3.9.28 \(\int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx\) [828]

Optimal. Leaf size=78 \[ -\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \sin ^{-1}(x) \]

[Out]

11/2*arcsin(x)+2/3*x^3*(1+x)^(1/2)/(1-x)^(3/2)-13/3*x^2*(1+x)^(1/2)/(1-x)^(1/2)-1/6*(52+33*x)*(1-x)^(1/2)*(1+x
)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {99, 155, 152, 41, 222} \begin {gather*} \frac {11 \text {ArcSin}(x)}{2}+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}-\frac {13 \sqrt {x+1} x^2}{3 \sqrt {1-x}}-\frac {1}{6} \sqrt {1-x} \sqrt {x+1} (33 x+52) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*Sqrt[1 + x])/(1 - x)^(5/2),x]

[Out]

(-13*x^2*Sqrt[1 + x])/(3*Sqrt[1 - x]) + (2*x^3*Sqrt[1 + x])/(3*(1 - x)^(3/2)) - (Sqrt[1 - x]*Sqrt[1 + x]*(52 +
33*x))/6 + (11*ArcSin[x])/2

Rule 41

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

Rule 99

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

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 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx &=\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {x^2 \left (3+\frac {7 x}{2}\right )}{(1-x)^{3/2} \sqrt {1+x}} \, dx\\ &=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {\left (-13-\frac {33 x}{2}\right ) x}{\sqrt {1-x} \sqrt {1+x}} \, dx\\ &=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx\\ &=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx\\ &=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11}{2} \sin ^{-1}(x)\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 56, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {1+x} \left (52-71 x+12 x^2+3 x^3\right )}{6 (1-x)^{3/2}}+11 \tan ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sqrt[1 + x])/(1 - x)^(5/2),x]

[Out]

-1/6*(Sqrt[1 + x]*(52 - 71*x + 12*x^2 + 3*x^3))/(1 - x)^(3/2) + 11*ArcTan[Sqrt[1 + x]/Sqrt[1 - x]]

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Maple [A]
time = 0.09, size = 97, normalized size = 1.24

method	result	size
risch	\(\frac {\left (3 x^{4}+15 x^{3}-59 x^{2}-19 x +52\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{6 \left (-1+x \right ) \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {11 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {1-x}\, \sqrt {1+x}}\)	\(89\)
default	\(\frac {\left (-3 x^{3} \sqrt {-x^{2}+1}+33 \arcsin \left (x \right ) x^{2}-12 x^{2} \sqrt {-x^{2}+1}-66 \arcsin \left (x \right ) x +71 x \sqrt {-x^{2}+1}+33 \arcsin \left (x \right )-52 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{6 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(1+x)^(1/2)/(1-x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(-3*x^3*(-x^2+1)^(1/2)+33*arcsin(x)*x^2-12*x^2*(-x^2+1)^(1/2)-66*arcsin(x)*x+71*x*(-x^2+1)^(1/2)+33*arcsin
(x)-52*(-x^2+1)^(1/2))*(1-x)^(1/2)*(1+x)^(1/2)/(-1+x)^2/(-x^2+1)^(1/2)

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 1.49, size = 80, normalized size = 1.03 \begin {gather*} -\frac {52 \, x^{2} + {\left (3 \, x^{3} + 12 \, x^{2} - 71 \, x + 52\right )} \sqrt {x + 1} \sqrt {-x + 1} + 66 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 104 \, x + 52}{6 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(52*x^2 + (3*x^3 + 12*x^2 - 71*x + 52)*sqrt(x + 1)*sqrt(-x + 1) + 66*(x^2 - 2*x + 1)*arctan((sqrt(x + 1)*
sqrt(-x + 1) - 1)/x) - 104*x + 52)/(x^2 - 2*x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {x + 1}}{\left (1 - x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(1+x)**(1/2)/(1-x)**(5/2),x)

[Out]

Integral(x**3*sqrt(x + 1)/(1 - x)**(5/2), x)

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Giac [A]
time = 1.14, size = 49, normalized size = 0.63 \begin {gather*} -\frac {{\left ({\left (3 \, {\left (x + 2\right )} {\left (x + 1\right )} - 86\right )} {\left (x + 1\right )} + 132\right )} \sqrt {x + 1} \sqrt {-x + 1}}{6 \, {\left (x - 1\right )}^{2}} + 11 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="giac")

[Out]

-1/6*((3*(x + 2)*(x + 1) - 86)*(x + 1) + 132)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 11*arcsin(1/2*sqrt(2)*sqrt(
x + 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\sqrt {x+1}}{{\left (1-x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(x + 1)^(1/2))/(1 - x)^(5/2),x)

[Out]

int((x^3*(x + 1)^(1/2))/(1 - x)^(5/2), x)

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