3.3.90 \(\int \frac {x}{\sqrt {a x^2+b x^5}} \, dx\) [290]

Optimal. Leaf size=212 \[ \frac {2 \sqrt {2+\sqrt {3}} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \]

[Out]

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)/b^(1/3)/(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 = 0.05, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2057, 224} \begin {gather*} \frac {2 \sqrt {2+\sqrt {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 (\text {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 )}{\sqrt [4]{3} \sqrt [3]{b} \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 {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[2 + Sqrt[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]])/(3^(1/4)*b^(1/3)*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 224

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]*Sq
rt[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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 {x}{\sqrt {a x^2+b x^5}} \, dx &=\frac {\left (x \sqrt {a+b x^3}\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{\sqrt {a x^2+b x^5}}\\ &=\frac {2 \sqrt {2+\sqrt {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 )}{\sqrt [4]{3} \sqrt [3]{b} \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 52, normalized size = 0.25 \begin {gather*} \frac {x^2 \sqrt {1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a}\right )}{\sqrt {x^2 \left (a+b x^3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a*x^2 + b*x^5],x]

[Out]

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

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Maple [A]
time = 0.37, size = 231, normalized size = 1.09

method	result	size
default	\(-\frac {i x \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {2 \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right )}{3 \sqrt {b \,x^{5}+a \,x^{2}}\, b}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(b*x^5 + a*x^2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.38, size = 14, normalized size = 0.07 \begin {gather*} \frac {2 \, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(b*x^5 + a*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{\sqrt {b\,x^5+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x^2 + b*x^5)^(1/2),x)

[Out]

int(x/(a*x^2 + b*x^5)^(1/2), x)

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