3.736 \(\int x^3 \sqrt{\frac{-1+x}{1+x}} \, dx\)

Optimal. Leaf size=69 \[ \frac{1}{4} (x-1)^{3/2} \sqrt{x+1} x^2+\frac{1}{24} (7-2 x) (x-1)^{3/2} \sqrt{x+1}-\frac{3}{8} \sqrt{x-1} \sqrt{x+1}+\frac{3}{8} \cosh ^{-1}(x) \]

[Out]

(-3*Sqrt[-1 + x]*Sqrt[1 + x])/8 + ((7 - 2*x)*(-1 + x)^(3/2)*Sqrt[1 + x])/24 + ((-1 + x)^(3/2)*x^2*Sqrt[1 + x])
/4 + (3*ArcCosh[x])/8

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Rubi [A]  time = 0.0248089, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {1958, 100, 147, 50, 52} \[ \frac{1}{4} (x-1)^{3/2} \sqrt{x+1} x^2+\frac{1}{24} (7-2 x) (x-1)^{3/2} \sqrt{x+1}-\frac{3}{8} \sqrt{x-1} \sqrt{x+1}+\frac{3}{8} \cosh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[(-1 + x)/(1 + x)],x]

[Out]

(-3*Sqrt[-1 + x]*Sqrt[1 + x])/8 + ((7 - 2*x)*(-1 + x)^(3/2)*Sqrt[1 + x])/24 + ((-1 + x)^(3/2)*x^2*Sqrt[1 + x])
/4 + (3*ArcCosh[x])/8

Rule 1958

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[(u*(e*(a + b*x
^n))^p)/(c + d*x^n)^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - (a*d)/b, 0]

Rule 100

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

Rule 147

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 x^3 \sqrt{\frac{-1+x}{1+x}} \, dx &=\int \frac{\sqrt{-1+x} x^3}{\sqrt{1+x}} \, dx\\ &=\frac{1}{4} (-1+x)^{3/2} x^2 \sqrt{1+x}+\frac{1}{4} \int \frac{(2-x) \sqrt{-1+x} x}{\sqrt{1+x}} \, dx\\ &=\frac{1}{24} (7-2 x) (-1+x)^{3/2} \sqrt{1+x}+\frac{1}{4} (-1+x)^{3/2} x^2 \sqrt{1+x}-\frac{3}{8} \int \frac{\sqrt{-1+x}}{\sqrt{1+x}} \, dx\\ &=-\frac{3}{8} \sqrt{-1+x} \sqrt{1+x}+\frac{1}{24} (7-2 x) (-1+x)^{3/2} \sqrt{1+x}+\frac{1}{4} (-1+x)^{3/2} x^2 \sqrt{1+x}+\frac{3}{8} \int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx\\ &=-\frac{3}{8} \sqrt{-1+x} \sqrt{1+x}+\frac{1}{24} (7-2 x) (-1+x)^{3/2} \sqrt{1+x}+\frac{1}{4} (-1+x)^{3/2} x^2 \sqrt{1+x}+\frac{3}{8} \cosh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0376549, size = 76, normalized size = 1.1 \[ \frac{\sqrt{\frac{x-1}{x+1}} \left (6 x^5-8 x^4+3 x^3-8 x^2-18 \sqrt{1-x^2} \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right )-9 x+16\right )}{24 (x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(-1 + x)/(1 + x)]*(16 - 9*x - 8*x^2 + 3*x^3 - 8*x^4 + 6*x^5 - 18*Sqrt[1 - x^2]*ArcSin[Sqrt[1 - x]/Sqrt[2
]]))/(24*(-1 + x))

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Maple [A]  time = 0.009, size = 79, normalized size = 1.1 \begin{align*}{\frac{1+x}{24}\sqrt{{\frac{x-1}{1+x}}} \left ( 6\,x \left ({x}^{2}-1 \right ) ^{3/2}-8\, \left ( \left ( x-1 \right ) \left ( 1+x \right ) \right ) ^{3/2}+15\,x\sqrt{{x}^{2}-1}-24\,\sqrt{{x}^{2}-1}+9\,\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{ \left ( x-1 \right ) \left ( 1+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((x-1)/(1+x))^(1/2)*(1+x)*(6*x*(x^2-1)^(3/2)-8*((x-1)*(1+x))^(3/2)+15*x*(x^2-1)^(1/2)-24*(x^2-1)^(1/2)+9*
ln(x+(x^2-1)^(1/2)))/((x-1)*(1+x))^(1/2)

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Maxima [B]  time = 1.02151, size = 186, normalized size = 2.7 \begin{align*} -\frac{39 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{7}{2}} - 31 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{5}{2}} + 49 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{3}{2}} - 9 \, \sqrt{\frac{x - 1}{x + 1}}}{12 \,{\left (\frac{4 \,{\left (x - 1\right )}}{x + 1} - \frac{6 \,{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac{4 \,{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac{{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} + \frac{3}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{3}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(39*((x - 1)/(x + 1))^(7/2) - 31*((x - 1)/(x + 1))^(5/2) + 49*((x - 1)/(x + 1))^(3/2) - 9*sqrt((x - 1)/(
x + 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1)^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) + 3/8*l
og(sqrt((x - 1)/(x + 1)) + 1) - 3/8*log(sqrt((x - 1)/(x + 1)) - 1)

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Fricas [A]  time = 1.67558, size = 182, normalized size = 2.64 \begin{align*} \frac{1}{24} \,{\left (6 \, x^{4} - 2 \, x^{3} + x^{2} - 7 \, x - 16\right )} \sqrt{\frac{x - 1}{x + 1}} + \frac{3}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{3}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*x^4 - 2*x^3 + x^2 - 7*x - 16)*sqrt((x - 1)/(x + 1)) + 3/8*log(sqrt((x - 1)/(x + 1)) + 1) - 3/8*log(sqr
t((x - 1)/(x + 1)) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12025, size = 84, normalized size = 1.22 \begin{align*} -\frac{3}{8} \, \log \left ({\left | -x + \sqrt{x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (x + 1\right ) + \frac{1}{24} \,{\left ({\left (2 \,{\left (3 \, x \mathrm{sgn}\left (x + 1\right ) - 4 \, \mathrm{sgn}\left (x + 1\right )\right )} x + 9 \, \mathrm{sgn}\left (x + 1\right )\right )} x - 16 \, \mathrm{sgn}\left (x + 1\right )\right )} \sqrt{x^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*log(abs(-x + sqrt(x^2 - 1)))*sgn(x + 1) + 1/24*((2*(3*x*sgn(x + 1) - 4*sgn(x + 1))*x + 9*sgn(x + 1))*x -
16*sgn(x + 1))*sqrt(x^2 - 1)