Integrand size = 16, antiderivative size = 49 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {b x^{3/2}}{3 c}-\frac {b \text {arctanh}\left (c x^{3/2}\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \] Output:
1/3*b*x^(3/2)/c-1/3*b*arctanh(c*x^(3/2))/c^2+1/3*x^3*(a+b*arctanh(c*x^(3/2 )))
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {b x^{3/2}}{3 c}+\frac {a x^3}{3}+\frac {1}{3} b x^3 \text {arctanh}\left (c x^{3/2}\right )+\frac {b \log \left (1-c x^{3/2}\right )}{6 c^2}-\frac {b \log \left (1+c x^{3/2}\right )}{6 c^2} \] Input:
Integrate[x^2*(a + b*ArcTanh[c*x^(3/2)]),x]
Output:
(b*x^(3/2))/(3*c) + (a*x^3)/3 + (b*x^3*ArcTanh[c*x^(3/2)])/3 + (b*Log[1 - c*x^(3/2)])/(6*c^2) - (b*Log[1 + c*x^(3/2)])/(6*c^2)
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6452, 843, 851, 807, 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.
\(\displaystyle \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )-\frac {1}{2} b c \int \frac {x^{7/2}}{1-c^2 x^3}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )-\frac {1}{2} b c \left (\frac {\int \frac {\sqrt {x}}{1-c^2 x^3}dx}{c^2}-\frac {2 x^{3/2}}{3 c^2}\right )\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )-\frac {1}{2} b c \left (\frac {2 \int \frac {x}{1-c^2 x^3}d\sqrt {x}}{c^2}-\frac {2 x^{3/2}}{3 c^2}\right )\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )-\frac {1}{2} b c \left (\frac {2 \int \frac {1}{1-c^2 x}dx^{3/2}}{3 c^2}-\frac {2 x^{3/2}}{3 c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )-\frac {1}{2} b c \left (\frac {2 \text {arctanh}\left (c x^{3/2}\right )}{3 c^3}-\frac {2 x^{3/2}}{3 c^2}\right )\) |
Input:
Int[x^2*(a + b*ArcTanh[c*x^(3/2)]),x]
Output:
(x^3*(a + b*ArcTanh[c*x^(3/2)]))/3 - (b*c*((-2*x^(3/2))/(3*c^2) + (2*ArcTa nh[c*x^(3/2)])/(3*c^3)))/2
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[(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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && FractionQ[m] && IntBinomialQ[a, b, c, n, 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 = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12
method | result | size |
parts | \(\frac {a \,x^{3}}{3}+\frac {2 b \left (\frac {c^{2} x^{3} \operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{2}+\frac {c \,x^{\frac {3}{2}}}{2}+\frac {\ln \left (c \,x^{\frac {3}{2}}-1\right )}{4}-\frac {\ln \left (c \,x^{\frac {3}{2}}+1\right )}{4}\right )}{3 c^{2}}\) | \(55\) |
derivativedivides | \(\frac {\frac {a \,c^{2} x^{3}}{3}+\frac {2 b \left (\frac {c^{2} x^{3} \operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{2}+\frac {c \,x^{\frac {3}{2}}}{2}+\frac {\ln \left (c \,x^{\frac {3}{2}}-1\right )}{4}-\frac {\ln \left (c \,x^{\frac {3}{2}}+1\right )}{4}\right )}{3}}{c^{2}}\) | \(59\) |
default | \(\frac {\frac {a \,c^{2} x^{3}}{3}+\frac {2 b \left (\frac {c^{2} x^{3} \operatorname {arctanh}\left (c \,x^{\frac {3}{2}}\right )}{2}+\frac {c \,x^{\frac {3}{2}}}{2}+\frac {\ln \left (c \,x^{\frac {3}{2}}-1\right )}{4}-\frac {\ln \left (c \,x^{\frac {3}{2}}+1\right )}{4}\right )}{3}}{c^{2}}\) | \(59\) |
Input:
