*x^3 + 1)^2 - log(c*x^3 - 1)^2 - 4*log(c*x^3 - 1) + 24*log(x))*c^2 + 4*(c*log(c*x^3 + 1) - c*log(c*x^3 - 1) - 2/x^3)*c*arctanh(c*x^3))*a*b^2 - 1/48*b^3*(((c^2*x ^6 - 1)*log(-c*x^3 + 1)^3 + 3*(2*c*x^3 - (c^2*x^6 - 1)*log(c*x^3 + 1))*log (-c*x^3 + 1)^2)/x^6 + 6*integrate(-((c*x^3 - 1)*log(c*x^3 + 1)^3 + 3*(2*c^ 2*x^6 - (c*x^3 - 1)*log(c*x^3 + 1)^2 - (c^3*x^9 - c*x^3)*log(c*x^3 + 1))*l og(-c*x^3 + 1))/(c*x^10 - x^7), x)) - 1/2*a*b^2*arctanh(c*x^3)^2/x^6 - 1/6 *a^3/x^6
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{7}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^7} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^3}{x^7} \,d x \]