3.9.29 \(\int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx\) [829]

Optimal. Leaf size=61 \[ -\frac {4 \sqrt {1+x}}{\sqrt {1-x}}-\sqrt {1-x} \sqrt {1+x}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}+3 \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {91, 21, 49, 52, 41, 222} \begin {gather*} 3 \text {ArcSin}(x)-\frac {2 (x+1)^{3/2}}{\sqrt {1-x}}+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-3 \sqrt {1-x} \sqrt {x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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

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Mathematica [A]
time = 0.07, size = 51, normalized size = 0.84 \begin {gather*} -\frac {\sqrt {1+x} \left (14-19 x+3 x^2\right )}{3 (1-x)^{3/2}}+6 \tan ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[1 + x]*(14 - 19*x + 3*x^2))/(1 - x)^(3/2) + 6*ArcTan[Sqrt[1 + x]/Sqrt[1 - x]]

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Maple [A]
time = 0.09, size = 83, normalized size = 1.36

method	result	size
default	\(\frac {\left (9 \arcsin \left (x \right ) x^{2}-3 x^{2} \sqrt {-x^{2}+1}-18 \arcsin \left (x \right ) x +19 x \sqrt {-x^{2}+1}+9 \arcsin \left (x \right )-14 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{3 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\)	\(83\)
risch	\(\frac {\left (3 x^{3}-16 x^{2}-5 x +14\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (-1+x \right ) \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {3 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(9*arcsin(x)*x^2-3*x^2*(-x^2+1)^(1/2)-18*arcsin(x)*x+19*x*(-x^2+1)^(1/2)+9*arcsin(x)-14*(-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^2*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 1.33, size = 75, normalized size = 1.23 \begin {gather*} -\frac {14 \, x^{2} + {\left (3 \, x^{2} - 19 \, x + 14\right )} \sqrt {x + 1} \sqrt {-x + 1} + 18 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 28 \, x + 14}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(14*x^2 + (3*x^2 - 19*x + 14)*sqrt(x + 1)*sqrt(-x + 1) + 18*(x^2 - 2*x + 1)*arctan((sqrt(x + 1)*sqrt(-x +
1) - 1)/x) - 28*x + 14)/(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^{2} \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**2*(1+x)**(1/2)/(1-x)**(5/2),x)

[Out]

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

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Giac [A]
time = 2.97, size = 44, normalized size = 0.72 \begin {gather*} -\frac {{\left ({\left (3 \, x - 22\right )} {\left (x + 1\right )} + 36\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} + 6 \, \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^2*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="giac")

[Out]

-1/3*((3*x - 22)*(x + 1) + 36)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 6*arcsin(1/2*sqrt(2)*sqrt(x + 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2\,\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^2*(x + 1)^(1/2))/(1 - x)^(5/2),x)

[Out]

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

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