Integrand size = 14, antiderivative size = 50 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} b c^3 x+\frac {1}{12} b c x^3+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )-\frac {1}{4} b c^4 \text {arctanh}\left (\frac {x}{c}\right ) \] Output:
1/4*b*c^3*x+1/12*b*c*x^3+1/4*x^4*(a+b*arctanh(c/x))-1/4*b*c^4*arctanh(x/c)
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.34 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} b c^3 x+\frac {1}{12} b c x^3+\frac {a x^4}{4}+\frac {1}{4} b x^4 \text {arctanh}\left (\frac {c}{x}\right )+\frac {1}{8} b c^4 \log (-c+x)-\frac {1}{8} b c^4 \log (c+x) \] Input:
Integrate[x^3*(a + b*ArcTanh[c/x]),x]
Output:
(b*c^3*x)/4 + (b*c*x^3)/12 + (a*x^4)/4 + (b*x^4*ArcTanh[c/x])/4 + (b*c^4*L og[-c + x])/8 - (b*c^4*Log[c + x])/8
Time = 0.38 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6452, 795, 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.
\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} b c \int \frac {x^2}{1-\frac {c^2}{x^2}}dx+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{4} b c \int \frac {x^4}{x^2-c^2}dx+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{4} b c \int \left (\frac {c^4}{x^2-c^2}+c^2+x^2\right )dx+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right )+\frac {1}{4} b c \left (c^3 \left (-\text {arctanh}\left (\frac {x}{c}\right )\right )+c^2 x+\frac {x^3}{3}\right )\) |
Input:
Int[x^3*(a + b*ArcTanh[c/x]),x]
Output:
(x^4*(a + b*ArcTanh[c/x]))/4 + (b*c*(c^2*x + x^3/3 - c^3*ArcTanh[x/c]))/4
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[((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 = 0.56 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right ) b}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x}\right ) b \,c^{4}}{4}+\frac {x^{4} a}{4}+\frac {b c \,x^{3}}{12}+\frac {b \,c^{3} x}{4}\) | \(46\) |
parts | \(\frac {x^{4} a}{4}-b \,c^{4} \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\) | \(63\) |
derivativedivides | \(-c^{4} \left (-\frac {a \,x^{4}}{4 c^{4}}+b \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\right )\) | \(67\) |
default | \(-c^{4} \left (-\frac {a \,x^{4}}{4 c^{4}}+b \left (-\frac {x^{4} \operatorname {arctanh}\left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\ln \left (1+\frac {c}{x}\right )}{8}-\frac {\ln \left (\frac {c}{x}-1\right )}{8}-\frac {x^{3}}{12 c^{3}}-\frac {x}{4 c}\right )\right )\) | \(67\) |
orering | \(-\frac {\left (2 c^{4}-c^{2} x^{2}-x^{4}\right ) \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )\right )}{2}+\frac {\left (3 c^{2}+x^{2}\right ) \left (c -x \right ) \left (x +c \right ) \left (3 x^{2} \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x}\right )\right )-\frac {x b c}{1-\frac {c^{2}}{x^{2}}}\right )}{12 x^{2}}\) | \(88\) |
risch | \(\frac {b \,x^{4} \ln \left (x +c \right )}{8}-\frac {b \,x^{4} \ln \left (c -x \right )}{8}-\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )}{16}-\frac {i \pi b \,x^{4} \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{16}+\frac {i \pi b \,x^{4} \operatorname {csgn}\left (i \left (x +c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{16}+\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{8}-\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{2}}{16}-\frac {i \pi b \,x^{4}}{8}+\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (c -x \right )\right ) \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )}{16}-\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i \left (c -x \right )}{x}\right )^{3}}{16}+\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{2}}{16}-\frac {i \pi b \,x^{4} \operatorname {csgn}\left (\frac {i \left (x +c \right )}{x}\right )^{3}}{16}+\frac {x^{4} a}{4}+\frac {b \,c^{3} x}{4}+\frac {b c \,x^{3}}{12}-\frac {b \,c^{4} \ln \left (x +c \right )}{8}+\frac {b \,c^{4} \ln \left (x -c \right )}{8}\) | \(320\) |
Input:
int(x^3*(a+b*arctanh(c/x)),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*arctanh(c/x)*b-1/4*arctanh(c/x)*b*c^4+1/4*x^4*a+1/12*b*c*x^3+1/4*b *c^3*x
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} \, b c^{3} x + \frac {1}{12} \, b c x^{3} + \frac {1}{4} \, a x^{4} - \frac {1}{8} \, {\left (b c^{4} - b x^{4}\right )} \log \left (-\frac {c + x}{c - x}\right ) \] Input:
integrate(x^3*(a+b*arctanh(c/x)),x, algorithm="fricas")
Output:
1/4*b*c^3*x + 1/12*b*c*x^3 + 1/4*a*x^4 - 1/8*(b*c^4 - b*x^4)*log(-(c + x)/ (c - x))
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {a x^{4}}{4} - \frac {b c^{4} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{4} + \frac {b c^{3} x}{4} + \frac {b c x^{3}}{12} + \frac {b x^{4} \operatorname {atanh}{\left (\frac {c}{x} \right )}}{4} \] Input:
integrate(x**3*(a+b*atanh(c/x)),x)
Output:
a*x**4/4 - b*c**4*atanh(c/x)/4 + b*c**3*x/4 + b*c*x**3/12 + b*x**4*atanh(c /x)/4
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (\frac {c}{x}\right ) - {\left (3 \, c^{3} \log \left (c + x\right ) - 3 \, c^{3} \log \left (-c + x\right ) - 6 \, c^{2} x - 2 \, x^{3}\right )} c\right )} b \] Input:
integrate(x^3*(a+b*arctanh(c/x)),x, algorithm="maxima")
Output:
1/4*a*x^4 + 1/24*(6*x^4*arctanh(c/x) - (3*c^3*log(c + x) - 3*c^3*log(-c + x) - 6*c^2*x - 2*x^3)*c)*b
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (42) = 84\).
