\(\int \frac {x^4}{\sqrt {1+x^3}} \, dx\) [246]

Optimal result

Integrand size = 13, antiderivative size = 246 \[ \int \frac {x^4}{\sqrt {1+x^3}} \, dx=\frac {2}{7} x^2 \sqrt {1+x^3}-\frac {8 \sqrt {1+x^3}}{7 \left (1+\sqrt {3}+x\right )}+\frac {4 \sqrt [4]{3} \sqrt {2-\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{7 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}-\frac {8 \sqrt {2} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{7 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

2/7*x^2*(x^3+1)^(1/2)-8*(x^3+1)^(1/2)/(7+7*3^(1/2)+7*x)+4/7*3^(1/4)*(1/2*6 
^(1/2)-1/2*2^(1/2))*(1+x)*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*EllipticE((1+x 
-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)/((1+x)/(1+x+3^(1/2))^2)^(1/2)/(x^3+ 
1)^(1/2)-8/21*2^(1/2)*(1+x)*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*EllipticF((1 
+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*3^(3/4)/((1+x)/(1+x+3^(1/2))^2)^( 
1/2)/(x^3+1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.14 \[ \int \frac {x^4}{\sqrt {1+x^3}} \, dx=\frac {2}{7} x^2 \left (\sqrt {1+x^3}-\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-x^3\right )\right ) \] Input:

Integrate[x^4/Sqrt[1 + x^3],x]
 

Output:

(2*x^2*(Sqrt[1 + x^3] - Hypergeometric2F1[1/2, 2/3, 5/3, -x^3]))/7
 

Rubi [A] (warning: unable to verify)

Time = 0.56 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {843, 832, 759, 2416}

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.

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

\(\Big \downarrow \) 843

\(\displaystyle \frac {2}{7} x^2 \sqrt {x^3+1}-\frac {4}{7} \int \frac {x}{\sqrt {x^3+1}}dx\)

\(\Big \downarrow \) 832

\(\displaystyle \frac {2}{7} x^2 \sqrt {x^3+1}-\frac {4}{7} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3+1}}dx\right )\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {2}{7} x^2 \sqrt {x^3+1}-\frac {4}{7} \left (\int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )\)

\(\Big \downarrow \) 2416

\(\displaystyle \frac {2}{7} x^2 \sqrt {x^3+1}-\frac {4}{7} \left (-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {2 \sqrt {x^3+1}}{x+\sqrt {3}+1}\right )\)

Input:

Int[x^4/Sqrt[1 + x^3],x]
 

Output:

(2*x^2*Sqrt[1 + x^3])/7 - (4*((2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) - (3^(1/ 
4)*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Ellip 
ticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sqrt[( 
1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (2*(1 - Sqrt[3])*Sqrt[2 + Sqr 
t[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 
- Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(1/4)*Sqrt[(1 + x)/ 
(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/7
 

Defintions of rubi rules used

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 832

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 843

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]
 

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.93 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.07

method	result	size
meijerg	\(\frac {x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x^{3}\right )}{5}\)	\(17\)
default	\(\frac {2 x^{2} \sqrt {x^{3}+1}}{7}-\frac {8 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{7 \sqrt {x^{3}+1}}\)	\(186\)
risch	\(\frac {2 x^{2} \sqrt {x^{3}+1}}{7}-\frac {8 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{7 \sqrt {x^{3}+1}}\)	\(186\)
elliptic	\(\frac {2 x^{2} \sqrt {x^{3}+1}}{7}-\frac {8 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{7 \sqrt {x^{3}+1}}\)	\(186\)

Input:

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

Output:

1/5*x^5*hypergeom([1/2,5/3],[8/3],-x^3)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.09 \[ \int \frac {x^4}{\sqrt {1+x^3}} \, dx=\frac {2}{7} \, \sqrt {x^{3} + 1} x^{2} + \frac {8}{7} \, {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) \] Input:

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

Output:

2/7*sqrt(x^3 + 1)*x^2 + 8/7*weierstrassZeta(0, -4, weierstrassPInverse(0, 
-4, x))
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.12 \[ \int \frac {x^4}{\sqrt {1+x^3}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \] Input:

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

Output:

x**5*gamma(5/3)*hyper((1/2, 5/3), (8/3,), x**3*exp_polar(I*pi))/(3*gamma(8 
/3))
 

Maxima [F]

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

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

Output:

integrate(x^4/sqrt(x^3 + 1), x)
 

Giac [F]

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

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

Output:

integrate(x^4/sqrt(x^3 + 1), x)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.92 \[ \int \frac {x^4}{\sqrt {1+x^3}} \, dx=\frac {2\,x^2\,\sqrt {x^3+1}}{7}+\frac {8\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{7\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

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

Output:

(2*x^2*(x^3 + 1)^(1/2))/7 + (8*(((3^(1/2)*1i)/2 - 1/2)*ellipticF(asin(((x 
+ 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i) 
/2 - 3/2)) - ((3^(1/2)*1i)/2 - 3/2)*ellipticE(asin(((x + 1)/((3^(1/2)*1i)/ 
2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))*((3^(1/ 
2)*1i)/2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)* 
((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/ 
2)*1i)/2 + 3/2))^(1/2))/(7*(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/ 
2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2))
 

Reduce [F]

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

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

Output:

(2*(sqrt(x**3 + 1)*x**2 - 2*int((sqrt(x**3 + 1)*x)/(x**3 + 1),x)))/7