Integrand size = 31, antiderivative size = 115 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=-\frac {3 \left (1+10 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {1}{2} \text {RootSum}\left [6-4 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-6 \log (x)+6 \log \left (\sqrt [3]{x+2 x^3}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{x+2 x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-2 \text {$\#$1}^2+\text {$\#$1}^5}\&\right ] \] Output:
Unintegrable
Time = 3.14 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.31 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\frac {\sqrt [3]{x+2 x^3} \left (-9 \sqrt [3]{1+2 x^2} \left (1+10 x^2\right )+4 x^{8/3} \text {RootSum}\left [6-4 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-12 \log (x)+18 \log \left (\sqrt [3]{1+2 x^2}-x^{2/3} \text {$\#$1}\right )+2 \log (x) \text {$\#$1}^3-3 \log \left (\sqrt [3]{1+2 x^2}-x^{2/3} \text {$\#$1}\right ) \text {$\#$1}^3}{-2 \text {$\#$1}^2+\text {$\#$1}^5}\&\right ]\right )}{24 x^3 \sqrt [3]{1+2 x^2}} \] Input:
Integrate[((1 + 2*x^2)*(x + 2*x^3)^(1/3))/(x^4*(1 + 2*x^4)),x]
Output:
((x + 2*x^3)^(1/3)*(-9*(1 + 2*x^2)^(1/3)*(1 + 10*x^2) + 4*x^(8/3)*RootSum[ 6 - 4*#1^3 + #1^6 & , (-12*Log[x] + 18*Log[(1 + 2*x^2)^(1/3) - x^(2/3)*#1] + 2*Log[x]*#1^3 - 3*Log[(1 + 2*x^2)^(1/3) - x^(2/3)*#1]*#1^3)/(-2*#1^2 + #1^5) & ]))/(24*x^3*(1 + 2*x^2)^(1/3))
Result contains complex when optimal does not.
Time = 1.48 (sec) , antiderivative size = 706, normalized size of antiderivative = 6.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2467, 1593, 1815, 1845, 27, 796, 7276, 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.
\(\displaystyle \int \frac {\left (2 x^2+1\right ) \sqrt [3]{2 x^3+x}}{x^4 \left (2 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [3]{2 x^3+x} \int \frac {\left (2 x^2+1\right )^{4/3}}{x^{11/3} \left (2 x^4+1\right )}dx}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 1593 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \int \frac {\left (2 x^2+1\right )^{4/3}}{x^3 \left (2 x^4+1\right )}d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 1815 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \int \frac {(2 x+1)^{4/3}}{x^{5/3} \left (2 x^2+1\right )}dx^{2/3}}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 1845 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \left (\int \frac {\sqrt [3]{2 x+1}}{x^{5/3}}dx^{2/3}+\int \frac {2 (1-x) \sqrt [3]{2 x+1}}{x^{2/3} \left (2 x^2+1\right )}dx^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \left (\int \frac {\sqrt [3]{2 x+1}}{x^{5/3}}dx^{2/3}+2 \int \frac {(1-x) \sqrt [3]{2 x+1}}{x^{2/3} \left (2 x^2+1\right )}dx^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \left (2 \int \frac {(1-x) \sqrt [3]{2 x+1}}{x^{2/3} \left (2 x^2+1\right )}dx^{2/3}-\frac {(2 x+1)^{4/3}}{4 x^{4/3}}\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \left (2 \int \left (\frac {x^{2/3} \sqrt [3]{2 x+1} (-2 x-1)}{2 x^2+1}+\frac {\sqrt [3]{2 x+1}}{x^{2/3}}\right )dx^{2/3}-\frac {(2 x+1)^{4/3}}{4 x^{4/3}}\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \sqrt [3]{2 x^3+x} \left (-\frac {(2 x+1)^{4/3}}{4 x^{4/3}}+2 \left (\frac {\left (3+2 i \sqrt {2}\right ) \arctan \left (\frac {\frac {2 \sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x+1}}+1}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}+\frac {\left (3-2 i \sqrt {2}\right ) \arctan \left (\frac {\frac {2 \sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x+1}}+1}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}-\frac {\sqrt [3]{2} \arctan \left (\frac {\frac {2 \sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\left (1-2 i \sqrt {2}\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{2-i \sqrt {2}} x^{2/3}}{\sqrt [3]{2 x+1}}}{\sqrt {3}}\right )}{2 \sqrt {3} \left (2-i \sqrt {2}\right )^{2/3}}-\frac {\left (1+2 i \sqrt {2}\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{2+i \sqrt {2}} x^{2/3}}{\sqrt [3]{2 x+1}}}{\sqrt {3}}\right )}{2 \sqrt {3} \left (2+i \sqrt {2}\right )^{2/3}}+\frac {\left (3+2 i \sqrt {2}\right ) \log \left (\sqrt [3]{2} x^{2/3}-\sqrt [3]{2 x+1}\right )}{6\ 2^{2/3}}+\frac {\left (3-2 i \sqrt {2}\right ) \log \left (\sqrt [3]{2} x^{2/3}-\sqrt [3]{2 x+1}\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2} x^{2/3}-\sqrt [3]{2 x+1}\right )}{2^{2/3}}-\frac {\left (1-2 i \sqrt {2}\right ) \log \left (-\sqrt [3]{2 x+1}+\sqrt [3]{2-i \sqrt {2}} x^{2/3}\right )}{4 \left (2-i \sqrt {2}\right )^{2/3}}-\frac {\left (1+2 i \sqrt {2}\right ) \log \left (-\sqrt [3]{2 x+1}+\sqrt [3]{2+i \sqrt {2}} x^{2/3}\right )}{4 \left (2+i \sqrt {2}\right )^{2/3}}-\frac {\sqrt [3]{2 x+1}}{\sqrt [3]{x}}+\frac {\left (1-2 i \sqrt {2}\right ) \log \left (2 x-i \sqrt {2}\right )}{12 \left (2-i \sqrt {2}\right )^{2/3}}+\frac {\left (1+2 i \sqrt {2}\right ) \log \left (2 x+i \sqrt {2}\right )}{12 \left (2+i \sqrt {2}\right )^{2/3}}\right )\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
Input:
Int[((1 + 2*x^2)*(x + 2*x^3)^(1/3))/(x^4*(1 + 2*x^4)),x]
Output:
(3*(x + 2*x^3)^(1/3)*(-1/4*(1 + 2*x)^(4/3)/x^(4/3) + 2*(-((1 + 2*x)^(1/3)/ x^(1/3)) - (2^(1/3)*ArcTan[(1 + (2*2^(1/3)*x^(2/3))/(1 + 2*x)^(1/3))/Sqrt[ 3]])/Sqrt[3] + ((3 - (2*I)*Sqrt[2])*ArcTan[(1 + (2*2^(1/3)*x^(2/3))/(1 + 2 *x)^(1/3))/Sqrt[3]])/(3*2^(2/3)*Sqrt[3]) + ((3 + (2*I)*Sqrt[2])*ArcTan[(1 + (2*2^(1/3)*x^(2/3))/(1 + 2*x)^(1/3))/Sqrt[3]])/(3*2^(2/3)*Sqrt[3]) - ((1 - (2*I)*Sqrt[2])*ArcTan[(1 + (2*(2 - I*Sqrt[2])^(1/3)*x^(2/3))/(1 + 2*x)^ (1/3))/Sqrt[3]])/(2*Sqrt[3]*(2 - I*Sqrt[2])^(2/3)) - ((1 + (2*I)*Sqrt[2])* ArcTan[(1 + (2*(2 + I*Sqrt[2])^(1/3)*x^(2/3))/(1 + 2*x)^(1/3))/Sqrt[3]])/( 2*Sqrt[3]*(2 + I*Sqrt[2])^(2/3)) + ((1 - (2*I)*Sqrt[2])*Log[(-I)*Sqrt[2] + 2*x])/(12*(2 - I*Sqrt[2])^(2/3)) + ((1 + (2*I)*Sqrt[2])*Log[I*Sqrt[2] + 2 *x])/(12*(2 + I*Sqrt[2])^(2/3)) - Log[2^(1/3)*x^(2/3) - (1 + 2*x)^(1/3)]/2 ^(2/3) + ((3 - (2*I)*Sqrt[2])*Log[2^(1/3)*x^(2/3) - (1 + 2*x)^(1/3)])/(6*2 ^(2/3)) + ((3 + (2*I)*Sqrt[2])*Log[2^(1/3)*x^(2/3) - (1 + 2*x)^(1/3)])/(6* 2^(2/3)) - ((1 - (2*I)*Sqrt[2])*Log[(2 - I*Sqrt[2])^(1/3)*x^(2/3) - (1 + 2 *x)^(1/3)])/(4*(2 - I*Sqrt[2])^(2/3)) - ((1 + (2*I)*Sqrt[2])*Log[(2 + I*Sq rt[2])^(1/3)*x^(2/3) - (1 + 2*x)^(1/3)])/(4*(2 + I*Sqrt[2])^(2/3)))))/(2*x ^(1/3)*(1 + 2*x^2)^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_ ), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/f Subst[Int[x^(k*(m + 1 ) - 1)*(d + e*(x^(2*k)/f))^q*(a + c*(x^(4*k)/f))^p, x], x, (f*x)^(1/k)], x] ] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_ .), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/ k - 1)*(d + e*x^(n/k))^q*(a + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m ]
Int[(((f_.)*(x_))^(m_)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.)), x_Symbol] :> Simp[d/a Int[(f*x)^m*(d + e*x^n)^(q - 1), x], x] + S imp[1/(a*f^n) Int[(f*x)^(m + n)*(d + e*x^n)^(q - 1)*(Simp[a*e - c*d*x^n, x]/(a + c*x^(2*n))), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && !IntegerQ[q] && GtQ[q, 0] && LtQ[m, 0]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 180.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {\left (-30 x^{2}-3\right ) \left (2 x^{3}+x \right )^{\frac {1}{3}}-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-4 \textit {\_Z}^{3}+6\right )}{\sum }\frac {\left (\textit {\_R}^{3}-6\right ) \ln \left (\frac {-\textit {\_R} x +\left (2 x^{3}+x \right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R}^{2} \left (\textit {\_R}^{3}-2\right )}\right ) x^{3}}{8 x^{3}}\) | \(78\) |
