3.231 \(\int \frac{1}{(9+3 x-5 x^2+x^3)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{15 (x+1) (3-x)^3}{256 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{5 (x+1) (3-x)^2}{64 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{(x+1) (3-x)}{8 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{15 (x+1)^{3/2} (3-x)^3 \tanh ^{-1}\left (\frac{\sqrt{x+1}}{2}\right )}{512 \left (x^3-5 x^2+3 x+9\right )^{3/2}} \]

[Out]

((3 - x)*(1 + x))/(8*(9 + 3*x - 5*x^2 + x^3)^(3/2)) + (5*(3 - x)^2*(1 + x))/(64*(9 + 3*x - 5*x^2 + x^3)^(3/2))
- (15*(3 - x)^3*(1 + x))/(256*(9 + 3*x - 5*x^2 + x^3)^(3/2)) + (15*(3 - x)^3*(1 + x)^(3/2)*ArcTanh[Sqrt[1 + x
]/2])/(512*(9 + 3*x - 5*x^2 + x^3)^(3/2))

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Rubi [A]  time = 0.126954, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2067, 2064, 51, 63, 206} \[ -\frac{15 (x+1) (3-x)^3}{256 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{5 (x+1) (3-x)^2}{64 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{(x+1) (3-x)}{8 \left (x^3-5 x^2+3 x+9\right )^{3/2}}+\frac{15 (x+1)^{3/2} (3-x)^3 \tanh ^{-1}\left (\frac{\sqrt{x+1}}{2}\right )}{512 \left (x^3-5 x^2+3 x+9\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(9 + 3*x - 5*x^2 + x^3)^(-3/2),x]

[Out]

((3 - x)*(1 + x))/(8*(9 + 3*x - 5*x^2 + x^3)^(3/2)) + (5*(3 - x)^2*(1 + x))/(64*(9 + 3*x - 5*x^2 + x^3)^(3/2))
- (15*(3 - x)^3*(1 + x))/(256*(9 + 3*x - 5*x^2 + x^3)^(3/2)) + (15*(3 - x)^3*(1 + x)^(3/2)*ArcTanh[Sqrt[1 + x
]/2])/(512*(9 + 3*x - 5*x^2 + x^3)^(3/2))

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rule 2064

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{27}-\frac{16 x}{3}+x^3\right )^{3/2}} \, dx,x,-\frac{5}{3}+x\right )\\ &=\frac{\left (2097152 (3-x)^3 (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^3 \left (\frac{128}{9}+\frac{16 x}{3}\right )^{3/2}} \, dx,x,-\frac{5}{3}+x\right )}{81 \sqrt{3} \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ &=\frac{(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{\left (20480 (3-x)^3 (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right )^2 \left (\frac{128}{9}+\frac{16 x}{3}\right )^{3/2}} \, dx,x,-\frac{5}{3}+x\right )}{27 \sqrt{3} \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ &=\frac{(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{\left (80 (3-x)^3 (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right ) \left (\frac{128}{9}+\frac{16 x}{3}\right )^{3/2}} \, dx,x,-\frac{5}{3}+x\right )}{3 \sqrt{3} \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ &=\frac{(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac{15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{\left (5 (3-x)^3 (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right ) \sqrt{\frac{128}{9}+\frac{16 x}{3}}} \, dx,x,-\frac{5}{3}+x\right )}{4 \sqrt{3} \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ &=\frac{(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac{15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{\left (5 \sqrt{3} (3-x)^3 (1+x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{128}{3}-2 x^2} \, dx,x,\frac{4 \sqrt{1+x}}{\sqrt{3}}\right )}{32 \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ &=\frac{(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac{15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac{15 (3-x)^3 (1+x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1+x}}{2}\right )}{512 \left (9+3 x-5 x^2+x^3\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0075373, size = 35, normalized size = 0.25 \[ \frac{(x-3) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{x+1}{4}\right )}{32 \sqrt{(x-3)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(9 + 3*x - 5*x^2 + x^3)^(-3/2),x]

[Out]

((-3 + x)*Hypergeometric2F1[-1/2, 3, 1/2, (1 + x)/4])/(32*Sqrt[(-3 + x)^2*(1 + x)])

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Maple [A]  time = 0.016, size = 144, normalized size = 1. \begin{align*} -{\frac{ \left ( -3+x \right ) ^{3} \left ( 1+x \right ) }{1024} \left ( 15\,\ln \left ( \sqrt{1+x}+2 \right ) \left ( 1+x \right ) ^{5/2}-15\,\ln \left ( \sqrt{1+x}-2 \right ) \left ( 1+x \right ) ^{5/2}-120\,\ln \left ( \sqrt{1+x}+2 \right ) \left ( 1+x \right ) ^{3/2}+120\,\ln \left ( \sqrt{1+x}-2 \right ) \left ( 1+x \right ) ^{3/2}+240\,\ln \left ( \sqrt{1+x}+2 \right ) \sqrt{1+x}-240\,\ln \left ( \sqrt{1+x}-2 \right ) \sqrt{1+x}-60\,{x}^{2}+280\,x-172 \right ) \left ({x}^{3}-5\,{x}^{2}+3\,x+9 \right ) ^{-{\frac{3}{2}}} \left ( \sqrt{1+x}+2 \right ) ^{-2} \left ( \sqrt{1+x}-2 \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-5*x^2+3*x+9)^(3/2),x)

[Out]

-1/1024*(-3+x)^3*(1+x)*(15*ln((1+x)^(1/2)+2)*(1+x)^(5/2)-15*ln((1+x)^(1/2)-2)*(1+x)^(5/2)-120*ln((1+x)^(1/2)+2
)*(1+x)^(3/2)+120*ln((1+x)^(1/2)-2)*(1+x)^(3/2)+240*ln((1+x)^(1/2)+2)*(1+x)^(1/2)-240*ln((1+x)^(1/2)-2)*(1+x)^
(1/2)-60*x^2+280*x-172)/(x^3-5*x^2+3*x+9)^(3/2)/((1+x)^(1/2)+2)^2/((1+x)^(1/2)-2)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(3/2),x, algorithm="maxima")

[Out]

integrate((x^3 - 5*x^2 + 3*x + 9)^(-3/2), x)

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Fricas [A]  time = 1.74365, size = 355, normalized size = 2.55 \begin{align*} -\frac{15 \,{\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (\frac{2 \, x + \sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 15 \,{\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (-\frac{2 \, x - \sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 4 \, \sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9}{\left (15 \, x^{2} - 70 \, x + 43\right )}}{1024 \,{\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(3/2),x, algorithm="fricas")

[Out]

-1/1024*(15*(x^4 - 8*x^3 + 18*x^2 - 27)*log((2*x + sqrt(x^3 - 5*x^2 + 3*x + 9) - 6)/(x - 3)) - 15*(x^4 - 8*x^3
+ 18*x^2 - 27)*log(-(2*x - sqrt(x^3 - 5*x^2 + 3*x + 9) - 6)/(x - 3)) - 4*sqrt(x^3 - 5*x^2 + 3*x + 9)*(15*x^2
- 70*x + 43))/(x^4 - 8*x^3 + 18*x^2 - 27)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-5*x**2+3*x+9)**(3/2),x)

[Out]

Integral((x**3 - 5*x**2 + 3*x + 9)**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(3/2),x, algorithm="giac")

[Out]

sage0*x