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

Optimal. Leaf size=292 \[ \frac{\left (1+\sqrt{3}\right ) \sqrt{x^3+1} \sqrt{a x^3}}{x \left (\left (1+\sqrt{3}\right ) x+1\right )}-\frac{\left (1-\sqrt{3}\right ) (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{a x^3} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2 \sqrt [4]{3} x \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}}-\frac{\sqrt [4]{3} (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{a x^3} E\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{x \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}} \]

[Out]

((1 + Sqrt[3])*Sqrt[a*x^3]*Sqrt[1 + x^3])/(x*(1 + (1 + Sqrt[3])*x)) - (3^(1/4)*Sqrt[a*x^3]*(1 + x)*Sqrt[(1 - x
+ x^2)/(1 + (1 + Sqrt[3])*x)^2]*EllipticE[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3])/
4])/(x*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3]) - ((1 - Sqrt[3])*Sqrt[a*x^3]*(1 + x)*Sqrt[(1 -
x + x^2)/(1 + (1 + Sqrt[3])*x)^2]*EllipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3]
)/4])/(2*3^(1/4)*x*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.231685, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {15, 329, 308, 225, 1881} \[ \frac{\left (1+\sqrt{3}\right ) \sqrt{x^3+1} \sqrt{a x^3}}{x \left (\left (1+\sqrt{3}\right ) x+1\right )}-\frac{\left (1-\sqrt{3}\right ) (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{a x^3} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2 \sqrt [4]{3} x \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}}-\frac{\sqrt [4]{3} (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{a x^3} E\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{x \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^3]/Sqrt[1 + x^3],x]

[Out]

((1 + Sqrt[3])*Sqrt[a*x^3]*Sqrt[1 + x^3])/(x*(1 + (1 + Sqrt[3])*x)) - (3^(1/4)*Sqrt[a*x^3]*(1 + x)*Sqrt[(1 - x
+ x^2)/(1 + (1 + Sqrt[3])*x)^2]*EllipticE[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3])/
4])/(x*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3]) - ((1 - Sqrt[3])*Sqrt[a*x^3]*(1 + x)*Sqrt[(1 -
x + x^2)/(1 + (1 + Sqrt[3])*x)^2]*EllipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3]
)/4])/(2*3^(1/4)*x*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[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 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 308

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(
(Sqrt[3] - 1)*s^2)/(2*r^2), Int[1/Sqrt[a + b*x^6], x], x] - Dist[1/(2*r^2), Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4
)/Sqrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, 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]

Rule 1881

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/
a, 3]]}, Simp[((1 + Sqrt[3])*d*s^3*x*Sqrt[a + b*x^6])/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2)), x] - Simp[(3^(1/4)*
d*s*x*(s + r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticE[ArcCos[(s + (1 - Sqrt[
3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*
x^2)^2]*Sqrt[a + b*x^6]), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{a x^3}}{\sqrt{1+x^3}} \, dx &=\frac{\sqrt{a x^3} \int \frac{x^{3/2}}{\sqrt{1+x^3}} \, dx}{x^{3/2}}\\ &=\frac{\left (2 \sqrt{a x^3}\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^6}} \, dx,x,\sqrt{x}\right )}{x^{3/2}}\\ &=-\frac{\sqrt{a x^3} \operatorname{Subst}\left (\int \frac{-1+\sqrt{3}-2 x^4}{\sqrt{1+x^6}} \, dx,x,\sqrt{x}\right )}{x^{3/2}}+\frac{\left (\left (-1+\sqrt{3}\right ) \sqrt{a x^3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^6}} \, dx,x,\sqrt{x}\right )}{x^{3/2}}\\ &=\frac{\left (1+\sqrt{3}\right ) \sqrt{a x^3} \sqrt{1+x^3}}{x \left (1+\left (1+\sqrt{3}\right ) x\right )}-\frac{\sqrt [4]{3} \sqrt{a x^3} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} E\left (\cos ^{-1}\left (\frac{1+\left (1-\sqrt{3}\right ) x}{1+\left (1+\sqrt{3}\right ) x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{x \sqrt{\frac{x (1+x)}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} \sqrt{1+x^3}}-\frac{\left (1-\sqrt{3}\right ) \sqrt{a x^3} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} F\left (\cos ^{-1}\left (\frac{1+\left (1-\sqrt{3}\right ) x}{1+\left (1+\sqrt{3}\right ) x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{2 \sqrt [4]{3} x \sqrt{\frac{x (1+x)}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0041012, size = 29, normalized size = 0.1 \[ \frac{2}{5} x \sqrt{a x^3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};-x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[a*x^3]*Hypergeometric2F1[1/2, 5/6, 11/6, -x^3])/5

________________________________________________________________________________________

Maple [C]  time = 0.234, size = 1521, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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