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

Optimal. Leaf size=312 \[ -2 \sqrt {x^3+1} x \sqrt {\frac {a}{x^3}}+\frac {2 \left (1+\sqrt {3}\right ) \sqrt {x^3+1} x^2 \sqrt {\frac {a}{x^3}}}{\left (1+\sqrt {3}\right ) x+1}-\frac {\left (1-\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} x^2 \sqrt {\frac {a}{x^3}} F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}-\frac {2 \sqrt [4]{3} (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} x^2 \sqrt {\frac {a}{x^3}} E\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}} \]

[Out]

-2*x*(a/x^3)^(1/2)*(x^3+1)^(1/2)+2*x^2*(1+3^(1/2))*(a/x^3)^(1/2)*(x^3+1)^(1/2)/(1+x*(1+3^(1/2)))-2*3^(1/4)*x^2
*(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^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^(1/2)))^2)^(1/2)-1/3*x^2*(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^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 = 312, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {15, 325, 329, 308, 225, 1881} \[ \frac {2 \left (1+\sqrt {3}\right ) \sqrt {x^3+1} x^2 \sqrt {\frac {a}{x^3}}}{\left (1+\sqrt {3}\right ) x+1}-2 \sqrt {x^3+1} x \sqrt {\frac {a}{x^3}}-\frac {\left (1-\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} x^2 \sqrt {\frac {a}{x^3}} F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}-\frac {2 \sqrt [4]{3} (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} x^2 \sqrt {\frac {a}{x^3}} E\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a/x^3]/Sqrt[1 + x^3],x]

[Out]

-2*Sqrt[a/x^3]*x*Sqrt[1 + x^3] + (2*(1 + Sqrt[3])*Sqrt[a/x^3]*x^2*Sqrt[1 + x^3])/(1 + (1 + Sqrt[3])*x) - (2*3^
(1/4)*Sqrt[a/x^3]*x^2*(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])/(Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3]) - ((1 -
Sqrt[3])*Sqrt[a/x^3]*x^2*(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])/(3^(1/4)*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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 {\frac {a}{x^3}}}{\sqrt {1+x^3}} \, dx &=\left (\sqrt {\frac {a}{x^3}} x^{3/2}\right ) \int \frac {1}{x^{3/2} \sqrt {1+x^3}} \, dx\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^3}+\left (2 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \int \frac {x^{3/2}}{\sqrt {1+x^3}} \, dx\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^3}+\left (4 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^3}-\left (2 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {-1+\sqrt {3}-2 x^4}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )+\left (2 \left (-1+\sqrt {3}\right ) \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^3}+\frac {2 \left (1+\sqrt {3}\right ) \sqrt {\frac {a}{x^3}} x^2 \sqrt {1+x^3}}{1+\left (1+\sqrt {3}\right ) x}-\frac {2 \sqrt [4]{3} \sqrt {\frac {a}{x^3}} x^2 (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 )}{\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 {\frac {a}{x^3}} x^2 (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 )}{\sqrt [4]{3} \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 = 27, normalized size = 0.09 \[ -2 x \sqrt {\frac {a}{x^3}} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};-x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{3} + 1}}, x\right ) \]

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 \[ \int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{3} + 1}}\,{d x} \]

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 = 0.14, size = 1784, normalized size = 5.72 \[ \text {result too large to display} \]

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)*(-2*I*3^(1/2)*((x^3+1)*x)^(1/2)*x-2*I*3^(1/2)*((x^3+1)*x)^(1/2)*x^3+I*3^(1/2)
*(-(x+1)*(2*x+I*3^(1/2)-1)*(-2*x+I*3^(1/2)+1)*x)^(1/2)-4*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*
x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*Elliptic
F(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(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+6*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^
(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(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+1)*(2*x+I*3^(1/2)-1)*(-2*x+I
*3^(1/2)+1)*x)^(1/2)*x^3+4*I*3^(1/2)*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(
1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))/(1+
I*3^(1/2))/(x+1)*x)^(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-8*((x^3+1)*x)^(1
/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)
+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticF(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(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+12*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((
2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(
1/2))/(1+I*3^(1/2))/(x+1)*x)^(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+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/
2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2),((-3+I
*3^(1/2))*(1+I*3^(1/2))/(I*3^(1/2)-1)/(3+I*3^(1/2)))^(1/2))-4*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(
x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*Ell
ipticF(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2),((-3+I*3^(1/2))*(1+I*3^(1/2))/(I*3^(1/2)-1)/(3+I*3^(1/2)))^
(1/2))+6*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))
^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x+1))^(1/2)*EllipticE(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(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*I*3^(1/2)*((x^3+1)*x)^(1/2)*((3+I*3^(1/2))/(
1+I*3^(1/2))/(x+1)*x)^(1/2)*((2*x+I*3^(1/2)-1)/(I*3^(1/2)-1)/(x+1))^(1/2)*((-2*x+I*3^(1/2)+1)/(1+I*3^(1/2))/(x
+1))^(1/2)*EllipticE(((3+I*3^(1/2))/(1+I*3^(1/2))/(x+1)*x)^(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^2-6*((x^3+1)*x)^(1/2)*x^3+3*(-(x+1)*(2*x+I*3^(1/2)-1)*(-2*x+I*3^(1/2)+1)*x)^(1/2)*x^3+2
*I*3^(1/2)*((x^3+1)*x)^(1/2)*x^2+6*((x^3+1)*x)^(1/2)*x^2-6*((x^3+1)*x)^(1/2)*x+3*(-(x+1)*(2*x+I*3^(1/2)-1)*(-2
*x+I*3^(1/2)+1)*x)^(1/2))/(3+I*3^(1/2))/(-(x+1)*(2*x+I*3^(1/2)-1)*(-2*x+I*3^(1/2)+1)*x)^(1/2)

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

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 \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {x^3+1}} \,d x \]

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 \[ \int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \]

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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