3.455 \(\int \frac{1}{x^5 \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=262 \[ -\frac{5 (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{4 \sqrt{2} \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}+\frac{5 \sqrt{x^3+1}}{8 x}-\frac{5 \sqrt{x^3+1}}{8 \left (x+\sqrt{3}+1\right )}-\frac{\sqrt{x^3+1}}{4 x^4}+\frac{5 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{16 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

[Out]

-Sqrt[1 + x^3]/(4*x^4) + (5*Sqrt[1 + x^3])/(8*x) - (5*Sqrt[1 + x^3])/(8*(1 + Sqrt[3] + x)) + (5*3^(1/4)*Sqrt[2
- Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] +
x)], -7 - 4*Sqrt[3]])/(16*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (5*(1 + x)*Sqrt[(1 - x + x^2)/(1
+ Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(4*Sqrt[2]*3^(1/4)*S
qrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])

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Rubi [A]  time = 0.0660223, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {325, 303, 218, 1877} \[ \frac{5 \sqrt{x^3+1}}{8 x}-\frac{5 \sqrt{x^3+1}}{8 \left (x+\sqrt{3}+1\right )}-\frac{\sqrt{x^3+1}}{4 x^4}-\frac{5 (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{4 \sqrt{2} \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}+\frac{5 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{16 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[1 + x^3]),x]

[Out]

-Sqrt[1 + x^3]/(4*x^4) + (5*Sqrt[1 + x^3])/(8*x) - (5*Sqrt[1 + x^3])/(8*(1 + Sqrt[3] + x)) + (5*3^(1/4)*Sqrt[2
- Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] +
x)], -7 - 4*Sqrt[3]])/(16*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (5*(1 + x)*Sqrt[(1 - x + x^2)/(1
+ Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(4*Sqrt[2]*3^(1/4)*S
qrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])

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 303

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

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin{align*} \int \frac{1}{x^5 \sqrt{1+x^3}} \, dx &=-\frac{\sqrt{1+x^3}}{4 x^4}-\frac{5}{8} \int \frac{1}{x^2 \sqrt{1+x^3}} \, dx\\ &=-\frac{\sqrt{1+x^3}}{4 x^4}+\frac{5 \sqrt{1+x^3}}{8 x}-\frac{5}{16} \int \frac{x}{\sqrt{1+x^3}} \, dx\\ &=-\frac{\sqrt{1+x^3}}{4 x^4}+\frac{5 \sqrt{1+x^3}}{8 x}-\frac{5}{16} \int \frac{1-\sqrt{3}+x}{\sqrt{1+x^3}} \, dx-\frac{1}{8} \left (5 \sqrt{\frac{1}{2} \left (2-\sqrt{3}\right )}\right ) \int \frac{1}{\sqrt{1+x^3}} \, dx\\ &=-\frac{\sqrt{1+x^3}}{4 x^4}+\frac{5 \sqrt{1+x^3}}{8 x}-\frac{5 \sqrt{1+x^3}}{8 \left (1+\sqrt{3}+x\right )}+\frac{5 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\sqrt{3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac{1-\sqrt{3}+x}{1+\sqrt{3}+x}\right )|-7-4 \sqrt{3}\right )}{16 \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1+x^3}}-\frac{5 (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\sqrt{3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac{1-\sqrt{3}+x}{1+\sqrt{3}+x}\right )|-7-4 \sqrt{3}\right )}{4 \sqrt{2} \sqrt [4]{3} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0027303, size = 22, normalized size = 0.08 \[ -\frac{\, _2F_1\left (-\frac{4}{3},\frac{1}{2};-\frac{1}{3};-x^3\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[1 + x^3]),x]

[Out]

-Hypergeometric2F1[-4/3, 1/2, -1/3, -x^3]/(4*x^4)

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Maple [A]  time = 0.018, size = 198, normalized size = 0.8 \begin{align*} -{\frac{1}{4\,{x}^{4}}\sqrt{{x}^{3}+1}}+{\frac{5}{8\,x}\sqrt{{x}^{3}+1}}-{\frac{{\frac{15}{2}}-{\frac{5\,i}{2}}\sqrt{3}}{8}\sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}}\sqrt{{\frac{1}{-{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }}\sqrt{{\frac{1}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) }} \left ( \left ( -{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3} \right ){\it EllipticE} \left ( \sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}},\sqrt{{\frac{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}{-{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}} \right ) + \left ({\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ){\it EllipticF} \left ( \sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}},\sqrt{{\frac{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}{-{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}} \right ) \right ){\frac{1}{\sqrt{{x}^{3}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(x^3+1)^(1/2),x)

[Out]

-1/4*(x^3+1)^(1/2)/x^4+5/8*(x^3+1)^(1/2)/x-5/8*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1
/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*((-
3/2-1/2*I*3^(1/2))*EllipticE(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/
2))+(1/2+1/2*I*3^(1/2))*EllipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2))
)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{3} + 1} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^3 + 1)*x^5), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{3} + 1}}{x^{8} + x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^3 + 1)/(x^8 + x^5), x)

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Sympy [A]  time = 1.02435, size = 36, normalized size = 0.14 \begin{align*} \frac{\Gamma \left (- \frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, \frac{1}{2} \\ - \frac{1}{3} \end{matrix}\middle |{x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(x**3+1)**(1/2),x)

[Out]

gamma(-4/3)*hyper((-4/3, 1/2), (-1/3,), x**3*exp_polar(I*pi))/(3*x**4*gamma(-1/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{3} + 1} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^3 + 1)*x^5), x)