3.4.1 \(\int \frac {\sqrt {x}}{\sqrt {a x^2+b x^5}} \, dx\) [301]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 203, normalized size of antiderivative = 1.00, 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 = {2057, 335, 231} \begin {gather*} \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 (\text {ArcCos}\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}} \end {gather*}

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 231

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

Rule 335

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 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 {\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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 55, normalized size = 0.27 \begin {gather*} \frac {2 x^{3/2} \sqrt {1+\frac {b x^3}{a}} \, _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 )}} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.52, size = 437, normalized size = 2.15

method	result	size
default	\(-\frac {4 x^{\frac {3}{2}} \left (b \,x^{3}+a \right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right ) \left (i \sqrt {3}\, b^{2} x^{2}-2 i \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {3}\, b x +i \left (-a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}-b^{2} x^{2}+2 \left (-a \,b^{2}\right )^{\frac {1}{3}} b x -\left (-a \,b^{2}\right )^{\frac {2}{3}}\right )}{\sqrt {b \,x^{5}+a \,x^{2}}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} b \sqrt {x \left (b \,x^{3}+a \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{b^{2}}}}\)	\(437\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(b*x^5+a*x^2)^(1/2)*x^(3/2)*(b*x^3+a)/(-a*b^2)^(1/3)/b*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1
/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^
(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-
3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(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)*b^2*x^2-2*I*(-a*b^2)^(1/3)*3^(1/2)*b*x+I*(-a*b^2)^(2/3)*3^(1/2)-b^2*x^2+2*(-a*b^2)^(1/3)*b*x
-(-a*b^2)^(2/3))/(x*(b*x^3+a))^(1/2)/(I*3^(1/2)-3)/(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*
b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(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^(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.43, size = 16, normalized size = 0.08 \begin {gather*} -\frac {2 \, {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right )}{\sqrt {a}} \end {gather*}

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]

-2*weierstrassPInverse(0, -4*b/a, 1/x)/sqrt(a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {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**(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 = 0.00, 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^(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)

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

[Out]

int(x^(1/2)/(a*x^2 + b*x^5)^(1/2), x)

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