3.387 \(\int \frac {\sqrt {a x^2}}{\sqrt {1+x^3}} \, dx\)

Optimal. Leaf size=260 \[ \frac {2 \sqrt {x^3+1} \sqrt {a x^2}}{x \left (x+\sqrt {3}+1\right )}+\frac {2 \sqrt {2} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {a x^2} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {a x^2} E\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{x \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {15, 303, 218, 1877} \[ \frac {2 \sqrt {x^3+1} \sqrt {a x^2}}{x \left (x+\sqrt {3}+1\right )}+\frac {2 \sqrt {2} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {a x^2} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {a x^2} E\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{x \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2]/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[a*x^2]*Sqrt[1 + x^3])/(x*(1 + Sqrt[3] + x)) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*Sqrt[a*x^2]*(1 + x)*Sqrt[(1 -
x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(x*Sqrt
[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) + (2*Sqrt[2]*Sqrt[a*x^2]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3]
+ x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(1/4)*x*Sqrt[(1 + x)/(1 + S
qrt[3] + x)^2]*Sqrt[1 + x^3])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 303

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

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {a x^2}}{\sqrt {1+x^3}} \, dx &=\frac {\sqrt {a x^2} \int \frac {x}{\sqrt {1+x^3}} \, dx}{x}\\ &=\frac {\sqrt {a x^2} \int \frac {1-\sqrt {3}+x}{\sqrt {1+x^3}} \, dx}{x}+\frac {\left (\sqrt {2 \left (2-\sqrt {3}\right )} \sqrt {a x^2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{x}\\ &=\frac {2 \sqrt {a x^2} \sqrt {1+x^3}}{x \left (1+\sqrt {3}+x\right )}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {a x^2} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{x \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}+\frac {2 \sqrt {2} \sqrt {a x^2} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 29, normalized size = 0.11 \[ \frac {1}{2} x \sqrt {a x^2} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2]/Sqrt[1 + x^3],x]

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a x^{2}}}{\sqrt {x^{3} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2)^(1/2)/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*x^2)/sqrt(x^3 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x^{2}}}{\sqrt {x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2)^(1/2)/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*x^2)/sqrt(x^3 + 1), x)

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maple [A]  time = 0.06, size = 270, normalized size = 1.04 \[ \frac {\sqrt {a \,x^{2}}\, \left (-3+i \sqrt {3}\right ) \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{3+i \sqrt {3}}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \left (i \sqrt {3}\, \EllipticE \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{3+i \sqrt {3}}}\right )+3 \EllipticE \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{3+i \sqrt {3}}}\right )-i \sqrt {3}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{3+i \sqrt {3}}}\right )-\EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{3+i \sqrt {3}}}\right )\right )}{2 \sqrt {x^{3}+1}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2)^(1/2)/(x^3+1)^(1/2),x)

[Out]

1/2*(a*x^2)^(1/2)*(-3+I*3^(1/2))*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((-2*x+I*3^(1/2)+1)/(3+I*3^(1/2)))^(1/2)*((2*
x+I*3^(1/2)-1)/(-3+I*3^(1/2)))^(1/2)*(I*EllipticE((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(3+I*3^(1/2
)))^(1/2))*3^(1/2)-I*EllipticF((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(3+I*3^(1/2)))^(1/2))*3^(1/2)+
3*EllipticE((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(3+I*3^(1/2)))^(1/2))-EllipticF((-2*(x+1)/(-3+I*3
^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(3+I*3^(1/2)))^(1/2)))/x/(x^3+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x^{2}}}{\sqrt {x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2)^(1/2)/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^2)/sqrt(x^3 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a\,x^2}}{\sqrt {x^3+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2)^(1/2)/(x^3 + 1)^(1/2),x)

[Out]

int((a*x^2)^(1/2)/(x^3 + 1)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**2)**(1/2)/(x**3+1)**(1/2),x)

[Out]

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

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