3.841 \(\int \sqrt{-1+x} x^2 \sqrt{1+x} \, dx\)

Optimal. Leaf size=45 \[ \frac{1}{4} (x-1)^{3/2} x (x+1)^{3/2}+\frac{1}{8} \sqrt{x-1} x \sqrt{x+1}-\frac{1}{8} \cosh ^{-1}(x) \]

[Out]

(Sqrt[-1 + x]*x*Sqrt[1 + x])/8 + ((-1 + x)^(3/2)*x*(1 + x)^(3/2))/4 - ArcCosh[x]/8

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Rubi [A]  time = 0.0061312, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {90, 38, 52} \[ \frac{1}{4} (x-1)^{3/2} x (x+1)^{3/2}+\frac{1}{8} \sqrt{x-1} x \sqrt{x+1}-\frac{1}{8} \cosh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x]*x^2*Sqrt[1 + x],x]

[Out]

(Sqrt[-1 + x]*x*Sqrt[1 + x])/8 + ((-1 + x)^(3/2)*x*(1 + x)^(3/2))/4 - ArcCosh[x]/8

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \sqrt{-1+x} x^2 \sqrt{1+x} \, dx &=\frac{1}{4} (-1+x)^{3/2} x (1+x)^{3/2}+\frac{1}{4} \int \sqrt{-1+x} \sqrt{1+x} \, dx\\ &=\frac{1}{8} \sqrt{-1+x} x \sqrt{1+x}+\frac{1}{4} (-1+x)^{3/2} x (1+x)^{3/2}-\frac{1}{8} \int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx\\ &=\frac{1}{8} \sqrt{-1+x} x \sqrt{1+x}+\frac{1}{4} (-1+x)^{3/2} x (1+x)^{3/2}-\frac{1}{8} \cosh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0205533, size = 63, normalized size = 1.4 \[ \frac{x \sqrt{x+1} \left (2 x^3-2 x^2-x+1\right )+2 \sqrt{1-x} \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right )}{8 \sqrt{x-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x]*x^2*Sqrt[1 + x],x]

[Out]

(x*Sqrt[1 + x]*(1 - x - 2*x^2 + 2*x^3) + 2*Sqrt[1 - x]*ArcSin[Sqrt[1 - x]/Sqrt[2]])/(8*Sqrt[-1 + x])

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Maple [A]  time = 0.007, size = 52, normalized size = 1.2 \begin{align*} -{\frac{1}{8}\sqrt{-1+x}\sqrt{1+x} \left ( -2\,{x}^{3}\sqrt{{x}^{2}-1}+x\sqrt{{x}^{2}-1}+\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-1+x)^(1/2)*(1+x)^(1/2)*(-2*x^3*(x^2-1)^(1/2)+x*(x^2-1)^(1/2)+ln(x+(x^2-1)^(1/2)))/(x^2-1)^(1/2)

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Maxima [A]  time = 3.37858, size = 50, normalized size = 1.11 \begin{align*} \frac{1}{4} \,{\left (x^{2} - 1\right )}^{\frac{3}{2}} x + \frac{1}{8} \, \sqrt{x^{2} - 1} x - \frac{1}{8} \, \log \left (2 \, x + 2 \, \sqrt{x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^2 - 1)^(3/2)*x + 1/8*sqrt(x^2 - 1)*x - 1/8*log(2*x + 2*sqrt(x^2 - 1))

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Fricas [A]  time = 1.62247, size = 108, normalized size = 2.4 \begin{align*} \frac{1}{8} \,{\left (2 \, x^{3} - x\right )} \sqrt{x + 1} \sqrt{x - 1} + \frac{1}{8} \, \log \left (\sqrt{x + 1} \sqrt{x - 1} - x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^3 - x)*sqrt(x + 1)*sqrt(x - 1) + 1/8*log(sqrt(x + 1)*sqrt(x - 1) - x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{x - 1} \sqrt{x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x - 1)*sqrt(x + 1), x)

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Giac [A]  time = 1.40156, size = 62, normalized size = 1.38 \begin{align*} \frac{1}{8} \,{\left ({\left (2 \,{\left (x + 1\right )}{\left (x - 2\right )} + 5\right )}{\left (x + 1\right )} - 1\right )} \sqrt{x + 1} \sqrt{x - 1} + \frac{1}{4} \, \log \left ({\left | -\sqrt{x + 1} + \sqrt{x - 1} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((2*(x + 1)*(x - 2) + 5)*(x + 1) - 1)*sqrt(x + 1)*sqrt(x - 1) + 1/4*log(abs(-sqrt(x + 1) + sqrt(x - 1)))