3.9.5 \(\int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ \frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-\frac {2 \sqrt {x+1}}{\sqrt {1-x}}+\sin ^{-1}(x) \]

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 47, 41, 216} \begin {gather*} \frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-\frac {2 \sqrt {x+1}}{\sqrt {1-x}}+\sin ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[1 + x])/(1 - x)^(5/2),x]

[Out]

(-2*Sqrt[1 + x])/Sqrt[1 - x] + (1 + x)^(3/2)/(3*(1 - x)^(3/2)) + ArcSin[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 47

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 78

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

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

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Mathematica [C]  time = 0.04, size = 47, normalized size = 1.15 \begin {gather*} -\frac {(x+1)^{3/2}-4 \sqrt {2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {1-x}{2}\right )}{3 (1-x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((1 + x)^(3/2) - 4*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - x)/2])/(1 - x)^(3/2)

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IntegrateAlgebraic [A]  time = 0.05, size = 54, normalized size = 1.32 \begin {gather*} \frac {\sqrt {x+1} \left (\frac {x+1}{1-x}-6\right )}{3 \sqrt {1-x}}+2 \tan ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {1-x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.57, size = 71, normalized size = 1.73 \begin {gather*} -\frac {5 \, x^{2} - {\left (7 \, x - 5\right )} \sqrt {x + 1} \sqrt {-x + 1} + 6 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 10 \, x + 5}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(5*x^2 - (7*x - 5)*sqrt(x + 1)*sqrt(-x + 1) + 6*(x^2 - 2*x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)
- 10*x + 5)/(x^2 - 2*x + 1)

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giac [A]  time = 1.28, size = 38, normalized size = 0.93 \begin {gather*} \frac {{\left (7 \, x - 5\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} + 2 \, \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*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="giac")

[Out]

1/3*(7*x - 5)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 2*arcsin(1/2*sqrt(2)*sqrt(x + 1))

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maple [B]  time = 0.01, size = 69, normalized size = 1.68 \begin {gather*} \frac {\left (3 x^{2} \arcsin \relax (x )-6 x \arcsin \relax (x )+7 \sqrt {-x^{2}+1}\, x +3 \arcsin \relax (x )-5 \sqrt {-x^{2}+1}\right ) \sqrt {-x +1}\, \sqrt {x +1}}{3 \left (x -1\right )^{2} \sqrt {-x^{2}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^2*arcsin(x)-6*x*arcsin(x)+7*(-x^2+1)^(1/2)*x+3*arcsin(x)-5*(-x^2+1)^(1/2))*(-x+1)^(1/2)*(x+1)^(1/2)/(
x-1)^2/(-x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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*(1+x)**(1/2)/(1-x)**(5/2),x)

[Out]

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

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