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

Optimal. Leaf size=203 \[ \frac{x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{\sqrt [4]{3} \sqrt [3]{a} \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a x^2+b x^5}} \]

[Out]

(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])

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Rubi [A]  time = 0.165711, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2032, 329, 225} \[ \frac{x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{\sqrt [4]{3} \sqrt [3]{a} \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a x^2+b x^5}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[a*x^2 + b*x^5],x]

[Out]

(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])

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]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[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 225

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]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(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]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\sqrt{a x^2+b x^5}} \, dx &=\frac{\left (x \sqrt{a+b x^3}\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^3}} \, dx}{\sqrt{a x^2+b x^5}}\\ &=\frac{\left (2 x \sqrt{a+b x^3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^6}} \, dx,x,\sqrt{x}\right )}{\sqrt{a x^2+b x^5}}\\ &=\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}} F\left (\cos ^{-1}\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}}\\ \end{align*}

Mathematica [C]  time = 0.0135442, size = 55, normalized size = 0.27 \[ \frac{2 x^{3/2} \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{b x^3}{a}\right )}{\sqrt{x^2 \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)]

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Maple [C]  time = 0.126, size = 437, normalized size = 2.2 \begin{align*} -4\,{\frac{{x}^{3/2} \left ( b{x}^{3}+a \right ) \left ( i\sqrt{3}{x}^{2}{b}^{2}-2\,i\sqrt [3]{-{b}^{2}a}\sqrt{3}xb+i \left ( -{b}^{2}a \right ) ^{2/3}\sqrt{3}-{b}^{2}{x}^{2}+2\,\sqrt [3]{-{b}^{2}a}xb- \left ( -{b}^{2}a \right ) ^{2/3} \right ) }{\sqrt{b{x}^{5}+a{x}^{2}}\sqrt [3]{-{b}^{2}a}b\sqrt{ \left ( b{x}^{3}+a \right ) x} \left ( i\sqrt{3}-3 \right ) }\sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xb}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{b}^{2}a}+2\,bx+\sqrt [3]{-{b}^{2}a}}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{b}^{2}a}-2\,bx-\sqrt [3]{-{b}^{2}a}}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}}{\it EllipticF} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xb}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}},\sqrt{{\frac{ \left ( i\sqrt{3}+3 \right ) \left ( -1+i\sqrt{3} \right ) }{ \left ( i\sqrt{3}-3 \right ) \left ( 1+i\sqrt{3} \right ) }}} \right ){\frac{1}{\sqrt{{\frac{x \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{-{b}^{2}a}+2\,bx+\sqrt [3]{-{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{-{b}^{2}a}-2\,bx-\sqrt [3]{-{b}^{2}a} \right ) }{{b}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\sqrt{b x^{5} + a x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x}}{\sqrt{b x^{5} + a x^{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\sqrt{b x^{5} + a x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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