Integrand size = 12, antiderivative size = 48 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=-\frac {b c}{12 x^3}-\frac {b c^3}{4 x}+\frac {1}{4} b c^4 \text {arctanh}(c x)-\frac {a+b \text {arctanh}(c x)}{4 x^4} \] Output:
-1/12*b*c/x^3-1/4*b*c^3/x+1/4*b*c^4*arctanh(c*x)-1/4*(a+b*arctanh(c*x))/x^ 4
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.46 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {b c}{12 x^3}-\frac {b c^3}{4 x}-\frac {b \text {arctanh}(c x)}{4 x^4}-\frac {1}{8} b c^4 \log (1-c x)+\frac {1}{8} b c^4 \log (1+c x) \] Input:
Integrate[(a + b*ArcTanh[c*x])/x^5,x]
Output:
-1/4*a/x^4 - (b*c)/(12*x^3) - (b*c^3)/(4*x) - (b*ArcTanh[c*x])/(4*x^4) - ( b*c^4*Log[1 - c*x])/8 + (b*c^4*Log[1 + c*x])/8
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6452, 264, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} b c \int \frac {1}{x^4 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \left (c^2 \int \frac {1}{1-c^2 x^2}dx-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} b c \left (c^2 \left (c \text {arctanh}(c x)-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}(c x)}{4 x^4}\) |
Input:
Int[(a + b*ArcTanh[c*x])/x^5,x]
Output:
-1/4*(a + b*ArcTanh[c*x])/x^4 + (b*c*(-1/3*1/x^3 + c^2*(-x^(-1) + c*ArcTan h[c*x])))/4
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[((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 = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(-\frac {-3 b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}+3 b \,c^{3} x^{3}+b c x +3 b \,\operatorname {arctanh}\left (c x \right )+3 a}{12 x^{4}}\) | \(43\) |
parts | \(-\frac {a}{4 x^{4}}+b \,c^{4} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\) | \(58\) |
derivativedivides | \(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\right )\) | \(62\) |
default | \(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\ln \left (c x +1\right )}{8}-\frac {\ln \left (c x -1\right )}{8}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{4 c x}\right )\right )\) | \(62\) |
risch | \(-\frac {b \ln \left (c x +1\right )}{8 x^{4}}+\frac {3 b \,c^{4} \ln \left (-c x -1\right ) x^{4}-3 b \,x^{4} \ln \left (-c x +1\right ) c^{4}-6 b \,c^{3} x^{3}-2 b c x +3 b \ln \left (-c x +1\right )-6 a}{24 x^{4}}\) | \(79\) |
orering | \(\frac {\left (\frac {3}{2} c^{4} x^{5}-\frac {5}{6} c^{2} x^{3}-\frac {2}{3} x \right ) \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )}{x^{5}}+\frac {\left (3 c^{2} x^{2}+1\right ) \left (c x -1\right ) \left (c x +1\right ) x^{2} \left (\frac {b c}{\left (-c^{2} x^{2}+1\right ) x^{5}}-\frac {5 \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )}{x^{6}}\right )}{12}\) | \(91\) |
Input:
int((a+b*arctanh(c*x))/x^5,x,method=_RETURNVERBOSE)
Output:
-1/12*(-3*b*arctanh(c*x)*x^4*c^4+3*b*c^3*x^3+b*c*x+3*b*arctanh(c*x)+3*a)/x ^4
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=-\frac {6 \, b c^{3} x^{3} + 2 \, b c x - 3 \, {\left (b c^{4} x^{4} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a}{24 \, x^{4}} \] Input:
integrate((a+b*arctanh(c*x))/x^5,x, algorithm="fricas")
Output:
-1/24*(6*b*c^3*x^3 + 2*b*c*x - 3*(b*c^4*x^4 - b)*log(-(c*x + 1)/(c*x - 1)) + 6*a)/x^4
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=- \frac {a}{4 x^{4}} + \frac {b c^{4} \operatorname {atanh}{\left (c x \right )}}{4} - \frac {b c^{3}}{4 x} - \frac {b c}{12 x^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} \] Input:
integrate((a+b*atanh(c*x))/x**5,x)
Output:
-a/(4*x**4) + b*c**4*atanh(c*x)/4 - b*c**3/(4*x) - b*c/(12*x**3) - b*atanh (c*x)/(4*x**4)
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=\frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \] Input:
integrate((a+b*arctanh(c*x))/x^5,x, algorithm="maxima")
Output:
1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b - 1/4*a/x^4
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (40) = 80\).
Time = 0.13 (sec) , antiderivative size = 292, normalized size of antiderivative = 6.08 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=\frac {1}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c x + 1\right )}^{3} b c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {{\left (c x + 1\right )} b c^{3}}{c x - 1}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {6 \, {\left (c x + 1\right )}^{3} a c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a c^{3}}{c x - 1} + \frac {3 \, {\left (c x + 1\right )}^{3} b c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} b c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b c^{3}}{c x - 1} + 2 \, b c^{3}}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} \] Input:
integrate((a+b*arctanh(c*x))/x^5,x, algorithm="giac")
Output:
1/3*c*(3*((c*x + 1)^3*b*c^3/(c*x - 1)^3 + (c*x + 1)*b*c^3/(c*x - 1))*log(- (c*x + 1)/(c*x - 1))/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1) + (6*(c*x + 1)^3* a*c^3/(c*x - 1)^3 + 6*(c*x + 1)*a*c^3/(c*x - 1) + 3*(c*x + 1)^3*b*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*b*c^3/(c*x - 1)^2 + 5*(c*x + 1)*b*c^3/(c*x - 1) + 2*b*c^3)/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1 )^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1))
Time = 3.99 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=\frac {b\,\ln \left (1-c\,x\right )}{8\,x^4}-\frac {b\,\ln \left (c\,x+1\right )}{8\,x^4}-\frac {b\,c^3\,x^3+\frac {b\,c\,x}{3}+a}{4\,x^4}-\frac {b\,c^4\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \] Input:
int((a + b*atanh(c*x))/x^5,x)
Output:
(b*log(1 - c*x))/(8*x^4) - (b*c^4*atan(c*x*1i)*1i)/4 - (b*log(c*x + 1))/(8 *x^4) - (a + b*c^3*x^3 + (b*c*x)/3)/(4*x^4)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \text {arctanh}(c x)}{x^5} \, dx=\frac {3 \mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-3 \mathit {atanh} \left (c x \right ) b -3 a -3 b \,c^{3} x^{3}-b c x}{12 x^{4}} \] Input:
int((a+b*atanh(c*x))/x^5,x)
Output:
(3*atanh(c*x)*b*c**4*x**4 - 3*atanh(c*x)*b - 3*a - 3*b*c**3*x**3 - b*c*x)/ (12*x**4)