3.8.36 \(\int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx\) [736]

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

[Out]

3/8*arccosh(x)+1/24*(7-2*x)*(-1+x)^(3/2)*(1+x)^(1/2)+1/4*(-1+x)^(3/2)*x^2*(1+x)^(1/2)-3/8*(-1+x)^(1/2)*(1+x)^(
1/2)

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Rubi [A]
time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 = {1978, 102, 152, 52, 54} \begin {gather*} \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) \end {gather*}

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 52

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 54

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]

Rule 102

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 152

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 1978

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*((a*e + b*e*
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]

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*}

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Mathematica [A]
time = 0.00, size = 74, normalized size = 1.07 \begin {gather*} \frac {\sqrt {\frac {-1+x}{1+x}} \left (\sqrt {-1+x} \left (-16-7 x+x^2-2 x^3+6 x^4\right )+18 \sqrt {1+x} \tanh ^{-1}\left (\frac {1}{\sqrt {\frac {-1+x}{1+x}}}\right )\right )}{24 \sqrt {-1+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(-1 + x)/(1 + x)]*(Sqrt[-1 + x]*(-16 - 7*x + x^2 - 2*x^3 + 6*x^4) + 18*Sqrt[1 + x]*ArcTanh[1/Sqrt[(-1 +
x)/(1 + x)]]))/(24*Sqrt[-1 + x])

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Maple [A]
time = 0.10, size = 79, normalized size = 1.14

method	result	size
risch	\(\frac {\left (6 x^{3}-8 x^{2}+9 x -16\right ) \left (1+x \right ) \sqrt {\frac {-1+x}{1+x}}}{24}+\frac {3 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}}{8 \left (-1+x \right )}\)	\(70\)
trager	\(\frac {\left (1+x \right ) \left (6 x^{3}-8 x^{2}+9 x -16\right ) \sqrt {-\frac {1-x}{1+x}}}{24}+\frac {3 \ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{8}\)	\(71\)
default	\(-\frac {\sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right ) \left (-6 x \left (x^{2}-1\right )^{\frac {3}{2}}+8 \left (\left (1+x \right ) \left (-1+x \right )\right )^{\frac {3}{2}}-15 x \sqrt {x^{2}-1}+24 \sqrt {x^{2}-1}-9 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{24 \sqrt {\left (1+x \right ) \left (-1+x \right )}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (49) = 98\).
time = 0.27, size = 138, normalized size = 2.00 \begin {gather*} -\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 {gather*}

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 = 0.35, size = 64, normalized size = 0.93 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sqrt {\frac {x - 1}{x + 1}}\, dx \end {gather*}

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 = 2.88, size = 62, normalized size = 0.90 \begin {gather*} -\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 {gather*}

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)

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Mupad [B]
time = 0.05, size = 119, normalized size = 1.72 \begin {gather*} \frac {3\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4}-\frac {\frac {3\,\sqrt {\frac {x-1}{x+1}}}{4}-\frac {49\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{12}+\frac {31\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{12}-\frac {13\,{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*((x - 1)/(x + 1))^(1/2),x)

[Out]

(3*atanh(((x - 1)/(x + 1))^(1/2)))/4 - ((3*((x - 1)/(x + 1))^(1/2))/4 - (49*((x - 1)/(x + 1))^(3/2))/12 + (31*
((x - 1)/(x + 1))^(5/2))/12 - (13*((x - 1)/(x + 1))^(7/2))/4)/((6*(x - 1)^2)/(x + 1)^2 - (4*(x - 1))/(x + 1) -
(4*(x - 1)^3)/(x + 1)^3 + (x - 1)^4/(x + 1)^4 + 1)

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