Integrand size = 14, antiderivative size = 56 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=-\frac {b c}{12 x^4}-\frac {a+b \text {arctanh}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right ) \] Output:
-1/12*b*c/x^4-1/6*(a+b*arctanh(c*x^2))/x^6+1/3*b*c^3*ln(x)-1/12*b*c^3*ln(- c^2*x^4+1)
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=-\frac {a}{6 x^6}-\frac {b c}{12 x^4}-\frac {b \text {arctanh}\left (c x^2\right )}{6 x^6}+\frac {1}{3} b c^3 \log (x)-\frac {1}{12} b c^3 \log \left (1-c^2 x^4\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^2])/x^7,x]
Output:
-1/6*a/x^6 - (b*c)/(12*x^4) - (b*ArcTanh[c*x^2])/(6*x^6) + (b*c^3*Log[x])/ 3 - (b*c^3*Log[1 - c^2*x^4])/12
Time = 0.40 (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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} b c \int \frac {1}{x^5 \left (1-c^2 x^4\right )}dx-\frac {a+b \text {arctanh}\left (c x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{12} b c \int \frac {1}{x^8 \left (1-c^2 x^4\right )}dx^4-\frac {a+b \text {arctanh}\left (c x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{12} b c \int \left (-\frac {c^4}{c^2 x^4-1}+\frac {c^2}{x^4}+\frac {1}{x^8}\right )dx^4-\frac {a+b \text {arctanh}\left (c x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{12} b c \left (c^2 \log \left (x^4\right )-c^2 \log \left (1-c^2 x^4\right )-\frac {1}{x^4}\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{6 x^6}\) |
Input:
Int[(a + b*ArcTanh[c*x^2])/x^7,x]
Output:
-1/6*(a + b*ArcTanh[c*x^2])/x^6 + (b*c*(-x^(-4) + c^2*Log[x^4] - c^2*Log[1 - c^2*x^4]))/12
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 = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.12
method | result | size |
default | \(-\frac {a}{6 x^{6}}+b \left (-\frac {\operatorname {arctanh}\left (c \,x^{2}\right )}{6 x^{6}}+\frac {c \left (-\frac {1}{4 x^{4}}+c^{2} \ln \left (x \right )-\frac {c^{2} \ln \left (c \,x^{2}+1\right )}{4}-\frac {c^{2} \ln \left (c \,x^{2}-1\right )}{4}\right )}{3}\right )\) | \(63\) |
parts | \(-\frac {a}{6 x^{6}}+b \left (-\frac {\operatorname {arctanh}\left (c \,x^{2}\right )}{6 x^{6}}+\frac {c \left (-\frac {1}{4 x^{4}}+c^{2} \ln \left (x \right )-\frac {c^{2} \ln \left (c \,x^{2}+1\right )}{4}-\frac {c^{2} \ln \left (c \,x^{2}-1\right )}{4}\right )}{3}\right )\) | \(63\) |
risch | \(-\frac {b \ln \left (c \,x^{2}+1\right )}{12 x^{6}}+\frac {4 b \,c^{3} \ln \left (x \right ) x^{6}-b \,c^{3} \ln \left (c^{2} x^{4}-1\right ) x^{6}-b c \,x^{2}+b \ln \left (-c \,x^{2}+1\right )-2 a}{12 x^{6}}\) | \(73\) |
parallelrisch | \(\frac {4 b \,c^{3} \ln \left (x \right ) x^{6}-2 \ln \left (c \,x^{2}-1\right ) x^{6} b \,c^{3}-2 b \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{6} c^{3}-b \,c^{3} x^{6}-b c \,x^{2}-2 b \,\operatorname {arctanh}\left (c \,x^{2}\right )-2 a}{12 x^{6}}\) | \(78\) |
Input:
int((a+b*arctanh(c*x^2))/x^7,x,method=_RETURNVERBOSE)
Output:
-1/6*a/x^6+b*(-1/6/x^6*arctanh(c*x^2)+1/3*c*(-1/4/x^4+c^2*ln(x)-1/4*c^2*ln (c*x^2+1)-1/4*c^2*ln(c*x^2-1)))
Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=-\frac {b c^{3} x^{6} \log \left (c^{2} x^{4} - 1\right ) - 4 \, b c^{3} x^{6} \log \left (x\right ) + b c x^{2} + b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{12 \, x^{6}} \] Input:
integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="fricas")
Output:
-1/12*(b*c^3*x^6*log(c^2*x^4 - 1) - 4*b*c^3*x^6*log(x) + b*c*x^2 + b*log(- (c*x^2 + 1)/(c*x^2 - 1)) + 2*a)/x^6
Time = 7.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.73 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=\begin {cases} - \frac {a}{6 x^{6}} + \frac {b c^{3} \log {\left (x \right )}}{3} - \frac {b c^{3} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{6} - \frac {b c^{3} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{6} + \frac {b c^{3} \operatorname {atanh}{\left (c x^{2} \right )}}{6} - \frac {b c}{12 x^{4}} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{6 x^{6}} & \text {for}\: c \neq 0 \\- \frac {a}{6 x^{6}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c*x**2))/x**7,x)
Output:
Piecewise((-a/(6*x**6) + b*c**3*log(x)/3 - b*c**3*log(x - sqrt(-1/c))/6 - b*c**3*log(x + sqrt(-1/c))/6 + b*c**3*atanh(c*x**2)/6 - b*c/(12*x**4) - b* atanh(c*x**2)/(6*x**6), Ne(c, 0)), (-a/(6*x**6), True))
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=-\frac {1}{12} \, {\left ({\left (c^{2} \log \left (c^{2} x^{4} - 1\right ) - c^{2} \log \left (x^{4}\right ) + \frac {1}{x^{4}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{6}}\right )} b - \frac {a}{6 \, x^{6}} \] Input:
integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="maxima")
Output:
-1/12*((c^2*log(c^2*x^4 - 1) - c^2*log(x^4) + 1/x^4)*c + 2*arctanh(c*x^2)/ x^6)*b - 1/6*a/x^6
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=-\frac {1}{12} \, b c^{3} \log \left (c^{2} x^{4} - 1\right ) + \frac {1}{3} \, b c^{3} \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{12 \, x^{6}} - \frac {b c x^{2} + 2 \, a}{12 \, x^{6}} \] Input:
integrate((a+b*arctanh(c*x^2))/x^7,x, algorithm="giac")
Output:
-1/12*b*c^3*log(c^2*x^4 - 1) + 1/3*b*c^3*log(x) - 1/12*b*log(-(c*x^2 + 1)/ (c*x^2 - 1))/x^6 - 1/12*(b*c*x^2 + 2*a)/x^6
Time = 3.77 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=\frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^4-1\right )}{12}-\frac {a}{6\,x^6}-\frac {b\,c}{12\,x^4}-\frac {b\,\ln \left (c\,x^2+1\right )}{12\,x^6}+\frac {b\,\ln \left (1-c\,x^2\right )}{12\,x^6} \] Input:
int((a + b*atanh(c*x^2))/x^7,x)
Output:
(b*c^3*log(x))/3 - (b*c^3*log(c^2*x^4 - 1))/12 - a/(6*x^6) - (b*c)/(12*x^4 ) - (b*log(c*x^2 + 1))/(12*x^6) + (b*log(1 - c*x^2))/(12*x^6)
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.21 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^7} \, dx=\frac {2 \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{3} x^{6}-2 \mathit {atanh} \left (c \,x^{2}\right ) b -2 \,\mathrm {log}\left (c \,x^{2}+1\right ) b \,c^{3} x^{6}+4 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{6}-2 a -b c \,x^{2}}{12 x^{6}} \] Input:
int((a+b*atanh(c*x^2))/x^7,x)
Output:
(2*atanh(c*x**2)*b*c**3*x**6 - 2*atanh(c*x**2)*b - 2*log(c*x**2 + 1)*b*c** 3*x**6 + 4*log(x)*b*c**3*x**6 - 2*a - b*c*x**2)/(12*x**6)