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

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

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Rubi [A]  time = 0.03, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {100, 153, 147, 50, 41, 216} \begin {gather*} -\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) \end {gather*}

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 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 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 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 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 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*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.60 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.13, size = 128, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {1-x} \left (\frac {165 (1-x)^5}{(x+1)^5}+\frac {935 (1-x)^4}{(x+1)^4}+\frac {1986 (1-x)^3}{(x+1)^3}+\frac {3006 (1-x)^2}{(x+1)^2}+\frac {1305 (1-x)}{x+1}+795\right )}{120 \sqrt {x+1} \left (\frac {1-x}{x+1}+1\right )^6}-\frac {11}{8} \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*(Sqrt[1 - x]*(795 + (165*(1 - x)^5)/(1 + x)^5 + (935*(1 - x)^4)/(1 + x)^4 + (1986*(1 - x)^3)/(1 + x)^3
+ (3006*(1 - x)^2)/(1 + x)^2 + (1305*(1 - x))/(1 + x)))/(Sqrt[1 + x]*(1 + (1 - x)/(1 + x))^6) - (11*ArcTan[Sqr
t[1 - x]/Sqrt[1 + x]])/8

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fricas [A]  time = 1.20, size = 62, normalized size = 0.53 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 1.26, size = 59, normalized size = 0.50 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.02, size = 108, normalized size = 0.92 \begin {gather*} \frac {\sqrt {x +1}\, \sqrt {-x +1}\, \left (-40 \sqrt {-x^{2}+1}\, x^{5}-96 \sqrt {-x^{2}+1}\, x^{4}-110 \sqrt {-x^{2}+1}\, x^{3}-128 \sqrt {-x^{2}+1}\, x^{2}-165 \sqrt {-x^{2}+1}\, x +165 \arcsin \relax (x )-256 \sqrt {-x^{2}+1}\right )}{240 \sqrt {-x^{2}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x+1)^(3/2)/(-x+1)^(1/2),x)

[Out]

1/240*(x+1)^(1/2)*(-x+1)^(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 = 1.89, size = 84, normalized size = 0.71 \begin {gather*} -\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 \relax (x) \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(x + 1)^(3/2))/(1 - x)^(1/2),x)

[Out]

int((x^4*(x + 1)^(3/2))/(1 - x)^(1/2), x)

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sympy [F(-1)]  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**4*(1+x)**(3/2)/(1-x)**(1/2),x)

[Out]

Timed out

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