Time = 0.12 (sec) , antiderivative size = 262, normalized size of antiderivative = 5.24 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=-\frac {\frac {3 \, {\left (\frac {b {\left (c + x\right )}^{3} c^{5}}{{\left (c - x\right )}^{3}} + \frac {b {\left (c + x\right )} c^{5}}{c - x}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{4}}{{\left (c - x\right )}^{4}} + \frac {4 \, {\left (c + x\right )}^{3}}{{\left (c - x\right )}^{3}} + \frac {6 \, {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {4 \, {\left (c + x\right )}}{c - x} + 1} + \frac {2 \, b c^{5} + \frac {6 \, a {\left (c + x\right )}^{3} c^{5}}{{\left (c - x\right )}^{3}} + \frac {3 \, b {\left (c + x\right )}^{3} c^{5}}{{\left (c - x\right )}^{3}} + \frac {6 \, b {\left (c + x\right )}^{2} c^{5}}{{\left (c - x\right )}^{2}} + \frac {6 \, a {\left (c + x\right )} c^{5}}{c - x} + \frac {5 \, b {\left (c + x\right )} c^{5}}{c - x}}{\frac {{\left (c + x\right )}^{4}}{{\left (c - x\right )}^{4}} + \frac {4 \, {\left (c + x\right )}^{3}}{{\left (c - x\right )}^{3}} + \frac {6 \, {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {4 \, {\left (c + x\right )}}{c - x} + 1}}{3 \, c} \] Input:
integrate(x^3*(a+b*arctanh(c/x)),x, algorithm="giac")
Output:
-1/3*(3*(b*(c + x)^3*c^5/(c - x)^3 + b*(c + x)*c^5/(c - x))*log(-(c + x)/( c - x))/((c + x)^4/(c - x)^4 + 4*(c + x)^3/(c - x)^3 + 6*(c + x)^2/(c - x) ^2 + 4*(c + x)/(c - x) + 1) + (2*b*c^5 + 6*a*(c + x)^3*c^5/(c - x)^3 + 3*b *(c + x)^3*c^5/(c - x)^3 + 6*b*(c + x)^2*c^5/(c - x)^2 + 6*a*(c + x)*c^5/( c - x) + 5*b*(c + x)*c^5/(c - x))/((c + x)^4/(c - x)^4 + 4*(c + x)^3/(c - x)^3 + 6*(c + x)^2/(c - x)^2 + 4*(c + x)/(c - x) + 1))/c
Time = 3.62 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,c^4\,\mathrm {atanh}\left (\frac {c}{x}\right )}{4}+\frac {b\,x^4\,\mathrm {atanh}\left (\frac {c}{x}\right )}{4}+\frac {b\,c\,x^3}{12}+\frac {b\,c^3\,x}{4} \] Input:
int(x^3*(a + b*atanh(c/x)),x)
Output:
(a*x^4)/4 - (b*c^4*atanh(c/x))/4 + (b*x^4*atanh(c/x))/4 + (b*c*x^3)/12 + ( b*c^3*x)/4
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x}\right )\right ) \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b \,c^{4}}{4}+\frac {\mathit {atanh} \left (\frac {c}{x}\right ) b \,x^{4}}{4}+\frac {a \,x^{4}}{4}+\frac {b \,c^{3} x}{4}+\frac {b c \,x^{3}}{12} \] Input:
int(x^3*(a+b*atanh(c/x)),x)
Output:
( - 3*atanh(c/x)*b*c**4 + 3*atanh(c/x)*b*x**4 + 3*a*x**4 + 3*b*c**3*x + b* c*x**3)/12