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

Optimal. Leaf size=83 \[ \frac{2 \sqrt{x^2+1} \sqrt{a x^3}}{3 x}-\frac{(x+1) \sqrt{\frac{x^2+1}{(x+1)^2}} \sqrt{a x^3} F\left (2 \tan ^{-1}\left (\sqrt{x}\right )|\frac{1}{2}\right )}{3 x^{3/2} \sqrt{x^2+1}} \]

[Out]

(2*Sqrt[a*x^3]*Sqrt[1 + x^2])/(3*x) - (Sqrt[a*x^3]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x
]], 1/2])/(3*x^(3/2)*Sqrt[1 + x^2])

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Rubi [A]  time = 0.0258057, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {15, 321, 329, 220} \[ \frac{2 \sqrt{x^2+1} \sqrt{a x^3}}{3 x}-\frac{(x+1) \sqrt{\frac{x^2+1}{(x+1)^2}} \sqrt{a x^3} F\left (2 \tan ^{-1}\left (\sqrt{x}\right )|\frac{1}{2}\right )}{3 x^{3/2} \sqrt{x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^3]/Sqrt[1 + x^2],x]

[Out]

(2*Sqrt[a*x^3]*Sqrt[1 + x^2])/(3*x) - (Sqrt[a*x^3]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x
]], 1/2])/(3*x^(3/2)*Sqrt[1 + x^2])

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{\sqrt{a x^3}}{\sqrt{1+x^2}} \, dx &=\frac{\sqrt{a x^3} \int \frac{x^{3/2}}{\sqrt{1+x^2}} \, dx}{x^{3/2}}\\ &=\frac{2 \sqrt{a x^3} \sqrt{1+x^2}}{3 x}-\frac{\sqrt{a x^3} \int \frac{1}{\sqrt{x} \sqrt{1+x^2}} \, dx}{3 x^{3/2}}\\ &=\frac{2 \sqrt{a x^3} \sqrt{1+x^2}}{3 x}-\frac{\left (2 \sqrt{a x^3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\sqrt{x}\right )}{3 x^{3/2}}\\ &=\frac{2 \sqrt{a x^3} \sqrt{1+x^2}}{3 x}-\frac{\sqrt{a x^3} (1+x) \sqrt{\frac{1+x^2}{(1+x)^2}} F\left (2 \tan ^{-1}\left (\sqrt{x}\right )|\frac{1}{2}\right )}{3 x^{3/2} \sqrt{1+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0080856, size = 43, normalized size = 0.52 \[ \frac{2 \sqrt{a x^3} \left (\sqrt{x^2+1}-\, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-x^2\right )\right )}{3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.021, size = 76, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{x}^{2}}\sqrt{a{x}^{3}} \left ( i\sqrt{2}\sqrt{-i \left ( x+i \right ) }\sqrt{-i \left ( -x+i \right ) }\sqrt{ix}{\it EllipticF} \left ( \sqrt{-i \left ( x+i \right ) },{\frac{\sqrt{2}}{2}} \right ) -2\,{x}^{3}-2\,x \right ){\frac{1}{\sqrt{{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a x^{3}}}{\sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a x^{3}}}{\sqrt{x^{2} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a x^{3}}}{\sqrt{x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a x^{3}}}{\sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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