trager | \(\text {Expression too large to display}\) | \(10613\) |
risch | \(\text {Expression too large to display}\) | \(16959\) |
Input:
int((2*x^2+1)*(2*x^3+x)^(1/3)/x^4/(2*x^4+1),x,method=_RETURNVERBOSE)
Output:
1/8*((-30*x^2-3)*(2*x^3+x)^(1/3)-4*sum((_R^3-6)*ln((-_R*x+(2*x^3+x)^(1/3)) /x)/_R^2/(_R^3-2),_R=RootOf(_Z^6-4*_Z^3+6))*x^3)/x^3
Exception generated. \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((2*x^2+1)*(2*x^3+x)^(1/3)/x^4/(2*x^4+1),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
Not integrable
Time = 2.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.25 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\int \frac {\sqrt [3]{x \left (2 x^{2} + 1\right )} \left (2 x^{2} + 1\right )}{x^{4} \cdot \left (2 x^{4} + 1\right )}\, dx \] Input:
integrate((2*x**2+1)*(2*x**3+x)**(1/3)/x**4/(2*x**4+1),x)
Output:
Integral((x*(2*x**2 + 1))**(1/3)*(2*x**2 + 1)/(x**4*(2*x**4 + 1)), x)
Not integrable
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.27 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\int { \frac {{\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} {\left (2 \, x^{2} + 1\right )}}{{\left (2 \, x^{4} + 1\right )} x^{4}} \,d x } \] Input:
integrate((2*x^2+1)*(2*x^3+x)^(1/3)/x^4/(2*x^4+1),x, algorithm="maxima")
Output:
integrate((2*x^3 + x)^(1/3)*(2*x^2 + 1)/((2*x^4 + 1)*x^4), x)
Exception generated. \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((2*x^2+1)*(2*x^3+x)^(1/3)/x^4/(2*x^4+1),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Invalid _EXT in replace_ext Error: Bad Argument ValueDone
Not integrable
Time = 9.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.27 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\int \frac {{\left (2\,x^3+x\right )}^{1/3}\,\left (2\,x^2+1\right )}{x^4\,\left (2\,x^4+1\right )} \,d x \] Input:
int(((x + 2*x^3)^(1/3)*(2*x^2 + 1))/(x^4*(2*x^4 + 1)),x)
Output:
int(((x + 2*x^3)^(1/3)*(2*x^2 + 1))/(x^4*(2*x^4 + 1)), x)
Not integrable
Time = 6.94 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.31 \[ \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx=\frac {-18 \left (2 x^{2}+1\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {2}\, x^{2}+3 \left (2 x^{2}+1\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {2}+72 \left (2 x^{2}+1\right )^{\frac {1}{3}} x^{2}-12 \left (2 x^{2}+1\right )^{\frac {1}{3}}+32 x^{\frac {8}{3}} \sqrt {2}\, \sqrt {2}\, \left (\int \frac {x^{\frac {1}{3}} \left (2 x^{2}+1\right )^{\frac {1}{3}}}{4 x^{8}+2 x^{6}+2 x^{4}+x^{2}}d x \right )+32 x^{\frac {8}{3}} \left (\int \frac {x^{\frac {1}{3}} \left (2 x^{2}+1\right )^{\frac {1}{3}}}{4 x^{6}+2 x^{4}+2 x^{2}+1}d x \right )}{16 x^{\frac {8}{3}}} \] Input:
int((2*x^2+1)*(2*x^3+x)^(1/3)/x^4/(2*x^4+1),x)
Output:
( - 18*(2*x**2 + 1)**(1/3)*sqrt(2)*2**(1/3)*2**(1/6)*x**2 + 3*(2*x**2 + 1) **(1/3)*sqrt(2)*2**(1/3)*2**(1/6) + 72*(2*x**2 + 1)**(1/3)*x**2 - 12*(2*x* *2 + 1)**(1/3) + 32*x**(2/3)*sqrt(2)*2**(1/3)*2**(1/6)*int((x**(1/3)*(2*x* *2 + 1)**(1/3))/(4*x**8 + 2*x**6 + 2*x**4 + x**2),x)*x**2 + 32*x**(2/3)*in t((x**(1/3)*(2*x**2 + 1)**(1/3))/(4*x**6 + 2*x**4 + 2*x**2 + 1),x)*x**2)/( 16*x**(2/3)*x**2)