Integrand size = 16, antiderivative size = 90 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\frac {1}{3} c \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{3 x^3}+\frac {2}{3} b c \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{3} b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \]
1/3*c*(a+b*arctanh(c*x^3))^2-1/3*(a+b*arctanh(c*x^3))^2/x^3+2/3*b*c*(a+b*a rctanh(c*x^3))*ln(2-2/(c*x^3+1))-1/3*b^2*c*polylog(2,-1+2/(c*x^3+1))
Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\frac {b^2 \left (-1+c x^3\right ) \text {arctanh}\left (c x^3\right )^2+2 b \text {arctanh}\left (c x^3\right ) \left (-a+b c x^3 \log \left (1-e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )-a \left (a-2 b c x^3 \log \left (c x^3\right )+b c x^3 \log \left (1-c^2 x^6\right )\right )-b^2 c x^3 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{3 x^3} \]
(b^2*(-1 + c*x^3)*ArcTanh[c*x^3]^2 + 2*b*ArcTanh[c*x^3]*(-a + b*c*x^3*Log[ 1 - E^(-2*ArcTanh[c*x^3])]) - a*(a - 2*b*c*x^3*Log[c*x^3] + b*c*x^3*Log[1 - c^2*x^6]) - b^2*c*x^3*PolyLog[2, E^(-2*ArcTanh[c*x^3])])/(3*x^3)
Time = 0.56 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6454, 6452, 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.
\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^6}dx^3\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} \left (2 b c \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (1-c^2 x^6\right )}dx^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )\) |
\(\Big \downarrow \) 6550 |
\(\displaystyle \frac {1}{3} \left (2 b c \left (\int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^3 \left (c x^3+1\right )}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {1}{3} \left (2 b c \left (-b c \int \frac {\log \left (2-\frac {2}{c x^3+1}\right )}{1-c^2 x^6}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {1}{3} \left (2 b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right )\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3}\right )\) |
(-((a + b*ArcTanh[c*x^3])^2/x^3) + 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))/3
3.2.21.3.1 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_.)/((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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.20 (sec) , antiderivative size = 2993, normalized size of antiderivative = 33.26
method | result | size |
default | \(\text {Expression too large to display}\) | \(2993\) |
parts | \(\text {Expression too large to display}\) | \(2993\) |
-1/3*a^2/x^3+b^2*(-1/3/x^3*arctanh(c*x^3)^2+2*c*(arctanh(c*x^3)*ln(x)-1/6* arctanh(c*x^3)*ln(c*x^3-1)-1/6*arctanh(c*x^3)*ln(c*x^3+1)-1/2*c*(Sum(1/6*( ln(x-_alpha)*ln(c*x^3-1)-3*c*(1/6/_alpha^2/c*ln(x-_alpha)^2-1/3*_alpha*ln( x-_alpha)*(2*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/Roo tOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha ^2,index=2)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)+6*ln((RootOf(_Z^2+ 3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha ^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)*_alpha+3*ln((Root Of(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+ 3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*_alpha+9* ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+3*_Z *_alpha+3*_alpha^2,index=1))*_alpha^2+2*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alp ha^2,index=2)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootO f(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2, index=1)+3*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x+_alpha)/RootO f(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2 ,index=2)*_alpha+6*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x+_alph a)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*RootOf(_Z^2+3*_Z*_alpha+3* _alpha^2,index=1)*_alpha+9*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2) -x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2))*_alpha^2)/(3*_a...
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
-1/3*(c*(log(c^2*x^6 - 1) - log(x^6)) + 2*arctanh(c*x^3)/x^3)*a*b - 1/12*b ^2*(log(-c*x^3 + 1)^2/x^3 + 3*integrate(-((c*x^3 - 1)*log(c*x^3 + 1)^2 + 2 *(c*x^3 - (c*x^3 - 1)*log(c*x^3 + 1))*log(-c*x^3 + 1))/(c*x^7 - x^4), x)) - 1/3*a^2/x^3
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2}{x^4} \,d x \]