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

Optimal. Leaf size=175 \[ \frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (x+1)^{3/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{10 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )}-\frac{\sqrt{x^2-x+1} \left (x^3+1\right ) \sqrt{x+1}}{2 x^2}+\frac{9}{10} x \sqrt{x^2-x+1} \sqrt{x+1} \]

[Out]

(9*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/10 - (Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3))/(2*x^2) + (9*3^(3/4)*Sqrt[2
+ Sqrt[3]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt
[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(10*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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Rubi [A]  time = 0.0584062, antiderivative size = 175, 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, 277, 195, 218} \[ -\frac{\sqrt{x^2-x+1} \left (x^3+1\right ) \sqrt{x+1}}{2 x^2}+\frac{9}{10} x \sqrt{x^2-x+1} \sqrt{x+1}+\frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (x+1)^{3/2} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{10 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/10 - (Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3))/(2*x^2) + (9*3^(3/4)*Sqrt[2
+ Sqrt[3]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt
[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(10*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx &=\frac{\left (\sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \frac{\left (1+x^3\right )^{3/2}}{x^3} \, dx}{\sqrt{1+x^3}}\\ &=-\frac{\sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac{\left (9 \sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \sqrt{1+x^3} \, dx}{4 \sqrt{1+x^3}}\\ &=\frac{9}{10} x \sqrt{1+x} \sqrt{1-x+x^2}-\frac{\sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac{\left (27 \sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \frac{1}{\sqrt{1+x^3}} \, dx}{20 \sqrt{1+x^3}}\\ &=\frac{9}{10} x \sqrt{1+x} \sqrt{1-x+x^2}-\frac{\sqrt{1+x} \sqrt{1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} (1+x)^{3/2} \sqrt{1-x+x^2} \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 )}{10 \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \left (1+x^3\right )}\\ \end{align*}

Mathematica [C]  time = 0.354324, size = 192, normalized size = 1.1 \[ \frac{\sqrt{x+1} \left (\frac{2 \left (x^2-x+1\right ) \left (4 x^3-5\right )}{x^2}-\frac{27 i \sqrt{2} \sqrt{\frac{-2 i x+\sqrt{3}+i}{\sqrt{3}+3 i}} \sqrt{\frac{2 i x+\sqrt{3}-i}{\sqrt{3}-3 i}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{-\frac{i (x+1)}{\sqrt{3}+3 i}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )}{\sqrt{-\frac{i (x+1)}{\sqrt{3}+3 i}}}\right )}{20 \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x]*((2*(1 - x + x^2)*(-5 + 4*x^3))/x^2 - ((27*I)*Sqrt[2]*Sqrt[(I + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3]
)]*Sqrt[(-I + Sqrt[3] + (2*I)*x)/(-3*I + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[((-I)*(1 + x))/(3*I + Sqrt
[3])]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3])]))/(20*Sqrt[1 - x + x^2])

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Maple [A]  time = 0.592, size = 264, normalized size = 1.5 \begin{align*} -{\frac{1}{ \left ( 20\,{x}^{3}+20 \right ){x}^{2}}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( 27\,i\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 ) \sqrt{3}{x}^{2}-81\,\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}-8\,{x}^{6}+2\,{x}^{3}+10 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(27*I*(-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))*3^(1/2)*x^2-81*(-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^2-8
*x^6+2*x^3+10)/(x^3+1)/x^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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)^(3/2)*(x^2-x+1)^(3/2)/x^3,x, algorithm="maxima")

[Out]

integrate((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{{\left (x^{3} + 1\right )} \sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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)^(3/2)*(x^2-x+1)^(3/2)/x^3,x, algorithm="giac")

[Out]

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