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

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 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])

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 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]

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 = 0.01, 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)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 3.62, size = 67, normalized size = 0.48 \[ \frac {(x-3) \sqrt {x+1} \left (\frac {15 x^2-70 x+43}{256 (x-3)^2 \sqrt {x+1}}-\frac {15}{512} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{2}\right )\right )}{\sqrt {(x-3)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3 + x)*Sqrt[1 + x]*((43 - 70*x + 15*x^2)/(256*(-3 + x)^2*Sqrt[1 + x]) - (15*ArcTanh[Sqrt[1 + x]/2])/512))/S
qrt[(-3 + x)^2*(1 + x)]

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 138, normalized size = 0.99 \[ -\frac {15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (\frac {2 \, x + \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (-\frac {2 \, x - \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 4 \, \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} {\left (15 \, x^{2} - 70 \, x + 43\right )}}{1024 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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

________________________________________________________________________________________

maple [A]  time = 0.11, size = 71, normalized size = 0.51

method	result	size
risch	\(\frac {15 x^{2}-70 x +43}{256 \left (-3+x \right ) \sqrt {\left (1+x \right ) \left (-3+x \right )^{2}}}+\frac {\left (-\frac {15 \ln \left (\sqrt {1+x}+2\right )}{1024}+\frac {15 \ln \left (\sqrt {1+x}-2\right )}{1024}\right ) \sqrt {1+x}\, \left (-3+x \right )}{\sqrt {\left (1+x \right ) \left (-3+x \right )^{2}}}\)	\(71\)
trager	\(\frac {\left (15 x^{2}-70 x +43\right ) \sqrt {x^{3}-5 x^{2}+3 x +9}}{256 \left (-3+x \right )^{3} \left (1+x \right )}+\frac {15 \ln \left (\frac {-x^{2}+4 \sqrt {x^{3}-5 x^{2}+3 x +9}-2 x +15}{\left (-3+x \right )^{2}}\right )}{1024}\)	\(75\)
default	\(\frac {\left (-3+x \right )^{3} \left (1+x \right ) \left (15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}-2\right )-15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}+2\right )-120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}-2\right )+120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}+2\right )+240 \ln \left (\sqrt {1+x}-2\right ) \sqrt {1+x}-240 \ln \left (\sqrt {1+x}+2\right ) \sqrt {1+x}+60 x^{2}-280 x +172\right )}{1024 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {3}{2}} \left (\sqrt {1+x}-2\right )^{2} \left (\sqrt {1+x}+2\right )^{2}}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(15*x^2-70*x+43)/(-3+x)/((1+x)*(-3+x)^2)^(1/2)+(-15/1024*ln((1+x)^(1/2)+2)+15/1024*ln((1+x)^(1/2)-2))/((
1+x)*(-3+x)^2)^(1/2)*(1+x)^(1/2)*(-3+x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {3}{2}}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (x^3-5\,x^2+3\,x+9\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {3}{2}}}\, dx \]

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)

________________________________________________________________________________________