\(\int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx\) [270]

Optimal result

Integrand size = 19, antiderivative size = 243 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=-\frac {\sqrt {a x^2+b x^5}}{2 a x^3}-\frac {\sqrt {2+\sqrt {3}} b^{2/3} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \] Output:

-1/2*(b*x^5+a*x^2)^(1/2)/a/x^3-1/6*(1/2*6^(1/2)+1/2*2^(1/2))*b^(2/3)*x*(a^ 
(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/((1+3^(1/2))*a^( 
1/3)+b^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*a^(1/3)+b^(1/3)*x)/((1+3^( 
1/2))*a^(1/3)+b^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/a/(a^(1/3)*(a^(1/3)+b^(1/3 
)*x)/((1+3^(1/2))*a^(1/3)+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 = 0.23 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=-\frac {\sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a}\right )}{2 x \sqrt {x^2 \left (a+b x^3\right )}} \] Input:

Integrate[1/(x^2*Sqrt[a*x^2 + b*x^5]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1931, 1938, 759}

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.

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

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {b \int \frac {x}{\sqrt {b x^5+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^5}}{2 a x^3}\)

\(\Big \downarrow \) 1938

\(\displaystyle -\frac {b x \sqrt {a+b x^3} \int \frac {1}{\sqrt {b x^3+a}}dx}{4 a \sqrt {a x^2+b x^5}}-\frac {\sqrt {a x^2+b x^5}}{2 a x^3}\)

\(\Big \downarrow \) 759

\(\displaystyle -\frac {\sqrt {2+\sqrt {3}} b^{2/3} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {\sqrt {a x^2+b x^5}}{2 a x^3}\)

Input:

Int[1/(x^2*Sqrt[a*x^2 + b*x^5]),x]
 

Output:

-1/2*Sqrt[a*x^2 + b*x^5]/(a*x^3) - (Sqrt[2 + Sqrt[3]]*b^(2/3)*x*(a^(1/3) + 
 b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3] 
)*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3 
)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(2*3^(1/4)*a*S 
qrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2] 
*Sqrt[a*x^2 + b*x^5])
 

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 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ 
m + j*p + 1, 0]
 

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 [A] (verified)

Time = 0.64 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {2 \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right ) x^{2}-6 b \,x^{3}-6 a}{12 x \sqrt {b \,x^{5}+a \,x^{2}}\, a}\)	\(248\)
risch	\(-\frac {b \,x^{3}+a}{2 a x \sqrt {x^{2} \left (b \,x^{3}+a \right )}}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right ) x}{6 a \sqrt {x^{2} \left (b \,x^{3}+a \right )}}\)	\(317\)

Input:

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

Output:

1/12/x*(I*3^(1/2)*(-a*b^2)^(1/3)*(-I*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b 
^2)^(1/3))*3^(1/2)/(-a*b^2)^(1/3))^(1/2)*(-2*(-b*x+(-a*b^2)^(1/3))/(-a*b^2 
)^(1/3)/(I*3^(1/2)-3))^(1/2)*(-I*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^ 
(1/3))*3^(1/2)/(-a*b^2)^(1/3))^(1/2)*EllipticF(1/6*3^(1/2)*2^(1/2)*(-I*(I* 
3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))*3^(1/2)/(-a*b^2)^(1/3))^(1/2) 
,2^(1/2)*(I*3^(1/2)/(I*3^(1/2)-3))^(1/2))*x^2-6*b*x^3-6*a)/(b*x^5+a*x^2)^( 
1/2)/a
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=-\frac {\sqrt {b} x^{3} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) + \sqrt {b x^{5} + a x^{2}}}{2 \, a x^{3}} \] Input:

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

Output:

-1/2*(sqrt(b)*x^3*weierstrassPInverse(0, -4*a/b, x) + sqrt(b*x^5 + a*x^2)) 
/(a*x^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=\int { \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=\int { \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{2}} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.78 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=-\frac {2\,\sqrt {\frac {a}{b\,x^3}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{6};\ \frac {13}{6};\ -\frac {a}{b\,x^3}\right )}{7\,x\,\sqrt {b\,x^5+a\,x^2}} \] Input:

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

Output:

-(2*(a/(b*x^3) + 1)^(1/2)*hypergeom([1/2, 7/6], 13/6, -a/(b*x^3)))/(7*x*(a 
*x^2 + b*x^5)^(1/2))
 

Reduce [F]

\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {b \,x^{3}+a}}{b \,x^{6}+a \,x^{3}}d x \] Input:

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

Output:

int(sqrt(a + b*x**3)/(a*x**3 + b*x**6),x)