\(\int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx\) [498]

Optimal result

Integrand size = 19, antiderivative size = 292 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\frac {\left (1+\sqrt {3}\right ) \sqrt {a x^3} \sqrt {1+x^3}}{x \left (1+\left (1+\sqrt {3}\right ) x\right )}-\frac {\sqrt [4]{3} \sqrt {a x^3} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} E\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{x \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {1+x^3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt {a x^3} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} x \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {1+x^3}} \] Output:

(1+3^(1/2))*(a*x^3)^(1/2)*(x^3+1)^(1/2)/x/(1+(1+3^(1/2))*x)-3^(1/4)*(a*x^3 
)^(1/2)*(1+x)*((x^2-x+1)/(1+(1+3^(1/2))*x)^2)^(1/2)*EllipticE((1-(1+(1-3^( 
1/2))*x)^2/(1+(1+3^(1/2))*x)^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))/x/(x*(1+x)/ 
(1+(1+3^(1/2))*x)^2)^(1/2)/(x^3+1)^(1/2)-1/6*(1-3^(1/2))*(a*x^3)^(1/2)*(1+ 
x)*((x^2-x+1)/(1+(1+3^(1/2))*x)^2)^(1/2)*InverseJacobiAM(arccos((1+(1-3^(1 
/2))*x)/(1+(1+3^(1/2))*x)),1/4*6^(1/2)+1/4*2^(1/2))*3^(3/4)/x/(x*(1+x)/(1+ 
(1+3^(1/2))*x)^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 = 29, normalized size of antiderivative = 0.10 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\frac {2}{5} x \sqrt {a x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},-x^3\right ) \] Input:

Integrate[Sqrt[a*x^3]/Sqrt[1 + x^3],x]
 

Output:

(2*x*Sqrt[a*x^3]*Hypergeometric2F1[1/2, 5/6, 11/6, -x^3])/5
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {34, 851, 837, 25, 766, 2420}

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.

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

\(\Big \downarrow \) 34

\(\displaystyle \frac {\sqrt {a x^3} \int \frac {x^{3/2}}{\sqrt {x^3+1}}dx}{x^{3/2}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {2 \sqrt {a x^3} \int \frac {x^2}{\sqrt {x^3+1}}d\sqrt {x}}{x^{3/2}}\)

\(\Big \downarrow \) 837

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 766

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

\(\Big \downarrow \) 2420

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

Input:

Int[Sqrt[a*x^3]/Sqrt[1 + x^3],x]
 

Output:

(2*Sqrt[a*x^3]*((((1 + Sqrt[3])*Sqrt[x]*Sqrt[1 + x^3])/(1 + (1 + Sqrt[3])* 
x) - (3^(1/4)*Sqrt[x]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3])*x)^2]* 
EllipticE[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3 
])/4])/(Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3]))/2 - ((1 
- Sqrt[3])*Sqrt[x]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3])*x)^2]*Ell 
ipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3])/ 
4])/(4*3^(1/4)*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3])))/ 
x^(3/2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 34

Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F 
racPart[p]/x^(m*FracPart[p]))   Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x 
] &&  !IntegerQ[p]
 

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 837

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 851

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]
 

rule 2420

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.08

method	result	size
meijerg	\(\frac {2 \sqrt {a \,x^{3}}\, x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{6}\right ], \left [\frac {11}{6}\right ], -x^{3}\right )}{5}\)	\(22\)
default	\(\text {Expression too large to display}\)	\(1521\)

Input:

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

Output:

2/5*(a*x^3)^(1/2)*x*hypergeom([1/2,5/6],[11/6],-x^3)
 

Fricas [F]

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

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

Output:

integral(sqrt(a*x^3)/sqrt(x^3 + 1), x)
 

Sympy [F]

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

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

Output:

Integral(sqrt(a*x**3)/sqrt((x + 1)*(x**2 - x + 1)), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(a*x^3)/sqrt(x^3 + 1), x)
 

Giac [F]

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

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

Output:

integrate(sqrt(a*x^3)/sqrt(x^3 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int \frac {\sqrt {a\,x^3}}{\sqrt {x^3+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*int((sqrt(x)*sqrt(x**3 + 1)*x)/(x**3 + 1),x)