3.723 \(\int \frac{x^4 (1+x)^{3/2}}{\sqrt{1-x}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{1}{6} \sqrt{1-x} (x+1)^{5/2} x^3-\frac{1}{15} \sqrt{1-x} (x+1)^{5/2} x^2-\frac{11}{48} \sqrt{1-x} (x+1)^{3/2}-\frac{1}{120} \sqrt{1-x} (x+1)^{5/2} (19 x+18)-\frac{11}{16} \sqrt{1-x} \sqrt{x+1}+\frac{11}{16} \sin ^{-1}(x) \]

[Out]

(-11*Sqrt[1 - x]*Sqrt[1 + x])/16 - (11*Sqrt[1 - x]*(1 + x)^(3/2))/48 - (Sqrt[1 - x]*x^2*(1 + x)^(5/2))/15 - (S
qrt[1 - x]*x^3*(1 + x)^(5/2))/6 - (Sqrt[1 - x]*(1 + x)^(5/2)*(18 + 19*x))/120 + (11*ArcSin[x])/16

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Rubi [A]  time = 0.029721, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {100, 153, 147, 50, 41, 216} \[ -\frac{1}{6} \sqrt{1-x} (x+1)^{5/2} x^3-\frac{1}{15} \sqrt{1-x} (x+1)^{5/2} x^2-\frac{11}{48} \sqrt{1-x} (x+1)^{3/2}-\frac{1}{120} \sqrt{1-x} (x+1)^{5/2} (19 x+18)-\frac{11}{16} \sqrt{1-x} \sqrt{x+1}+\frac{11}{16} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

(-11*Sqrt[1 - x]*Sqrt[1 + x])/16 - (11*Sqrt[1 - x]*(1 + x)^(3/2))/48 - (Sqrt[1 - x]*x^2*(1 + x)^(5/2))/15 - (S
qrt[1 - x]*x^3*(1 + x)^(5/2))/6 - (Sqrt[1 - x]*(1 + x)^(5/2)*(18 + 19*x))/120 + (11*ArcSin[x])/16

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 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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 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^4 (1+x)^{3/2}}{\sqrt{1-x}} \, dx &=-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{6} \int \frac{(-3-2 x) x^2 (1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}+\frac{1}{30} \int \frac{x (1+x)^{3/2} (4+19 x)}{\sqrt{1-x}} \, dx\\ &=-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{120} \sqrt{1-x} (1+x)^{5/2} (18+19 x)+\frac{11}{24} \int \frac{(1+x)^{3/2}}{\sqrt{1-x}} \, dx\\ &=-\frac{11}{48} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{120} \sqrt{1-x} (1+x)^{5/2} (18+19 x)+\frac{11}{16} \int \frac{\sqrt{1+x}}{\sqrt{1-x}} \, dx\\ &=-\frac{11}{16} \sqrt{1-x} \sqrt{1+x}-\frac{11}{48} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{120} \sqrt{1-x} (1+x)^{5/2} (18+19 x)+\frac{11}{16} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{11}{16} \sqrt{1-x} \sqrt{1+x}-\frac{11}{48} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{120} \sqrt{1-x} (1+x)^{5/2} (18+19 x)+\frac{11}{16} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{11}{16} \sqrt{1-x} \sqrt{1+x}-\frac{11}{48} \sqrt{1-x} (1+x)^{3/2}-\frac{1}{15} \sqrt{1-x} x^2 (1+x)^{5/2}-\frac{1}{6} \sqrt{1-x} x^3 (1+x)^{5/2}-\frac{1}{120} \sqrt{1-x} (1+x)^{5/2} (18+19 x)+\frac{11}{16} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0403713, size = 71, normalized size = 0.6 \[ \frac{\sqrt{x+1} \left (40 x^6+56 x^5+14 x^4+18 x^3+37 x^2+91 x-256\right )}{240 \sqrt{1-x}}-\frac{11}{8} \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

(Sqrt[1 + x]*(-256 + 91*x + 37*x^2 + 18*x^3 + 14*x^4 + 56*x^5 + 40*x^6))/(240*Sqrt[1 - x]) - (11*ArcSin[Sqrt[1
- x]/Sqrt[2]])/8

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Maple [A]  time = 0.013, size = 108, normalized size = 0.9 \begin{align*}{\frac{1}{240}\sqrt{1-x}\sqrt{1+x} \left ( -40\,{x}^{5}\sqrt{-{x}^{2}+1}-96\,{x}^{4}\sqrt{-{x}^{2}+1}-110\,{x}^{3}\sqrt{-{x}^{2}+1}-128\,{x}^{2}\sqrt{-{x}^{2}+1}-165\,x\sqrt{-{x}^{2}+1}+165\,\arcsin \left ( x \right ) -256\,\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^4*(1+x)^(3/2)/(1-x)^(1/2),x)

[Out]

1/240*(1+x)^(1/2)*(1-x)^(1/2)*(-40*x^5*(-x^2+1)^(1/2)-96*x^4*(-x^2+1)^(1/2)-110*x^3*(-x^2+1)^(1/2)-128*x^2*(-x
^2+1)^(1/2)-165*x*(-x^2+1)^(1/2)+165*arcsin(x)-256*(-x^2+1)^(1/2))/(-x^2+1)^(1/2)

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Maxima [A]  time = 1.81031, size = 113, normalized size = 0.96 \begin{align*} -\frac{1}{6} \, \sqrt{-x^{2} + 1} x^{5} - \frac{2}{5} \, \sqrt{-x^{2} + 1} x^{4} - \frac{11}{24} \, \sqrt{-x^{2} + 1} x^{3} - \frac{8}{15} \, \sqrt{-x^{2} + 1} x^{2} - \frac{11}{16} \, \sqrt{-x^{2} + 1} x - \frac{16}{15} \, \sqrt{-x^{2} + 1} + \frac{11}{16} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(-x^2 + 1)*x^5 - 2/5*sqrt(-x^2 + 1)*x^4 - 11/24*sqrt(-x^2 + 1)*x^3 - 8/15*sqrt(-x^2 + 1)*x^2 - 11/16*
sqrt(-x^2 + 1)*x - 16/15*sqrt(-x^2 + 1) + 11/16*arcsin(x)

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Fricas [A]  time = 1.73168, size = 180, normalized size = 1.53 \begin{align*} -\frac{1}{240} \,{\left (40 \, x^{5} + 96 \, x^{4} + 110 \, x^{3} + 128 \, x^{2} + 165 \, x + 256\right )} \sqrt{x + 1} \sqrt{-x + 1} - \frac{11}{8} \, \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^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="fricas")

[Out]

-1/240*(40*x^5 + 96*x^4 + 110*x^3 + 128*x^2 + 165*x + 256)*sqrt(x + 1)*sqrt(-x + 1) - 11/8*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**4*(1+x)**(3/2)/(1-x)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21644, size = 80, normalized size = 0.68 \begin{align*} -\frac{1}{240} \,{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, x - 8\right )}{\left (x + 1\right )} + 63\right )}{\left (x + 1\right )} - 13\right )}{\left (x + 1\right )} + 55\right )}{\left (x + 1\right )} + 165\right )} \sqrt{x + 1} \sqrt{-x + 1} + \frac{11}{8} \, \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^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="giac")

[Out]

-1/240*((2*((4*(5*x - 8)*(x + 1) + 63)*(x + 1) - 13)*(x + 1) + 55)*(x + 1) + 165)*sqrt(x + 1)*sqrt(-x + 1) + 1
1/8*arcsin(1/2*sqrt(2)*sqrt(x + 1))