Integrand size = 16, antiderivative size = 97 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=-\frac {a b}{2 c x^2}-\frac {b^2 \coth ^{-1}\left (\frac {x^2}{c}\right )}{2 c x^2}+\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{4 c^2}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{4 x^4}-\frac {b^2 \log \left (1-\frac {c^2}{x^4}\right )}{4 c^2} \]
-1/2*a*b/c/x^2-1/2*b^2*arccoth(x^2/c)/c/x^2+1/4*(a+b*arccoth(x^2/c))^2/c^2 -1/4*(a+b*arccoth(x^2/c))^2/x^4-1/4*b^2*ln(1-c^2/x^4)/c^2
Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=-\frac {a^2 c^2+2 a b c x^2+2 b c \left (a c+b x^2\right ) \text {arctanh}\left (\frac {c}{x^2}\right )+b^2 \left (c^2-x^4\right ) \text {arctanh}\left (\frac {c}{x^2}\right )^2-4 b^2 x^4 \log (x)+a b x^4 \log \left (-c+x^2\right )+b^2 x^4 \log \left (-c+x^2\right )-a b x^4 \log \left (c+x^2\right )+b^2 x^4 \log \left (c+x^2\right )}{4 c^2 x^4} \]
-1/4*(a^2*c^2 + 2*a*b*c*x^2 + 2*b*c*(a*c + b*x^2)*ArcTanh[c/x^2] + b^2*(c^ 2 - x^4)*ArcTanh[c/x^2]^2 - 4*b^2*x^4*Log[x] + a*b*x^4*Log[-c + x^2] + b^2 *x^4*Log[-c + x^2] - a*b*x^4*Log[c + x^2] + b^2*x^4*Log[c + x^2])/(c^2*x^4 )
Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6454, 6452, 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.
\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} \left (b c \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{\left (1-\frac {c^2}{x^4}\right ) x^4}d\frac {1}{x^2}-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 x^4}\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle \frac {1}{2} \left (b c \left (\frac {\int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )d\frac {1}{x^2}}{c^2}\right )-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 x^4}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (b c \left (\frac {\int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}}{c^2}-\frac {\frac {a}{x^2}+\frac {b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^2}+\frac {b \log \left (1-\frac {c^2}{x^4}\right )}{2 c}}{c^2}\right )-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 x^4}\right )\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle \frac {1}{2} \left (b c \left (\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b c^3}-\frac {\frac {a}{x^2}+\frac {b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^2}+\frac {b \log \left (1-\frac {c^2}{x^4}\right )}{2 c}}{c^2}\right )-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 x^4}\right )\) |
(-1/2*(a + b*ArcTanh[c/x^2])^2/x^4 + b*c*((a + b*ArcTanh[c/x^2])^2/(2*b*c^ 3) - (a/x^2 + (b*ArcTanh[c/x^2])/x^2 + (b*Log[1 - c^2/x^4])/(2*c))/c^2))/2
3.2.75.3.1 Defintions of rubi rules used
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]
Time = 242.62 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.42
method | result | size |
parallelrisch | \(\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2} b^{2} x^{4}+4 b^{2} \ln \left (x \right ) x^{4}-2 \ln \left (x^{2}-c \right ) x^{4} b^{2}+2 a b \,x^{4} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )-2 x^{4} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b^{2}-2 \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b^{2} c \,x^{2}-b^{2} c^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2}-2 a b c \,x^{2}-2 a b \,c^{2} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )-a^{2} c^{2}}{4 x^{4} c^{2}}\) | \(138\) |
parts | \(\text {Expression too large to display}\) | \(810\) |
derivativedivides | \(\text {Expression too large to display}\) | \(812\) |
default | \(\text {Expression too large to display}\) | \(812\) |
risch | \(\text {Expression too large to display}\) | \(682002\) |
1/4*(arctanh(c/x^2)^2*b^2*x^4+4*b^2*ln(x)*x^4-2*ln(x^2-c)*x^4*b^2+2*a*b*x^ 4*arctanh(c/x^2)-2*x^4*arctanh(c/x^2)*b^2-2*arctanh(c/x^2)*b^2*c*x^2-b^2*c ^2*arctanh(c/x^2)^2-2*a*b*c*x^2-2*a*b*c^2*arctanh(c/x^2)-a^2*c^2)/x^4/c^2
Time = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=\frac {16 \, b^{2} x^{4} \log \left (x\right ) + 4 \, {\left (a b - b^{2}\right )} x^{4} \log \left (x^{2} + c\right ) - 4 \, {\left (a b + b^{2}\right )} x^{4} \log \left (x^{2} - c\right ) - 8 \, a b c x^{2} - 4 \, a^{2} c^{2} + {\left (b^{2} x^{4} - b^{2} c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )^{2} - 4 \, {\left (b^{2} c x^{2} + a b c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{16 \, c^{2} x^{4}} \]
1/16*(16*b^2*x^4*log(x) + 4*(a*b - b^2)*x^4*log(x^2 + c) - 4*(a*b + b^2)*x ^4*log(x^2 - c) - 8*a*b*c*x^2 - 4*a^2*c^2 + (b^2*x^4 - b^2*c^2)*log((x^2 + c)/(x^2 - c))^2 - 4*(b^2*c*x^2 + a*b*c^2)*log((x^2 + c)/(x^2 - c)))/(c^2* x^4)
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (78) = 156\).
