Integrand size = 14, antiderivative size = 45 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=-\frac {b}{4 c x^2}-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}+\frac {b \text {arctanh}\left (\frac {x^2}{c}\right )}{4 c^2} \]
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {b}{4 c x^2}-\frac {b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}-\frac {b \log \left (-c+x^2\right )}{8 c^2}+\frac {b \log \left (c+x^2\right )}{8 c^2} \]
-1/4*a/x^4 - b/(4*c*x^2) - (b*ArcTanh[c/x^2])/(4*x^4) - (b*Log[-c + x^2])/ (8*c^2) + (b*Log[c + x^2])/(8*c^2)
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6452, 795, 807, 25, 264, 219}
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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -\frac {1}{2} b c \int \frac {1}{\left (1-\frac {c^2}{x^4}\right ) x^7}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle -\frac {1}{2} b c \int \frac {1}{x^3 \left (x^4-c^2\right )}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{4} b c \int -\frac {1}{x^4 \left (c^2-x^4\right )}dx^2-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} b c \int \frac {1}{x^4 \left (c^2-x^4\right )}dx^2-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {1}{4} b c \left (\frac {1}{c^2 x^2}-\frac {\int \frac {1}{c^2-x^4}dx^2}{c^2}\right )-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{4 x^4}-\frac {1}{4} b c \left (\frac {1}{c^2 x^2}-\frac {\text {arctanh}\left (\frac {x^2}{c}\right )}{c^3}\right )\) |
3.2.63.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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]
Time = 0.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(-\frac {-\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,x^{4}+b c \,x^{2}+\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,c^{2}+a \,c^{2}}{4 x^{4} c^{2}}\) | \(44\) |
derivativedivides | \(-\frac {a}{4 x^{4}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{4 x^{4}}-\frac {b}{4 c \,x^{2}}+\frac {b \ln \left (1+\frac {c}{x^{2}}\right )}{8 c^{2}}-\frac {b \ln \left (\frac {c}{x^{2}}-1\right )}{8 c^{2}}\) | \(57\) |
default | \(-\frac {a}{4 x^{4}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{4 x^{4}}-\frac {b}{4 c \,x^{2}}+\frac {b \ln \left (1+\frac {c}{x^{2}}\right )}{8 c^{2}}-\frac {b \ln \left (\frac {c}{x^{2}}-1\right )}{8 c^{2}}\) | \(57\) |
parts | \(-\frac {a}{4 x^{4}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{4 x^{4}}-\frac {b}{4 c \,x^{2}}+\frac {b \ln \left (1+\frac {c}{x^{2}}\right )}{8 c^{2}}-\frac {b \ln \left (\frac {c}{x^{2}}-1\right )}{8 c^{2}}\) | \(57\) |
risch | \(-\frac {b \ln \left (x^{2}+c \right )}{8 x^{4}}-\frac {i \pi b \,c^{2} \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}-2 i \pi b \,c^{2}+i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}-i \pi b \,c^{2} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}-i \pi b \,c^{2} \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}-i \pi b \,c^{2} {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}+2 i \pi b \,c^{2} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}-i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )+i \pi b \,c^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )-2 b \ln \left (-x^{2}-c \right ) x^{4}+2 b \ln \left (-x^{2}+c \right ) x^{4}-2 b \ln \left (-x^{2}+c \right ) c^{2}+4 b c \,x^{2}+4 a \,c^{2}}{16 c^{2} x^{4}}\) | \(360\) |
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=-\frac {2 \, b c x^{2} + 2 \, a c^{2} - {\left (b x^{4} - b c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{8 \, c^{2} x^{4}} \]
Time = 5.62 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=\begin {cases} - \frac {a}{4 x^{4}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4 x^{4}} - \frac {b}{4 c x^{2}} + \frac {b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\- \frac {a}{4 x^{4}} & \text {otherwise} \end {cases} \]
Piecewise((-a/(4*x**4) - b*atanh(c/x**2)/(4*x**4) - b/(4*c*x**2) + b*atanh (c/x**2)/(4*c**2), Ne(c, 0)), (-a/(4*x**4), True))
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=\frac {1}{8} \, {\left (c {\left (\frac {\log \left (x^{2} + c\right )}{c^{3}} - \frac {\log \left (x^{2} - c\right )}{c^{3}} - \frac {2}{c^{2} x^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]
1/8*(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)*b - 1/4*a/x^4
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=\frac {b \log \left (x^{2} + c\right )}{8 \, c^{2}} - \frac {b \log \left (-x^{2} + c\right )}{8 \, c^{2}} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{8 \, x^{4}} - \frac {b x^{2} + a c}{4 \, c x^{4}} \]
1/8*b*log(x^2 + c)/c^2 - 1/8*b*log(-x^2 + c)/c^2 - 1/8*b*log((x^2 + c)/(x^ 2 - c))/x^4 - 1/4*(b*x^2 + a*c)/(c*x^4)
Time = 3.44 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.31 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^5} \, dx=\frac {\frac {b\,x^4\,\mathrm {atanh}\left (\frac {x^2}{c}\right )}{4}-\frac {b\,c\,x^2}{4}}{c^2\,x^4}-\frac {\frac {a}{4}-\frac {b\,\ln \left (x^2-c\right )}{8}+\frac {b\,\ln \left (x^2+c\right )}{8}}{x^4} \]