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

Optimal. Leaf size=243 \[ -\frac {\sqrt {2+\sqrt {3}} b^{2/3} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {\sqrt {a x^2+b x^5}}{2 a x^3} \]

[Out]

-1/2*(b*x^5+a*x^2)^(1/2)/a/x^3-1/6*b^(2/3)*x*(a^(1/3)+b^(1/3)*x)*EllipticF((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^
(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)
/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/a/(b*x^5+a*x^2)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*
x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2025, 2032, 218} \[ -\frac {\sqrt {2+\sqrt {3}} b^{2/3} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {\sqrt {a x^2+b x^5}}{2 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a*x^2 + b*x^5]),x]

[Out]

-Sqrt[a*x^2 + b*x^5]/(2*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*Sqrt[(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])

Rule 218

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

Rule 2025

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] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[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] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \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 {b \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx}{4 a}\\ &=-\frac {\sqrt {a x^2+b x^5}}{2 a x^3}-\frac {\left (b x \sqrt {a+b x^3}\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{4 a \sqrt {a x^2+b x^5}}\\ &=-\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}} F\left (\sin ^{-1}\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}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 55, normalized size = 0.23 \[ -\frac {\sqrt {\frac {b x^3}{a}+1} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};-\frac {b x^3}{a}\right )}{2 x \sqrt {x^2 \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-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)])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{5} + a x^{2}}}{b x^{7} + a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^5 + a*x^2)/(b*x^7 + a*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.52, size = 248, normalized size = 1.02 \[ \frac {-6 b \,x^{3}+i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i \left (-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {2 \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, x^{2} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right )-6 a}{12 \sqrt {b \,x^{5}+a \,x^{2}}\, a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/x*(I*3^(1/2)*(-a*b^2)^(1/3)*(-I*(-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-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*(2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-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*(-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{5} + a x^{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.71, size = 44, normalized size = 0.18 \[ -\frac {2\,\sqrt {\frac {a}{b\,x^3}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{6};\ \frac {13}{6};\ -\frac {a}{b\,x^3}\right )}{7\,x\,\sqrt {b\,x^5+a\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a*x^2 + b*x^5)^(1/2)),x)

[Out]

-(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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {x^{2} \left (a + b x^{3}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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