Integrand size = 30, antiderivative size = 28 \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=-\frac {2 \text {arctanh}\left (\frac {\left (1+x^3\right )^{3/2}}{\sqrt {3} x^3}\right )}{3 \sqrt {3}} \] Output:
-2/9*arctanh(1/3*(x^3+1)^(3/2)*3^(1/2)/x^3)*3^(1/2)
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {3} x^3}{\left (1+x^3\right )^{3/2}}\right )}{3 \sqrt {3}} \] Input:
Integrate[(x^2*(-2 + x^3)*Sqrt[1 + x^3])/(1 + 3*x^3 + x^9),x]
Output:
(-2*ArcTanh[(Sqrt[3]*x^3)/(1 + x^3)^(3/2)])/(3*Sqrt[3])
Time = 0.61 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {7266, 25, 7267, 2520, 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.
\(\displaystyle \int \frac {x^2 \left (x^3-2\right ) \sqrt {x^3+1}}{x^9+3 x^3+1} \, dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{3} \int -\frac {\left (2-x^3\right ) \sqrt {x^3+1}}{x^9+3 x^3+1}dx^3\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {\left (2-x^3\right ) \sqrt {x^3+1}}{x^9+3 x^3+1}dx^3\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {2}{3} \int \frac {x^6 \left (3-x^6\right )}{-x^{18}+3 x^{12}-6 x^6+3}d\sqrt {x^3+1}\) |
\(\Big \downarrow \) 2520 |
\(\displaystyle 2 \int \frac {1}{3-9 x^6}d\frac {x^9}{3 \left (1-x^6\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \text {arctanh}\left (\frac {x^9}{\sqrt {3} \left (1-x^6\right )}\right )}{3 \sqrt {3}}\) |
Input:
Int[(x^2*(-2 + x^3)*Sqrt[1 + x^3])/(1 + 3*x^3 + x^9),x]
Output:
(2*ArcTanh[x^9/(Sqrt[3]*(1 - x^6))])/(3*Sqrt[3])
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_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.) *(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_Symbol] :> Simp[A^2*((m - n + 1)/(m + 1) ) Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m - n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && EqQ[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(21)=42\).
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.29
method | result | size |
default | \(\frac {\sqrt {3}\, \left (\ln \left (-\sqrt {3}\, x^{3}+\sqrt {x^{3}+1}\, x^{3}+\sqrt {x^{3}+1}\right )-\ln \left (\sqrt {3}\, x^{3}+\sqrt {x^{3}+1}\, x^{3}+\sqrt {x^{3}+1}\right )\right )}{9}\) | \(64\) |
pseudoelliptic | \(\frac {\sqrt {3}\, \left (\ln \left (-\sqrt {3}\, x^{3}+\sqrt {x^{3}+1}\, x^{3}+\sqrt {x^{3}+1}\right )-\ln \left (\sqrt {3}\, x^{3}+\sqrt {x^{3}+1}\, x^{3}+\sqrt {x^{3}+1}\right )\right )}{9}\) | \(64\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{9}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{6}-6 \sqrt {x^{3}+1}\, x^{6}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{3}-6 \sqrt {x^{3}+1}\, x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{x^{9}+3 x^{3}+1}\right )}{9}\) | \(87\) |
elliptic | \(\frac {\sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{9}+3 \textit {\_Z}^{3}+1\right )}{\sum }\frac {\left (-\underline {\hspace {1.25 ex}}\alpha ^{6}+\underline {\hspace {1.25 ex}}\alpha ^{3}-1\right ) \left (-\underline {\hspace {1.25 ex}}\alpha ^{8}+\underline {\hspace {1.25 ex}}\alpha ^{7}-\underline {\hspace {1.25 ex}}\alpha ^{6}+\underline {\hspace {1.25 ex}}\alpha ^{5}-\underline {\hspace {1.25 ex}}\alpha ^{4}+\underline {\hspace {1.25 ex}}\alpha ^{3}-4 \underline {\hspace {1.25 ex}}\alpha ^{2}+4 \underline {\hspace {1.25 ex}}\alpha -4\right ) \left (3-i \sqrt {3}\right ) \sqrt {\frac {1+x}{3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x -i \sqrt {3}}{-3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x +i \sqrt {3}}{-3+i \sqrt {3}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {\underline {\hspace {1.25 ex}}\alpha ^{8}}{2}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{7}}{2}+\frac {\underline {\hspace {1.25 ex}}\alpha ^{6}}{2}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{5}}{2}+\frac {\underline {\hspace {1.25 ex}}\alpha ^{4}}{2}-\frac {\underline {\hspace {1.25 ex}}\alpha ^{3}}{2}+2 \underline {\hspace {1.25 ex}}\alpha ^{2}-2 \underline {\hspace {1.25 ex}}\alpha +2+\frac {i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{7}}{6}-\frac {2 i \sqrt {3}}{3}-\frac {i \underline {\hspace {1.25 ex}}\alpha ^{4} \sqrt {3}}{6}-\frac {i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{6}}{6}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {3}}{6}+\frac {i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{5}}{6}-\frac {2 i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{3}-\frac {i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{8}}{6}+\frac {2 i \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha }{3}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\right )}{27}\) | \(298\) |
Input:
int(x^2*(x^3-2)*(x^3+1)^(1/2)/(x^9+3*x^3+1),x,method=_RETURNVERBOSE)
Output:
1/9*3^(1/2)*(ln(-3^(1/2)*x^3+(x^3+1)^(1/2)*x^3+(x^3+1)^(1/2))-ln(3^(1/2)*x ^3+(x^3+1)^(1/2)*x^3+(x^3+1)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (21) = 42\).
