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+3^(1/2))*(a*x^3)^(1/2)*(x^3+1)^(1/2)/x/(1+x*(1+3^(1/2)))-3^(1/4)*(1+x)*((1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)
))^2)^(1/2)/(1+x*(1-3^(1/2)))*(1+x*(1+3^(1/2)))*EllipticE((1-(1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2),1/
4*6^(1/2)+1/4*2^(1/2))*(a*x^3)^(1/2)*((x^2-x+1)/(1+x*(1+3^(1/2)))^2)^(1/2)/x/(x^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^
(1/2)))^2)^(1/2)-1/6*(1+x)*((1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2)/(1+x*(1-3^(1/2)))*(1+x*(1+3^(1/2)))
*EllipticF((1-(1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(1-3^(1/2))*(a*x^3)^(1/2
)*((x^2-x+1)/(1+x*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/x/(x^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^(1/2)))^2)^(1/2)
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Rubi [A] time = 0.23, antiderivative size = 292, normalized size of antiderivative = 1.00,
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 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 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 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 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*}
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Mathematica [C] time = 0.01, size = 29, normalized size = 0.10 \[ \frac {2}{5} x \sqrt {a x^3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};-x^3\right ) \]
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
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}}, x\right ) \]
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)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}}\,{d x} \]
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)
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maple [C] time = 0.28, size = 1521, normalized size = 5.21 \[ \text {result too large to display} \]
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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(
I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2
))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))
/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-
1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^
(1/2),((-3+I*3^(1/2))*(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))/(x+1)
)^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticF(
((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*
x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*
3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticF(((3+I*3^(1/2
))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1
/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(
I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticF(((3+I*3^(1/2))*x/(1+I*3^(1/2
))/(x+1))^(1/2),((-3+I*3^(1/2))*(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))/(x+1))^(1/2)*((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(x+1))^(1/2)*
EllipticE(((3+I*3^(1/2))*x/(1+I*3^(1/2))/(x+1))^(1/2),((-3+I*3^(1/2))*(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*(x+1)*(I*3^(1/2)+2*x-1)*(I*3^(1
/2)-2*x+1))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}}\,{d x} \]
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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a\,x^3}}{\sqrt {x^3+1}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a*x^3)^(1/2)/(x^3 + 1)^(1/2),x)
[Out]
int((a*x^3)^(1/2)/(x^3 + 1)^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a x^{3}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a*x**3)**(1/2)/(x**3+1)**(1/2),x)
[Out]
Integral(sqrt(a*x**3)/sqrt((x + 1)*(x**2 - x + 1)), x)
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