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

Optimal. Leaf size=110 \[ -\frac{1}{5} \sqrt{1-x} x^2 (x+1)^{5/2}-\frac{1}{10} \sqrt{1-x} (x+1)^{7/2}-\frac{1}{10} \sqrt{1-x} (x+1)^{5/2}-\frac{1}{4} \sqrt{1-x} (x+1)^{3/2}-\frac{3}{4} \sqrt{1-x} \sqrt{x+1}+\frac{3}{4} \sin ^{-1}(x) \]

[Out]

(-3*Sqrt[1 - x]*Sqrt[1 + x])/4 - (Sqrt[1 - x]*(1 + x)^(3/2))/4 - (Sqrt[1 - x]*(1 + x)^(5/2))/10 - (Sqrt[1 - x]
*x^2*(1 + x)^(5/2))/5 - (Sqrt[1 - x]*(1 + x)^(7/2))/10 + (3*ArcSin[x])/4

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Rubi [A]  time = 0.0303069, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {100, 21, 80, 50, 41, 216} \[ -\frac{1}{5} \sqrt{1-x} x^2 (x+1)^{5/2}-\frac{1}{10} \sqrt{1-x} (x+1)^{7/2}-\frac{1}{10} \sqrt{1-x} (x+1)^{5/2}-\frac{1}{4} \sqrt{1-x} (x+1)^{3/2}-\frac{3}{4} \sqrt{1-x} \sqrt{x+1}+\frac{3}{4} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - x]*Sqrt[1 + x])/4 - (Sqrt[1 - x]*(1 + x)^(3/2))/4 - (Sqrt[1 - x]*(1 + x)^(5/2))/10 - (Sqrt[1 - x]
*x^2*(1 + x)^(5/2))/5 - (Sqrt[1 - x]*(1 + x)^(7/2))/10 + (3*ArcSin[x])/4

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 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 216

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 (1+x)^{3/2}}{\sqrt{1-x}} \, dx &=-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{5} \int \frac{(-2-2 x) x (1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}+\frac{2}{5} \int \frac{x (1+x)^{5/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{10} \int \frac{(1+x)^{5/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{10} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{1}{2} \int \frac{(1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{4} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{4} \int \frac{\sqrt{1+x}}{\sqrt{1-x}} \, dx\\ &=-\frac{3}{4} \sqrt{1-x} \sqrt{1+x}-\frac{1}{4} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{4} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{3}{4} \sqrt{1-x} \sqrt{1+x}-\frac{1}{4} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{4} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{3}{4} \sqrt{1-x} \sqrt{1+x}-\frac{1}{4} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{5/2}-\frac{1}{5} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{10} \sqrt{1-x} (1+x)^{7/2}+\frac{3}{4} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.057007, size = 56, normalized size = 0.51 \[ -\frac{1}{20} \sqrt{1-x^2} \left (4 x^4+10 x^3+12 x^2+15 x+24\right )-\frac{3}{2} \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - x^2]*(24 + 15*x + 12*x^2 + 10*x^3 + 4*x^4))/20 - (3*ArcSin[Sqrt[1 - x]/Sqrt[2]])/2

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Maple [A]  time = 0.008, size = 94, normalized size = 0.9 \begin{align*}{\frac{1}{20}\sqrt{1-x}\sqrt{1+x} \left ( -4\,{x}^{4}\sqrt{-{x}^{2}+1}-10\,{x}^{3}\sqrt{-{x}^{2}+1}-12\,{x}^{2}\sqrt{-{x}^{2}+1}-15\,x\sqrt{-{x}^{2}+1}+15\,\arcsin \left ( x \right ) -24\,\sqrt{-{x}^{2}+1} \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(1+x)^(1/2)*(1-x)^(1/2)*(-4*x^4*(-x^2+1)^(1/2)-10*x^3*(-x^2+1)^(1/2)-12*x^2*(-x^2+1)^(1/2)-15*x*(-x^2+1)^
(1/2)+15*arcsin(x)-24*(-x^2+1)^(1/2))/(-x^2+1)^(1/2)

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Maxima [A]  time = 1.72588, size = 95, normalized size = 0.86 \begin{align*} -\frac{1}{5} \, \sqrt{-x^{2} + 1} x^{4} - \frac{1}{2} \, \sqrt{-x^{2} + 1} x^{3} - \frac{3}{5} \, \sqrt{-x^{2} + 1} x^{2} - \frac{3}{4} \, \sqrt{-x^{2} + 1} x - \frac{6}{5} \, \sqrt{-x^{2} + 1} + \frac{3}{4} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(-x^2 + 1)*x^4 - 1/2*sqrt(-x^2 + 1)*x^3 - 3/5*sqrt(-x^2 + 1)*x^2 - 3/4*sqrt(-x^2 + 1)*x - 6/5*sqrt(-x
^2 + 1) + 3/4*arcsin(x)

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Fricas [A]  time = 1.85683, size = 158, normalized size = 1.44 \begin{align*} -\frac{1}{20} \,{\left (4 \, x^{4} + 10 \, x^{3} + 12 \, x^{2} + 15 \, x + 24\right )} \sqrt{x + 1} \sqrt{-x + 1} - \frac{3}{2} \, \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(4*x^4 + 10*x^3 + 12*x^2 + 15*x + 24)*sqrt(x + 1)*sqrt(-x + 1) - 3/2*arctan((sqrt(x + 1)*sqrt(-x + 1) -
1)/x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.28917, size = 70, normalized size = 0.64 \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \,{\left ({\left (2 \, x - 1\right )}{\left (x + 1\right )} + 3\right )}{\left (x + 1\right )} + 5\right )}{\left (x + 1\right )} + 15\right )} \sqrt{x + 1} \sqrt{-x + 1} + \frac{3}{2} \, \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*((2*((2*x - 1)*(x + 1) + 3)*(x + 1) + 5)*(x + 1) + 15)*sqrt(x + 1)*sqrt(-x + 1) + 3/2*arcsin(1/2*sqrt(2)
*sqrt(x + 1))