int(x^2*(a+b*arctanh(c*x^(3/2))),x,method=_RETURNVERBOSE)
Output:
1/3*a*x^3+2/3*b/c^2*(1/2*c^2*x^3*arctanh(c*x^(3/2))+1/2*c*x^(3/2)+1/4*ln(c *x^(3/2)-1)-1/4*ln(c*x^(3/2)+1))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.31 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {2 \, a c^{2} x^{3} + 2 \, b c x^{\frac {3}{2}} + {\left (b c^{2} x^{3} - b\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )}{6 \, c^{2}} \] Input:
integrate(x^2*(a+b*arctanh(c*x^(3/2))),x, algorithm="fricas")
Output:
1/6*(2*a*c^2*x^3 + 2*b*c*x^(3/2) + (b*c^2*x^3 - b)*log(-(c^2*x^3 + 2*c*x^( 3/2) + 1)/(c^2*x^3 - 1)))/c^2
Timed out. \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x**2*(a+b*atanh(c*x**(3/2))),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x^{\frac {3}{2}}\right ) + c {\left (\frac {2 \, x^{\frac {3}{2}}}{c^{2}} - \frac {\log \left (c x^{\frac {3}{2}} + 1\right )}{c^{3}} + \frac {\log \left (c x^{\frac {3}{2}} - 1\right )}{c^{3}}\right )}\right )} b \] Input:
integrate(x^2*(a+b*arctanh(c*x^(3/2))),x, algorithm="maxima")
Output:
1/3*a*x^3 + 1/6*(2*x^3*arctanh(c*x^(3/2)) + c*(2*x^(3/2)/c^2 - log(c*x^(3/ 2) + 1)/c^3 + log(c*x^(3/2) - 1)/c^3))*b
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (37) = 74\).
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.98 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {2}{3} \, b c {\left (\frac {1}{c^{3} {\left (\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1} - 1\right )}} + \frac {{\left (c x^{\frac {3}{2}} + 1\right )} \log \left (-\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1}\right )}{{\left (c x^{\frac {3}{2}} - 1\right )} c^{3} {\left (\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1} - 1\right )}^{2}}\right )} \] Input:
integrate(x^2*(a+b*arctanh(c*x^(3/2))),x, algorithm="giac")
Output:
1/3*a*x^3 + 2/3*b*c*(1/(c^3*((c*x^(3/2) + 1)/(c*x^(3/2) - 1) - 1)) + (c*x^ (3/2) + 1)*log(-(c*x^(3/2) + 1)/(c*x^(3/2) - 1))/((c*x^(3/2) - 1)*c^3*((c* x^(3/2) + 1)/(c*x^(3/2) - 1) - 1)^2))
Time = 4.42 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.24 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {a\,x^3}{3}+\frac {b\,x^{3/2}}{3\,c}+\frac {b\,\ln \left (\frac {c\,x^{3/2}-1}{c\,x^{3/2}+1}\right )}{6\,c^2}+\frac {b\,x^3\,\ln \left (c\,x^{3/2}+1\right )}{6}+\frac {b\,x^3\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,c^2\,x^3-2\right )}-\frac {b\,c^2\,x^6\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,c^2\,x^3-2\right )} \] Input:
int(x^2*(a + b*atanh(c*x^(3/2))),x)
Output:
(a*x^3)/3 + (b*x^(3/2))/(3*c) + (b*log((c*x^(3/2) - 1)/(c*x^(3/2) + 1)))/( 6*c^2) + (b*x^3*log(c*x^(3/2) + 1))/6 + (b*x^3*log(1 - c*x^(3/2)))/(3*(2*c ^2*x^3 - 2)) - (b*c^2*x^6*log(1 - c*x^(3/2)))/(3*(2*c^2*x^3 - 2))
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right ) \, dx=\frac {\mathit {atanh} \left (\sqrt {x}\, c x \right ) b \,c^{2} x^{3}-\mathit {atanh} \left (\sqrt {x}\, c x \right ) b +\sqrt {x}\, b c x +a \,c^{2} x^{3}}{3 c^{2}} \] Input:
int(x^2*(a+b*atanh(c*x^(3/2))),x)
Output:
(atanh(sqrt(x)*c*x)*b*c**2*x**3 - atanh(sqrt(x)*c*x)*b + sqrt(x)*b*c*x + a *c**2*x**3)/(3*c**2)