\(\int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx\) [1942]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {3}{8} \sqrt [3]{x^2+x^4}+\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{x^2+x^4}}{\sqrt {3}}\right )}{8\ 2^{2/3}}-\frac {\log \left (1+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{8\ 2^{2/3}}+\frac {\log \left (-1+2^{2/3} \sqrt [3]{x^2+x^4}-2 \sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{16\ 2^{2/3}} \] Output:

3/8*(x^4+x^2)^(1/3)-1/16*3^(1/2)*arctan(-1/3*3^(1/2)+2/3*2^(2/3)*(x^4+x^2) 
^(1/3)*3^(1/2))*2^(1/3)-1/16*ln(1+2^(2/3)*(x^4+x^2)^(1/3))*2^(1/3)+1/32*ln 
(-1+2^(2/3)*(x^4+x^2)^(1/3)-2*2^(1/3)*(x^4+x^2)^(2/3))*2^(1/3)
 

Mathematica [A] (verified)

Time = 10.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {1}{32} \left (12 \sqrt [3]{x^2+x^4}+2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1-2\ 2^{2/3} \sqrt [3]{x^2+x^4}}{\sqrt {3}}\right )-2 \sqrt [3]{2} \log \left (1+2^{2/3} \sqrt [3]{x^2+x^4}\right )+\sqrt [3]{2} \log \left (-1+2^{2/3} \sqrt [3]{x^2+x^4}-2 \sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )\right ) \] Input:

Integrate[(x*(x^2 + x^4)^(1/3))/(1 + 2*x^2),x]
 

Output:

(12*(x^2 + x^4)^(1/3) + 2*2^(1/3)*Sqrt[3]*ArcTan[(1 - 2*2^(2/3)*(x^2 + x^4 
)^(1/3))/Sqrt[3]] - 2*2^(1/3)*Log[1 + 2^(2/3)*(x^2 + x^4)^(1/3)] + 2^(1/3) 
*Log[-1 + 2^(2/3)*(x^2 + x^4)^(1/3) - 2*2^(1/3)*(x^2 + x^4)^(2/3)])/32
 

Rubi [A] (warning: unable to verify)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1940, 1118, 27, 243, 60, 70, 16, 1083, 217}

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.

\(\displaystyle \int \frac {x \sqrt [3]{x^4+x^2}}{2 x^2+1} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {\sqrt [3]{x^4+x^2}}{2 x^2+1}dx^2\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {1}{4} \int \frac {\sqrt [3]{x^4-1}}{2^{2/3} x^2}d\left (2 x^2+1\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt [3]{x^4-1}}{x^2}d\left (2 x^2+1\right )}{4\ 2^{2/3}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\int \frac {\sqrt [3]{2}}{\left (x^2\right )^{2/3}}dx^4}{8\ 2^{2/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {3 \sqrt [3]{2} \sqrt [3]{x^2}-\int \frac {1}{2^{2/3} \left (x^2\right )^{5/3}}dx^4}{8\ 2^{2/3}}\)

\(\Big \downarrow \) 70

\(\displaystyle \frac {-\frac {3}{2} \int \frac {1}{2 x^2+2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )-\frac {3}{2} \int \frac {1}{\sqrt [3]{2} \sqrt [3]{x^2}-2 x^2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}}{8\ 2^{2/3}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {-\frac {3}{2} \int \frac {1}{\sqrt [3]{2} \sqrt [3]{x^2}-2 x^2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {3 \int \frac {1}{-2 \sqrt [3]{2} \sqrt [3]{x^2}-2}d\left (2 \sqrt [3]{2} \sqrt [3]{x^2}-1\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{2} \sqrt [3]{x^2}-1}{\sqrt {3}}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\)

Input:

Int[(x*(x^2 + x^4)^(1/3))/(1 + 2*x^2),x]
 

Output:

(3*2^(1/3)*(x^2)^(1/3) - Sqrt[3]*ArcTan[(-1 + 2*2^(1/3)*(x^2)^(1/3))/Sqrt[ 
3]] + Log[x^4]/2 - (3*Log[2 + 2*x^2])/2)/(8*2^(2/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 70

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) 
, x] + (Simp[3/(2*b*q)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] + Simp[3/(2*b*q^2)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 17.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {3 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{8}-\frac {2^{\frac {1}{3}} \ln \left (2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+2^{\frac {1}{3}}\right )}{16}+\frac {2^{\frac {1}{3}} \ln \left (2^{\frac {2}{3}}-2 \,2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+4 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}\right )}{32}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, 2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{3}-\frac {\sqrt {3}}{3}\right )}{16}\)	\(111\)
risch	\(\text {Expression too large to display}\)	\(1968\)

Input:

int(x*(x^4+x^2)^(1/3)/(2*x^2+1),x,method=_RETURNVERBOSE)
 

Output:

3/8*(x^2*(x^2+1))^(1/3)-1/16*2^(1/3)*ln(2*(x^2*(x^2+1))^(1/3)+2^(1/3))+1/3 
2*2^(1/3)*ln(2^(2/3)-2*2^(1/3)*(x^2*(x^2+1))^(1/3)+4*(x^2*(x^2+1))^(2/3))- 
1/16*2^(1/3)*3^(1/2)*arctan(2/3*3^(1/2)*2^(2/3)*(x^2*(x^2+1))^(1/3)-1/3*3^ 
(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=-\frac {1}{16} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 4^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) + \frac {1}{64} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}}\right ) - \frac {1}{32} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} \] Input:

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="fricas")
 

Output:

-1/16*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*4^(2/3)*(x^4 + x^2)^(1 
/3) - 4^(1/3))) + 1/64*4^(2/3)*log(-4^(2/3)*(x^4 + x^2)^(1/3) + 4*(x^4 + x 
^2)^(2/3) + 4^(1/3)) - 1/32*4^(2/3)*log(4^(2/3) + 4*(x^4 + x^2)^(1/3)) + 3 
/8*(x^4 + x^2)^(1/3)
 

Sympy [F]

\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int \frac {x \sqrt [3]{x^{2} \left (x^{2} + 1\right )}}{2 x^{2} + 1}\, dx \] Input:

integrate(x*(x**4+x**2)**(1/3)/(2*x**2+1),x)
 

Output:

Integral(x*(x**2*(x**2 + 1))**(1/3)/(2*x**2 + 1), x)
 

Maxima [F]

\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1} \,d x } \] Input:

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="maxima")
 

Output:

integrate((x^4 + x^2)^(1/3)*x/(2*x^2 + 1), x)
 

Giac [F]

\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1} \,d x } \] Input:

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="giac")
 

Output:

integrate((x^4 + x^2)^(1/3)*x/(2*x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int \frac {x\,{\left (x^4+x^2\right )}^{1/3}}{2\,x^2+1} \,d x \] Input:

int((x*(x^2 + x^4)^(1/3))/(2*x^2 + 1),x)
 

Output:

int((x*(x^2 + x^4)^(1/3))/(2*x^2 + 1), x)
 

Reduce [F]

\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {3 x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{8}-\frac {\left (\int \frac {\left (x^{2}+1\right )^{\frac {1}{3}}}{2 x^{\frac {13}{3}}+3 x^{\frac {7}{3}}+x^{\frac {1}{3}}}d x \right )}{4} \] Input:

int(x*(x^4+x^2)^(1/3)/(2*x^2+1),x)
 

Output:

(3*x**(2/3)*(x**2 + 1)**(1/3) - 2*int((x**2 + 1)**(1/3)/(2*x**(1/3)*x**4 + 
 3*x**(1/3)*x**2 + x**(1/3)),x))/8