3.289 \(\int \frac {x^4}{\sqrt {a x^2+b x^5}} \, dx\)

Optimal. Leaf size=238 \[ \frac {2 \sqrt {a x^2+b x^5}}{5 b}-\frac {4 \sqrt {2+\sqrt {3}} a x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \]

[Out]

2/5*(b*x^5+a*x^2)^(1/2)/b-4/15*a*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^(4/3)/(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 = 0.14, antiderivative size = 238, 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 = {2024, 2032, 218} \[ \frac {2 \sqrt {a x^2+b x^5}}{5 b}-\frac {4 \sqrt {2+\sqrt {3}} a x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[a*x^2 + b*x^5])/(5*b) - (4*Sqrt[2 + Sqrt[3]]*a*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]])/(5*3^(1/4)*b^(4/3)*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 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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*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 {x^4}{\sqrt {a x^2+b x^5}} \, dx &=\frac {2 \sqrt {a x^2+b x^5}}{5 b}-\frac {(2 a) \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx}{5 b}\\ &=\frac {2 \sqrt {a x^2+b x^5}}{5 b}-\frac {\left (2 a x \sqrt {a+b x^3}\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{5 b \sqrt {a x^2+b x^5}}\\ &=\frac {2 \sqrt {a x^2+b x^5}}{5 b}-\frac {4 \sqrt {2+\sqrt {3}} a 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 )}{5 \sqrt [4]{3} b^{4/3} \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 = 0.03, size = 68, normalized size = 0.29 \[ \frac {2 x^2 \left (-a \sqrt {\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a}\right )+a+b x^3\right )}{5 b \sqrt {x^2 \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*x^2*(a + b*x^3 - a*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)]))/(5*b*Sqrt[x^2*(a +
b*x^3)])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{5} + a x^{2}} x^{2}}{b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^5 + a*x^2)*x^2/(b*x^3 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {b x^{5} + a x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(b*x^5 + a*x^2), x)

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maple [A]  time = 1.29, size = 248, normalized size = 1.04 \[ \frac {2 \left (3 b^{2} x^{4}+3 a b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i \left (-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {2 \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, a \EllipticF \left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (-2 b x +i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right )\right ) x}{15 \sqrt {b \,x^{5}+a \,x^{2}}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^5+a*x^2)^(1/2),x)

[Out]

2/15*x*(I*a*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))+3*b^2*x^4+3*a*b*x)/(b*x^5+
a*x^2)^(1/2)/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {b x^{5} + a x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(b*x^5 + a*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{\sqrt {b\,x^5+a\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x^2 + b*x^5)^(1/2),x)

[Out]

int(x^4/(a*x^2 + b*x^5)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {x^{2} \left (a + b x^{3}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**5+a*x**2)**(1/2),x)

[Out]

Integral(x**4/sqrt(x**2*(a + b*x**3)), x)

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