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]

2/3*(a*x^3)^(1/2)*(x^2+1)^(1/2)/x-1/3*(1+x)*(cos(2*arctan(x^(1/2)))^2)^(1/2)/cos(2*arctan(x^(1/2)))*EllipticF(
sin(2*arctan(x^(1/2))),1/2*2^(1/2))*(a*x^3)^(1/2)*((x^2+1)/(1+x)^2)^(1/2)/x^(3/2)/(x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 83, 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, 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 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]

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]

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 = 0.01, 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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [C]  time = 0.03, size = 76, normalized size = 0.92 \[ -\frac {\sqrt {a \,x^{3}}\, \left (-2 x^{3}-2 x +i \sqrt {-i \left (x +i\right )}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \sqrt {2}\, \EllipticF \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{3 \sqrt {x^{2}+1}\, x^{2}} \]

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*(-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^(1/2)-2*x^3-2*x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a\,x^3}}{\sqrt {x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3)^(1/2)/(x^2 + 1)^(1/2),x)

[Out]

int((a*x^3)^(1/2)/(x^2 + 1)^(1/2), x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________