Time = 6.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.77 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=\begin {cases} - \frac {a^{2}}{4 x^{4}} - \frac {a b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2 x^{4}} - \frac {a b}{2 c x^{2}} + \frac {a b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2 c^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x^{2}} \right )}}{4 x^{4}} - \frac {b^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2 c x^{2}} + \frac {b^{2} \log {\left (x \right )}}{c^{2}} - \frac {b^{2} \log {\left (x - \sqrt {- c} \right )}}{2 c^{2}} - \frac {b^{2} \log {\left (x + \sqrt {- c} \right )}}{2 c^{2}} + \frac {b^{2} \operatorname {atanh}^{2}{\left (\frac {c}{x^{2}} \right )}}{4 c^{2}} + \frac {b^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \]
Piecewise((-a**2/(4*x**4) - a*b*atanh(c/x**2)/(2*x**4) - a*b/(2*c*x**2) + a*b*atanh(c/x**2)/(2*c**2) - b**2*atanh(c/x**2)**2/(4*x**4) - b**2*atanh(c /x**2)/(2*c*x**2) + b**2*log(x)/c**2 - b**2*log(x - sqrt(-c))/(2*c**2) - b **2*log(x + sqrt(-c))/(2*c**2) + b**2*atanh(c/x**2)**2/(4*c**2) + b**2*ata nh(c/x**2)/(2*c**2), Ne(c, 0)), (-a**2/(4*x**4), True))
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (87) = 174\).
Time = 0.19 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=\frac {1}{4} \, {\left (c {\left (\frac {\log \left (x^{2} + c\right )}{c^{3}} - \frac {\log \left (x^{2} - c\right )}{c^{3}} - \frac {2}{c^{2} x^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x^{4}}\right )} a b - \frac {1}{16} \, {\left (c^{2} {\left (\frac {\log \left (x^{2} + c\right )^{2} - 2 \, {\left (\log \left (x^{2} + c\right ) - 2\right )} \log \left (x^{2} - c\right ) + \log \left (x^{2} - c\right )^{2} + 4 \, \log \left (x^{2} + c\right )}{c^{4}} - \frac {16 \, \log \left (x\right )}{c^{4}}\right )} - 4 \, c {\left (\frac {\log \left (x^{2} + c\right )}{c^{3}} - \frac {\log \left (x^{2} - c\right )}{c^{3}} - \frac {2}{c^{2} x^{2}}\right )} \operatorname {artanh}\left (\frac {c}{x^{2}}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (\frac {c}{x^{2}}\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]
1/4*(c*(log(x^2 + c)/c^3 - log(x^2 - c)/c^3 - 2/(c^2*x^2)) - 2*arctanh(c/x ^2)/x^4)*a*b - 1/16*(c^2*((log(x^2 + c)^2 - 2*(log(x^2 + c) - 2)*log(x^2 - c) + log(x^2 - c)^2 + 4*log(x^2 + c))/c^4 - 16*log(x)/c^4) - 4*c*(log(x^2 + c)/c^3 - log(x^2 - c)/c^3 - 2/(c^2*x^2))*arctanh(c/x^2))*b^2 - 1/4*b^2* arctanh(c/x^2)^2/x^4 - 1/4*a^2/x^4
\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x^{5}} \,d x } \]
Time = 3.93 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.70 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^5} \, dx=\frac {b^2\,{\ln \left (x^2+c\right )}^2}{16\,c^2}-\frac {b^2\,\ln \left (x^2-c\right )}{4\,c^2}-\frac {a^2}{4\,x^4}-\frac {b^2\,{\ln \left (x^2+c\right )}^2}{16\,x^4}+\frac {b^2\,{\ln \left (x^2-c\right )}^2}{16\,c^2}-\frac {b^2\,{\ln \left (x^2-c\right )}^2}{16\,x^4}+\frac {b^2\,\ln \left (x\right )}{c^2}-\frac {b^2\,\ln \left (x^2+c\right )}{4\,c^2}-\frac {a\,b\,\ln \left (x^2+c\right )}{4\,x^4}+\frac {b^2\,\ln \left (x^2-c\right )}{4\,c\,x^2}-\frac {a\,b\,\ln \left (x^2-c\right )}{4\,c^2}-\frac {b^2\,\ln \left (x^2+c\right )\,\ln \left (x^2-c\right )}{8\,c^2}+\frac {a\,b\,\ln \left (x^2-c\right )}{4\,x^4}+\frac {b^2\,\ln \left (x^2+c\right )\,\ln \left (x^2-c\right )}{8\,x^4}-\frac {a\,b}{2\,c\,x^2}-\frac {b^2\,\ln \left (x^2+c\right )}{4\,c\,x^2}+\frac {a\,b\,\ln \left (x^2+c\right )}{4\,c^2} \]
(b^2*log(c + x^2)^2)/(16*c^2) - (b^2*log(x^2 - c))/(4*c^2) - a^2/(4*x^4) - (b^2*log(c + x^2)^2)/(16*x^4) + (b^2*log(x^2 - c)^2)/(16*c^2) - (b^2*log( x^2 - c)^2)/(16*x^4) + (b^2*log(x))/c^2 - (b^2*log(c + x^2))/(4*c^2) - (a* b*log(c + x^2))/(4*x^4) + (b^2*log(x^2 - c))/(4*c*x^2) - (a*b*log(x^2 - c) )/(4*c^2) - (b^2*log(c + x^2)*log(x^2 - c))/(8*c^2) + (a*b*log(x^2 - c))/( 4*x^4) + (b^2*log(c + x^2)*log(x^2 - c))/(8*x^4) - (a*b)/(2*c*x^2) - (b^2* log(c + x^2))/(4*c*x^2) + (a*b*log(c + x^2))/(4*c^2)