\(\int \frac {\sqrt {\frac {a}{x^4}}}{\sqrt {1+x^3}} \, dx\) [504]

Optimal result

Integrand size = 19, antiderivative size = 281 \[ \int \frac {\sqrt {\frac {a}{x^4}}}{\sqrt {1+x^3}} \, dx=-\sqrt {\frac {a}{x^4}} x \sqrt {1+x^3}+\frac {\sqrt {\frac {a}{x^4}} x^2 \sqrt {1+x^3}}{1+\sqrt {3}+x}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {\frac {a}{x^4}} x^2 (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}+\frac {\sqrt {2} \sqrt {\frac {a}{x^4}} x^2 (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

-(a/x^4)^(1/2)*x*(x^3+1)^(1/2)+(a/x^4)^(1/2)*x^2*(x^3+1)^(1/2)/(1+x+3^(1/2 
))-1/2*3^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(a/x^4)^(1/2)*x^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)+1/3*2^(1/2)*(a/x^4)^(1/2)* 
x^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 = 27, normalized size of antiderivative = 0.10 \[ \int \frac {\sqrt {\frac {a}{x^4}}}{\sqrt {1+x^3}} \, dx=-\sqrt {\frac {a}{x^4}} x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},-x^3\right ) \] Input:

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

Output:

-(Sqrt[a/x^4]*x*Hypergeometric2F1[-1/3, 1/2, 2/3, -x^3])
 

Rubi [A] (warning: unable to verify)

Time = 0.59 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {34, 847, 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.

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

\(\Big \downarrow \) 34

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 832

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2416

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

Input:

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

Output:

Sqrt[a/x^4]*x^2*(-(Sqrt[1 + x^3]/x) + ((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]*EllipticE[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])*Sqr 
t[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[A 
rcSin[(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]))/2)
 

Defintions of rubi rules used

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 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 847

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] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.08

method	result	size
meijerg	\(-\sqrt {\frac {a}{x^{4}}}\, x \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )\)	\(22\)
risch	\(-\sqrt {\frac {a}{x^{4}}}\, x \sqrt {x^{3}+1}-\frac {i \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x +1}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )-\operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right ) \sqrt {\frac {a}{x^{4}}}\, x^{2} \sqrt {a \left (x^{3}+1\right )}}{3 \sqrt {a \,x^{3}+a}\, \sqrt {x^{3}+1}}\)	\(204\)
default	\(\frac {\sqrt {\frac {a}{x^{4}}}\, x \left (i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x -6 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticE}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x +3 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x -2 x^{3}-2\right )}{2 \sqrt {x^{3}+1}}\)	\(353\)

Input:

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

Output:

-(a/x^4)^(1/2)*x*hypergeom([-1/3,1/2],[2/3],-x^3)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.13 \[ \int \frac {\sqrt {\frac {a}{x^4}}}{\sqrt {1+x^3}} \, dx=-x^{2} \sqrt {\frac {a}{x^{4}}} {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) - \sqrt {x^{3} + 1} x \sqrt {\frac {a}{x^{4}}} \] Input:

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

Output:

-x^2*sqrt(a/x^4)*weierstrassZeta(0, -4, weierstrassPInverse(0, -4, x)) - s 
qrt(x^3 + 1)*x*sqrt(a/x^4)
 

Sympy [F]

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

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

Output:

Integral(sqrt(a/x**4)/sqrt((x + 1)*(x**2 - x + 1)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {\frac {a}{x^4}}}{\sqrt {1+x^3}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {x^{3}+1}}{x^{5}+x^{2}}d x \right ) \] Input:

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

Output:

sqrt(a)*int(sqrt(x**3 + 1)/(x**5 + x**2),x)