\(\int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx\) [280]

Optimal result

Integrand size = 21, antiderivative size = 203 \[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\frac {x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt [3]{a} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \] Output:

1/3*x^(3/2)*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/( 
a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)*InverseJacobiAM(arccos((a^(1/3)+(1 
-3^(1/2))*b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)),1/4*6^(1/2)+1/4*2^(1 
/2))*3^(3/4)/a^(1/3)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b 
^(1/3)*x)^2)^(1/2)/(b*x^5+a*x^2)^(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 = 55, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 x^{3/2} \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )}{\sqrt {x^2 \left (a+b x^3\right )}} \] Input:

Integrate[Sqrt[x]/Sqrt[a*x^2 + b*x^5],x]
 

Output:

(2*x^(3/2)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3)/ 
a)])/Sqrt[x^2*(a + b*x^3)]
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1938, 851, 766}

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.

\(\displaystyle \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {x \sqrt {a+b x^3} \int \frac {1}{\sqrt {x} \sqrt {b x^3+a}}dx}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {2 x \sqrt {a+b x^3} \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {x}}{\sqrt {a x^2+b x^5}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt [3]{a} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}\)

Input:

Int[Sqrt[x]/Sqrt[a*x^2 + b*x^5],x]
 

Output:

(x^(3/2)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3) 
*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + ( 
1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3] 
)/4])/(3^(1/4)*a^(1/3)*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + ( 
1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a*x^2 + b*x^5])
 

Defintions of rubi rules used

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 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 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.30 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.15

method	result	size
default	\(-\frac {4 x^{\frac {3}{2}} \left (b \,x^{3}+a \right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (i \sqrt {3}-1\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (i \sqrt {3}-1\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right ) \left (i \sqrt {3}\, b^{2} x^{2}-2 i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {2}{3}}-b^{2} x^{2}+2 \left (-a \,b^{2}\right )^{\frac {1}{3}} b x -\left (-a \,b^{2}\right )^{\frac {2}{3}}\right )}{\sqrt {b \,x^{5}+a \,x^{2}}\, b \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {x \left (b \,x^{3}+a \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{b^{2}}}}\)	\(437\)

Input:

int(x^(1/2)/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-4/(b*x^5+a*x^2)^(1/2)*x^(3/2)*(b*x^3+a)/b/(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)* 
x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+ 
2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/ 
2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3) 
))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)) 
)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*( 
I*3^(1/2)*b^2*x^2-2*I*(-a*b^2)^(1/3)*3^(1/2)*b*x+I*3^(1/2)*(-a*b^2)^(2/3)- 
b^2*x^2+2*(-a*b^2)^(1/3)*b*x-(-a*b^2)^(2/3))/(x*(b*x^3+a))^(1/2)/(I*3^(1/2 
)-3)/(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^ 
2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.08 \[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=-\frac {2 \, {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right )}{\sqrt {a}} \] Input:

integrate(x^(1/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2*weierstrassPInverse(0, -4*b/a, 1/x)/sqrt(a)
 

Sympy [F]

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

integrate(x**(1/2)/(b*x**5+a*x**2)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:

integrate(x^(1/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(x)/sqrt(b*x^5 + a*x^2), x)
 

Giac [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:

integrate(x^(1/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(x)/sqrt(b*x^5 + a*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {x}}{\sqrt {b\,x^5+a\,x^2}} \,d x \] Input:

int(x^(1/2)/(a*x^2 + b*x^5)^(1/2),x)
 

Output:

int(x^(1/2)/(a*x^2 + b*x^5)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}}{b \,x^{4}+a x}d x \] Input:

int(x^(1/2)/(b*x^5+a*x^2)^(1/2),x)
 

Output:

int((sqrt(x)*sqrt(a + b*x**3))/(a*x + b*x**4),x)