Integrand size = 14, antiderivative size = 54 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6+\frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{8} b c^4 \text {arctanh}\left (\frac {x^2}{c}\right ) \]
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6+\frac {a x^8}{8}+\frac {1}{8} b x^8 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{16} b c^4 \log \left (-c+x^2\right )-\frac {1}{16} b c^4 \log \left (c+x^2\right ) \]
(b*c^3*x^2)/8 + (b*c*x^6)/24 + (a*x^8)/8 + (b*x^8*ArcTanh[c/x^2])/8 + (b*c ^4*Log[-c + x^2])/16 - (b*c^4*Log[c + x^2])/16
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96, 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, 254, 2009}
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.
\(\displaystyle \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} b c \int \frac {x^5}{1-\frac {c^2}{x^4}}dx+\frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{4} b c \int \frac {x^9}{x^4-c^2}dx+\frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{8} b c \int -\frac {x^8}{c^2-x^4}dx^2+\frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{8} b c \int \frac {x^8}{c^2-x^4}dx^2\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{8} b c \int \left (\frac {c^4}{c^2-x^4}-c^2-x^4\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{8} b c \left (-c^3 \text {arctanh}\left (\frac {x^2}{c}\right )+c^2 x^2+\frac {x^6}{3}\right )\) |
3.2.57.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 0.95 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {x^{8} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b}{8}+\frac {x^{8} a}{8}+\frac {b c \,x^{6}}{24}+\frac {b \,c^{3} x^{2}}{8}-\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,c^{4}}{8}\) | \(48\) |
parts | \(\frac {x^{8} a}{8}+b \left (\frac {x^{8} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{8}-\frac {c \left (-\frac {x^{6}}{6}-\frac {c^{2} x^{2}}{2}+\frac {c^{3} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {c^{3} \ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )}{4}\right )\) | \(65\) |
derivativedivides | \(\frac {x^{8} a}{8}-b \left (-\frac {x^{8} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{8}+\frac {c \left (-\frac {x^{6}}{6}-\frac {c^{2} x^{2}}{2}+\frac {c^{3} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {c^{3} \ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )}{4}\right )\) | \(66\) |
default | \(\frac {x^{8} a}{8}-b \left (-\frac {x^{8} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{8}+\frac {c \left (-\frac {x^{6}}{6}-\frac {c^{2} x^{2}}{2}+\frac {c^{3} \ln \left (1+\frac {c}{x^{2}}\right )}{4}-\frac {c^{3} \ln \left (\frac {c}{x^{2}}-1\right )}{4}\right )}{4}\right )\) | \(66\) |
risch | \(\frac {x^{8} b \ln \left (x^{2}+c \right )}{16}-\frac {x^{8} b \ln \left (-x^{2}+c \right )}{16}-\frac {i \pi b \,x^{8} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{32}-\frac {i \pi b \,x^{8} \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{32}+\frac {i \pi b \,x^{8} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{32}+\frac {i \pi b \,x^{8} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{32}+\frac {i \pi b \,x^{8} \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{32}-\frac {i \pi b \,x^{8}}{16}-\frac {i \pi b \,x^{8} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}}{32}-\frac {i \pi b \,x^{8} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{32}+\frac {i \pi b \,x^{8} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{16}-\frac {i \pi b \,x^{8} {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}}{32}+\frac {x^{8} a}{8}+\frac {b c \,x^{6}}{24}+\frac {b \,c^{3} x^{2}}{8}-\frac {b \,c^{4} \ln \left (-x^{2}-c \right )}{16}+\frac {b \,c^{4} \ln \left (-x^{2}+c \right )}{16}\) | \(360\) |
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{8} \, a x^{8} + \frac {1}{24} \, b c x^{6} + \frac {1}{8} \, b c^{3} x^{2} + \frac {1}{16} \, {\left (b x^{8} - b c^{4}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) \]
Time = 3.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^{8}}{8} - \frac {b c^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{8} + \frac {b c^{3} x^{2}}{8} + \frac {b c x^{6}}{24} + \frac {b x^{8} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{8} \]
Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{8} \, a x^{8} + \frac {1}{48} \, {\left (6 \, x^{8} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (2 \, x^{6} + 6 \, c^{2} x^{2} - 3 \, c^{3} \log \left (x^{2} + c\right ) + 3 \, c^{3} \log \left (x^{2} - c\right )\right )} c\right )} b \]
1/8*a*x^8 + 1/48*(6*x^8*arctanh(c/x^2) + (2*x^6 + 6*c^2*x^2 - 3*c^3*log(x^ 2 + c) + 3*c^3*log(x^2 - c))*c)*b
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.31 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{16} \, b x^{8} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{8} \, a x^{8} + \frac {1}{24} \, b c x^{6} + \frac {1}{8} \, b c^{3} x^{2} - \frac {1}{16} \, b c^{4} \log \left (x^{2} + c\right ) + \frac {1}{16} \, b c^{4} \log \left (-x^{2} + c\right ) \]
1/16*b*x^8*log((x^2 + c)/(x^2 - c)) + 1/8*a*x^8 + 1/24*b*c*x^6 + 1/8*b*c^3 *x^2 - 1/16*b*c^4*log(x^2 + c) + 1/16*b*c^4*log(-x^2 + c)
Time = 3.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.22 \[ \int x^7 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a\,x^8}{8}+\frac {b\,c^3\,x^2}{8}+\frac {b\,x^8\,\ln \left (x^2+c\right )}{16}+\frac {b\,c\,x^6}{24}-\frac {b\,x^8\,\ln \left (x^2-c\right )}{16}+\frac {b\,c^4\,\mathrm {atan}\left (\frac {x^2\,1{}\mathrm {i}}{c}\right )\,1{}\mathrm {i}}{8} \]