Integrand size = 14, antiderivative size = 56 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=-\frac {b c}{18 x^6}-\frac {a+b \text {arctanh}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right ) \] Output:
-1/18*b*c/x^6-1/9*(a+b*arctanh(c*x^3))/x^9+1/3*b*c^3*ln(x)-1/18*b*c^3*ln(- c^2*x^6+1)
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=-\frac {a}{9 x^9}-\frac {b c}{18 x^6}-\frac {b \text {arctanh}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^3])/x^10,x]
Output:
-1/9*a/x^9 - (b*c)/(18*x^6) - (b*ArcTanh[c*x^3])/(9*x^9) + (b*c^3*Log[x])/ 3 - (b*c^3*Log[1 - c^2*x^6])/18
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6452, 798, 54, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} b c \int \frac {1}{x^7 \left (1-c^2 x^6\right )}dx-\frac {a+b \text {arctanh}\left (c x^3\right )}{9 x^9}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{18} b c \int \frac {1}{x^{12} \left (1-c^2 x^6\right )}dx^6-\frac {a+b \text {arctanh}\left (c x^3\right )}{9 x^9}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{18} b c \int \left (-\frac {c^4}{c^2 x^6-1}+\frac {c^2}{x^6}+\frac {1}{x^{12}}\right )dx^6-\frac {a+b \text {arctanh}\left (c x^3\right )}{9 x^9}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{18} b c \left (c^2 \log \left (x^6\right )-c^2 \log \left (1-c^2 x^6\right )-\frac {1}{x^6}\right )-\frac {a+b \text {arctanh}\left (c x^3\right )}{9 x^9}\) |
Input:
Int[(a + b*ArcTanh[c*x^3])/x^10,x]
Output:
-1/9*(a + b*ArcTanh[c*x^3])/x^9 + (b*c*(-x^(-6) + c^2*Log[x^6] - c^2*Log[1 - c^2*x^6]))/18
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.12
method | result | size |
default | \(-\frac {a}{9 x^{9}}+b \left (-\frac {\operatorname {arctanh}\left (c \,x^{3}\right )}{9 x^{9}}+\frac {c \left (-\frac {1}{6 x^{6}}+c^{2} \ln \left (x \right )-\frac {c^{2} \ln \left (c \,x^{3}-1\right )}{6}-\frac {c^{2} \ln \left (c \,x^{3}+1\right )}{6}\right )}{3}\right )\) | \(63\) |
parts | \(-\frac {a}{9 x^{9}}+b \left (-\frac {\operatorname {arctanh}\left (c \,x^{3}\right )}{9 x^{9}}+\frac {c \left (-\frac {1}{6 x^{6}}+c^{2} \ln \left (x \right )-\frac {c^{2} \ln \left (c \,x^{3}-1\right )}{6}-\frac {c^{2} \ln \left (c \,x^{3}+1\right )}{6}\right )}{3}\right )\) | \(63\) |
risch | \(-\frac {b \ln \left (c \,x^{3}+1\right )}{18 x^{9}}+\frac {6 b \,c^{3} \ln \left (x \right ) x^{9}-b \,c^{3} \ln \left (c^{2} x^{6}-1\right ) x^{9}-b c \,x^{3}+b \ln \left (-c \,x^{3}+1\right )-2 a}{18 x^{9}}\) | \(73\) |
parallelrisch | \(\frac {6 b \,c^{3} \ln \left (x \right ) x^{9}-2 \ln \left (c \,x^{3}-1\right ) x^{9} b \,c^{3}-2 b \,\operatorname {arctanh}\left (c \,x^{3}\right ) x^{9} c^{3}-b \,c^{3} x^{9}-b c \,x^{3}-2 b \,\operatorname {arctanh}\left (c \,x^{3}\right )-2 a}{18 x^{9}}\) | \(78\) |
Input:
int((a+b*arctanh(c*x^3))/x^10,x,method=_RETURNVERBOSE)
Output:
-1/9*a/x^9+b*(-1/9/x^9*arctanh(c*x^3)+1/3*c*(-1/6/x^6+c^2*ln(x)-1/6*c^2*ln (c*x^3-1)-1/6*c^2*ln(c*x^3+1)))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=-\frac {b c^{3} x^{9} \log \left (c^{2} x^{6} - 1\right ) - 6 \, b c^{3} x^{9} \log \left (x\right ) + b c x^{3} + b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a}{18 \, x^{9}} \] Input:
integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="fricas")
Output:
-1/18*(b*c^3*x^9*log(c^2*x^6 - 1) - 6*b*c^3*x^9*log(x) + b*c*x^3 + b*log(- (c*x^3 + 1)/(c*x^3 - 1)) + 2*a)/x^9
Timed out. \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=\text {Timed out} \] Input:
integrate((a+b*atanh(c*x**3))/x**10,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=-\frac {1}{18} \, {\left ({\left (c^{2} \log \left (c^{2} x^{6} - 1\right ) - c^{2} \log \left (x^{6}\right ) + \frac {1}{x^{6}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x^{3}\right )}{x^{9}}\right )} b - \frac {a}{9 \, x^{9}} \] Input:
integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="maxima")
Output:
-1/18*((c^2*log(c^2*x^6 - 1) - c^2*log(x^6) + 1/x^6)*c + 2*arctanh(c*x^3)/ x^9)*b - 1/9*a/x^9
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=-\frac {1}{18} \, b c^{3} \log \left (c^{2} x^{6} - 1\right ) + \frac {1}{3} \, b c^{3} \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{18 \, x^{9}} - \frac {b c x^{3} + 2 \, a}{18 \, x^{9}} \] Input:
integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="giac")
Output:
-1/18*b*c^3*log(c^2*x^6 - 1) + 1/3*b*c^3*log(x) - 1/18*b*log(-(c*x^3 + 1)/ (c*x^3 - 1))/x^9 - 1/18*(b*c*x^3 + 2*a)/x^9
Time = 3.44 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=\frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^6-1\right )}{18}-\frac {a}{9\,x^9}-\frac {b\,c}{18\,x^6}-\frac {b\,\ln \left (c\,x^3+1\right )}{18\,x^9}+\frac {b\,\ln \left (1-c\,x^3\right )}{18\,x^9} \] Input:
int((a + b*atanh(c*x^3))/x^10,x)
Output:
(b*c^3*log(x))/3 - (b*c^3*log(c^2*x^6 - 1))/18 - a/(9*x^9) - (b*c)/(18*x^6 ) - (b*log(c*x^3 + 1))/(18*x^9) + (b*log(1 - c*x^3))/(18*x^9)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.70 \[ \int \frac {a+b \text {arctanh}\left (c x^3\right )}{x^{10}} \, dx=\frac {2 \mathit {atanh} \left (c \,x^{3}\right ) b \,c^{3} x^{9}-2 \mathit {atanh} \left (c \,x^{3}\right ) b -2 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) b \,c^{3} x^{9}-2 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) b \,c^{3} x^{9}+6 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{9}-2 a -b c \,x^{3}}{18 x^{9}} \] Input:
int((a+b*atanh(c*x^3))/x^10,x)
Output:
(2*atanh(c*x**3)*b*c**3*x**9 - 2*atanh(c*x**3)*b - 2*log(c**(2/3)*x**2 - c **(1/3)*x + 1)*b*c**3*x**9 - 2*log(c**(2/3)*x + c**(1/3))*b*c**3*x**9 + 6* log(x)*b*c**3*x**9 - 2*a - b*c*x**3)/(18*x**9)