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=\frac {1}{9} \, \sqrt {3} \log \left (\frac {x^{9} + 6 \, x^{6} + 3 \, x^{3} - 2 \, \sqrt {3} {\left (x^{6} + x^{3}\right )} \sqrt {x^{3} + 1} + 1}{x^{9} + 3 \, x^{3} + 1}\right ) \] Input:
integrate(x^2*(x^3-2)*(x^3+1)^(1/2)/(x^9+3*x^3+1),x, algorithm="fricas")
Output:
1/9*sqrt(3)*log((x^9 + 6*x^6 + 3*x^3 - 2*sqrt(3)*(x^6 + x^3)*sqrt(x^3 + 1) + 1)/(x^9 + 3*x^3 + 1))
Timed out. \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=\text {Timed out} \] Input:
integrate(x**2*(x**3-2)*(x**3+1)**(1/2)/(x**9+3*x**3+1),x)
Output:
Timed out
\[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=\int { \frac {\sqrt {x^{3} + 1} {\left (x^{3} - 2\right )} x^{2}}{x^{9} + 3 \, x^{3} + 1} \,d x } \] Input:
integrate(x^2*(x^3-2)*(x^3+1)^(1/2)/(x^9+3*x^3+1),x, algorithm="maxima")
Output:
integrate(sqrt(x^3 + 1)*(x^3 - 2)*x^2/(x^9 + 3*x^3 + 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (21) = 42\).
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07 \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=-\frac {1}{9} \, \sqrt {3} \log \left ({\left | \sqrt {3} {\left (x^{3} + 1\right )} + {\left (x^{3} + 1\right )}^{\frac {3}{2}} - \sqrt {3} \right |}\right ) + \frac {1}{9} \, \sqrt {3} \log \left ({\left | -\sqrt {3} {\left (x^{3} + 1\right )} + {\left (x^{3} + 1\right )}^{\frac {3}{2}} + \sqrt {3} \right |}\right ) \] Input:
integrate(x^2*(x^3-2)*(x^3+1)^(1/2)/(x^9+3*x^3+1),x, algorithm="giac")
Output:
-1/9*sqrt(3)*log(abs(sqrt(3)*(x^3 + 1) + (x^3 + 1)^(3/2) - sqrt(3))) + 1/9 *sqrt(3)*log(abs(-sqrt(3)*(x^3 + 1) + (x^3 + 1)^(3/2) + sqrt(3)))
Time = 7.71 (sec) , antiderivative size = 236, normalized size of antiderivative = 8.43 \[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=\sum _{k=1}^9\frac {\sqrt {6}\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x+1\right )}\,\Pi \left (\frac {3+\sqrt {3}\,1{}\mathrm {i}}{2\,\left (\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )+1\right )};\mathrm {asin}\left (\frac {\sqrt {6}\,\sqrt {-\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x+1\right )}}{6}\right )\middle |\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-{\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )}^6+{\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )}^3+2\right )\,{\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )}^2\,\sqrt {3-3\,x+\sqrt {3}\,x\,1{}\mathrm {i}+\sqrt {3}\,1{}\mathrm {i}}\,\sqrt {3-3\,x-\sqrt {3}\,x\,1{}\mathrm {i}-\sqrt {3}\,1{}\mathrm {i}}}{162\,\left (\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )+1\right )\,\sqrt {x^3+1}\,\left ({\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )}^8+{\mathrm {root}\left (z^9+3\,z^3+1,z,k\right )}^2\right )} \] Input:
int((x^2*(x^3 + 1)^(1/2)*(x^3 - 2))/(3*x^3 + x^9 + 1),x)
Output:
symsum((6^(1/2)*((3^(1/2)*1i)/2 + 3/2)*(-(3^(1/2)*1i - 3)*(x + 1))^(1/2)*e llipticPi((3^(1/2)*1i + 3)/(2*(root(z^9 + 3*z^3 + 1, z, k) + 1)), asin((6^ (1/2)*(-(3^(1/2)*1i - 3)*(x + 1))^(1/2))/6), (3^(1/2)*1i)/2 + 1/2)*(root(z ^9 + 3*z^3 + 1, z, k)^3 - root(z^9 + 3*z^3 + 1, z, k)^6 + 2)*root(z^9 + 3* z^3 + 1, z, k)^2*(3^(1/2)*x*1i - 3*x + 3^(1/2)*1i + 3)^(1/2)*(3 - 3^(1/2)* x*1i - 3^(1/2)*1i - 3*x)^(1/2))/(162*(root(z^9 + 3*z^3 + 1, z, k) + 1)*(x^ 3 + 1)^(1/2)*(root(z^9 + 3*z^3 + 1, z, k)^2 + root(z^9 + 3*z^3 + 1, z, k)^ 8)), k, 1, 9)
\[ \int \frac {x^2 \left (-2+x^3\right ) \sqrt {1+x^3}}{1+3 x^3+x^9} \, dx=\int \frac {\sqrt {x^{3}+1}\, x^{5}}{x^{9}+3 x^{3}+1}d x -2 \left (\int \frac {\sqrt {x^{3}+1}\, x^{2}}{x^{9}+3 x^{3}+1}d x \right ) \] Input:
int(x^2*(x^3-2)*(x^3+1)^(1/2)/(x^9+3*x^3+1),x)
Output:
int((sqrt(x**3 + 1)*x**5)/(x**9 + 3*x**3 + 1),x) - 2*int((sqrt(x**3 + 1)*x **2)/(x**9 + 3*x**3 + 1),x)