3.516 \(\int \frac{1}{x^3 (1+x)^{3/2} (1-x+x^2)^{3/2}} \, dx\)

Optimal. Leaf size=170 \[ -\frac{7 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{6 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}}-\frac{7 \left (x^3+1\right )}{6 x^2 \sqrt{x+1} \sqrt{x^2-x+1}}+\frac{2}{3 x^2 \sqrt{x+1} \sqrt{x^2-x+1}} \]

[Out]

2/(3*x^2*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (7*(1 + x^3))/(6*x^2*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (7*Sqrt[2 + Sq
rt[3]]*Sqrt[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]])/(6*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2])

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Rubi [A]  time = 0.0652856, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {915, 290, 325, 218} \[ -\frac{7 \left (x^3+1\right )}{6 x^2 \sqrt{x+1} \sqrt{x^2-x+1}}+\frac{2}{3 x^2 \sqrt{x+1} \sqrt{x^2-x+1}}-\frac{7 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{6 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]

[Out]

2/(3*x^2*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (7*(1 + x^3))/(6*x^2*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (7*Sqrt[2 + Sq
rt[3]]*Sqrt[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]])/(6*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2])

Rule 915

Int[((g_.)*(x_))^(n_)*((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[((d
+ e*x)^FracPart[p]*(a + b*x + c*x^2)^FracPart[p])/(a*d + c*e*x^3)^FracPart[p], Int[(g*x)^n*(a*d + c*e*x^3)^p,
x], x] /; FreeQ[{a, b, c, d, e, g, m, n, p}, x] && EqQ[m - p, 0] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1+x^3} \int \frac{1}{x^3 \left (1+x^3\right )^{3/2}} \, dx}{\sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 x^2 \sqrt{1+x} \sqrt{1-x+x^2}}+\frac{\left (7 \sqrt{1+x^3}\right ) \int \frac{1}{x^3 \sqrt{1+x^3}} \, dx}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 x^2 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{7 \left (1+x^3\right )}{6 x^2 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{\left (7 \sqrt{1+x^3}\right ) \int \frac{1}{\sqrt{1+x^3}} \, dx}{12 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2}{3 x^2 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{7 \left (1+x^3\right )}{6 x^2 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{7 \sqrt{2+\sqrt{3}} \sqrt{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 )}{6 \sqrt [4]{3} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1-x+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.402872, size = 170, normalized size = 1. \[ \frac{-\frac{6 \left (7 x^3+3\right )}{x^2 \sqrt{x+1}}-\frac{7 i (x+1) \sqrt{1+\frac{6 i}{\left (\sqrt{3}-3 i\right ) (x+1)}} \sqrt{6-\frac{36 i}{\left (\sqrt{3}+3 i\right ) (x+1)}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{\sqrt{3}+3 i}}}{\sqrt{x+1}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )}{\sqrt{-\frac{i}{\sqrt{3}+3 i}}}}{36 \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]

[Out]

((-6*(3 + 7*x^3))/(x^2*Sqrt[1 + x]) - ((7*I)*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[6 - (36*I
)/((3*I + Sqrt[3])*(1 + x))]*EllipticF[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3
*I - Sqrt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(36*Sqrt[1 - x + x^2])

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Maple [A]  time = 1.35, size = 259, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( 12\,{x}^{3}+12 \right ){x}^{2}}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( 7\,i\sqrt{3}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}},\sqrt{-{\frac{i\sqrt{3}-3}{i\sqrt{3}+3}}} \right ){x}^{2}\sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{i\sqrt{3}+3}}}\sqrt{{\frac{2\,x-1+i\sqrt{3}}{i\sqrt{3}-3}}}-21\,\sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{i\sqrt{3}+3}}}\sqrt{{\frac{2\,x-1+i\sqrt{3}}{i\sqrt{3}-3}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}},\sqrt{-{\frac{i\sqrt{3}-3}{i\sqrt{3}+3}}} \right ){x}^{2}-14\,{x}^{3}-6 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(7*I*3^(1/2)*EllipticF((-2*(1+x)/(I*3^(1/2)-3))^(1/2),(-(I*3^(1/2)-3)/(I*3^(1
/2)+3))^(1/2))*x^2*(-2*(1+x)/(I*3^(1/2)-3))^(1/2)*((I*3^(1/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((2*x-1+I*3^(1/2))/(
I*3^(1/2)-3))^(1/2)-21*(-2*(1+x)/(I*3^(1/2)-3))^(1/2)*((I*3^(1/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((2*x-1+I*3^(1/2
))/(I*3^(1/2)-3))^(1/2)*EllipticF((-2*(1+x)/(I*3^(1/2)-3))^(1/2),(-(I*3^(1/2)-3)/(I*3^(1/2)+3))^(1/2))*x^2-14*
x^3-6)/(x^3+1)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x^{9} + 2 \, x^{6} + x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^2 - x + 1)*sqrt(x + 1)/(x^9 + 2*x^6 + x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)*x^3), x)