Integrand size = 14, antiderivative size = 43 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{4} b c x^2+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{4} b c^2 \text {arctanh}\left (\frac {x^2}{c}\right ) \] Output:
1/4*b*c*x^2+1/4*x^4*(a+b*arctanh(c/x^2))-1/4*b*c^2*arctanh(x^2/c)
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.44 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{4} b c x^2+\frac {a x^4}{4}+\frac {1}{4} b x^4 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{8} b c^2 \log \left (-c+x^2\right )-\frac {1}{8} b c^2 \log \left (c+x^2\right ) \] Input:
Integrate[x^3*(a + b*ArcTanh[c/x^2]),x]
Output:
(b*c*x^2)/4 + (a*x^4)/4 + (b*x^4*ArcTanh[c/x^2])/4 + (b*c^2*Log[-c + x^2]) /8 - (b*c^2*Log[c + x^2])/8
Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91, 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, 262, 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.
\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} b c \int \frac {x}{1-\frac {c^2}{x^4}}dx+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{2} b c \int \frac {x^5}{x^4-c^2}dx+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{4} b c \int -\frac {x^4}{c^2-x^4}dx^2+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{4} b c \int \frac {x^4}{c^2-x^4}dx^2\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{4} b c \left (x^2-c^2 \int \frac {1}{c^2-x^4}dx^2\right )+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} b c \left (x^2-c \text {arctanh}\left (\frac {x^2}{c}\right )\right )\) |
Input:
Int[x^3*(a + b*ArcTanh[c/x^2]),x]
Output:
(x^4*(a + b*ArcTanh[c/x^2]))/4 + (b*c*(x^2 - c*ArcTanh[x^2/c]))/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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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 = 0.55 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05
method | result | size |
parallelrisch | \(\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}+\frac {x^{4} a}{4}+\frac {b c \,x^{2}}{4}-\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,c^{2}}{4}+\frac {a \,c^{2}}{4}\) | \(45\) |
derivativedivides | \(\frac {x^{4} a}{4}+\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}+\frac {b c \,x^{2}}{4}+\frac {b \,c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{8}-\frac {b \,c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{8}\) | \(55\) |
default | \(\frac {x^{4} a}{4}+\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}+\frac {b c \,x^{2}}{4}+\frac {b \,c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{8}-\frac {b \,c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{8}\) | \(55\) |
parts | \(\frac {x^{4} a}{4}+\frac {\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}+\frac {b c \,x^{2}}{4}+\frac {b \,c^{2} \ln \left (\frac {c}{x^{2}}-1\right )}{8}-\frac {b \,c^{2} \ln \left (1+\frac {c}{x^{2}}\right )}{8}\) | \(55\) |
orering | \(-\frac {5 \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right ) \left (-x^{4}+c^{2}\right )}{8}+\frac {\left (x^{2}+c \right ) \left (-x^{2}+c \right ) \left (3 x^{2} \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right )-\frac {2 b c}{1-\frac {c^{2}}{x^{4}}}\right )}{8 x^{2}}\) | \(72\) |
risch | \(\text {Expression too large to display}\) | \(4322\) |
Input:
int(x^3*(a+b*arctanh(c/x^2)),x,method=_RETURNVERBOSE)
Output:
1/4*arctanh(c/x^2)*b*x^4+1/4*x^4*a+1/4*b*c*x^2-1/4*arctanh(c/x^2)*b*c^2+1/ 4*a*c^2
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{4} \, b c x^{2} + \frac {1}{8} \, {\left (b x^{4} - b c^{2}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) \] Input:
integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="fricas")
Output:
1/4*a*x^4 + 1/4*b*c*x^2 + 1/8*(b*x^4 - b*c^2)*log((x^2 + c)/(x^2 - c))
Time = 1.69 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a x^{4}}{4} - \frac {b c^{2} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4} + \frac {b c x^{2}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{4} \] Input:
integrate(x**3*(a+b*atanh(c/x**2)),x)
Output:
a*x**4/4 - b*c**2*atanh(c/x**2)/4 + b*c*x**2/4 + b*x**4*atanh(c/x**2)/4
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.14 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (2 \, x^{2} - c \log \left (x^{2} + c\right ) + c \log \left (x^{2} - c\right )\right )} c\right )} b \] Input:
integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="maxima")
Output:
1/4*a*x^4 + 1/8*(2*x^4*arctanh(c/x^2) + (2*x^2 - c*log(x^2 + c) + c*log(x^ 2 - c))*c)*b
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (37) = 74\).
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 3.77 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {\frac {{\left (x^{2} + c\right )} b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{{\left (x^{2} - c\right )} {\left (\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1\right )}} + \frac {\frac {2 \, {\left (x^{2} + c\right )} a c^{3}}{x^{2} - c} + \frac {{\left (x^{2} + c\right )} b c^{3}}{x^{2} - c} - b c^{3}}{\frac {{\left (x^{2} + c\right )}^{2}}{{\left (x^{2} - c\right )}^{2}} - \frac {2 \, {\left (x^{2} + c\right )}}{x^{2} - c} + 1}}{2 \, c} \] Input:
integrate(x^3*(a+b*arctanh(c/x^2)),x, algorithm="giac")
Output:
1/2*((x^2 + c)*b*c^3*log((x^2 + c)/(x^2 - c))/((x^2 - c)*((x^2 + c)^2/(x^2 - c)^2 - 2*(x^2 + c)/(x^2 - c) + 1)) + (2*(x^2 + c)*a*c^3/(x^2 - c) + (x^ 2 + c)*b*c^3/(x^2 - c) - b*c^3)/((x^2 + c)^2/(x^2 - c)^2 - 2*(x^2 + c)/(x^ 2 - c) + 1))/c
Time = 3.89 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.33 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a\,x^4}{4}+\frac {b\,x^4\,\ln \left (x^2+c\right )}{8}+\frac {b\,c\,x^2}{4}-\frac {b\,x^4\,\ln \left (x^2-c\right )}{8}+\frac {b\,c^2\,\mathrm {atan}\left (\frac {x^2\,1{}\mathrm {i}}{c}\right )\,1{}\mathrm {i}}{4} \] Input:
int(x^3*(a + b*atanh(c/x^2)),x)
Output:
(a*x^4)/4 + (b*x^4*log(c + x^2))/8 + (b*c^2*atan((x^2*1i)/c)*1i)/4 + (b*c* x^2)/4 - (b*x^4*log(x^2 - c))/8
Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int x^3 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=-\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,c^{2}}{4}+\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,x^{4}}{4}+\frac {a \,x^{4}}{4}+\frac {b c \,x^{2}}{4} \] Input:
int(x^3*(a+b*atanh(c/x^2)),x)
Output:
( - atanh(c/x**2)*b*c**2 + atanh(c/x**2)*b*x**4 + a*x**4 + b*c*x**2)/4