Integrand size = 17, antiderivative size = 92 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\frac {3 (3-x) (1+x)}{20 \left (9+3 x-5 x^2+x^3\right )^{4/3}}+\frac {9 (3-x)^2 (1+x)}{80 \left (9+3 x-5 x^2+x^3\right )^{4/3}}-\frac {27 (3-x)^3 (1+x)}{320 \left (9+3 x-5 x^2+x^3\right )^{4/3}} \] Output:
3/20*(3-x)*(1+x)/(x^3-5*x^2+3*x+9)^(4/3)+9/80*(3-x)^2*(1+x)/(x^3-5*x^2+3*x +9)^(4/3)-27/320*(3-x)^3*(1+x)/(x^3-5*x^2+3*x+9)^(4/3)
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\frac {3 (-3+x) (1+x) \left (29-42 x+9 x^2\right )}{320 \left ((-3+x)^2 (1+x)\right )^{4/3}} \] Input:
Integrate[(9 + 3*x - 5*x^2 + x^3)^(-4/3),x]
Output:
(3*(-3 + x)*(1 + x)*(29 - 42*x + 9*x^2))/(320*((-3 + x)^2*(1 + x))^(4/3))
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2480, 27, 55, 55, 48}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (x^3-5 x^2+3 x+9\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 2480 |
\(\displaystyle \frac {262144\ 2^{2/3} (x-3)^{8/3} (x+1)^{4/3} \int \frac {1}{262144\ 2^{2/3} (x-3)^{8/3} (x+1)^{4/3}}dx}{\left (x^3-5 x^2+3 x+9\right )^{4/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(x-3)^{8/3} (x+1)^{4/3} \int \frac {1}{(x-3)^{8/3} (x+1)^{4/3}}dx}{\left (x^3-5 x^2+3 x+9\right )^{4/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {(x-3)^{8/3} (x+1)^{4/3} \left (-\frac {3}{10} \int \frac {1}{(x-3)^{5/3} (x+1)^{4/3}}dx-\frac {3}{20 (x-3)^{5/3} \sqrt [3]{x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{4/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {(x-3)^{8/3} (x+1)^{4/3} \left (-\frac {3}{10} \left (-\frac {3}{8} \int \frac {1}{(x-3)^{2/3} (x+1)^{4/3}}dx-\frac {3}{8 (x-3)^{2/3} \sqrt [3]{x+1}}\right )-\frac {3}{20 (x-3)^{5/3} \sqrt [3]{x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{4/3}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {(x-3)^{8/3} (x+1)^{4/3} \left (-\frac {3}{10} \left (-\frac {9 \sqrt [3]{x-3}}{32 \sqrt [3]{x+1}}-\frac {3}{8 \sqrt [3]{x+1} (x-3)^{2/3}}\right )-\frac {3}{20 (x-3)^{5/3} \sqrt [3]{x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{4/3}}\) |
Input:
Int[(9 + 3*x - 5*x^2 + x^3)^(-4/3),x]
Output:
((-3 + x)^(8/3)*(1 + x)^(4/3)*(-3/(20*(-3 + x)^(5/3)*(1 + x)^(1/3)) - (3*( -3/(8*(-3 + x)^(2/3)*(1 + x)^(1/3)) - (9*(-3 + x)^(1/3))/(32*(1 + x)^(1/3) )))/10))/(9 + 3*x - 5*x^2 + x^3)^(4/3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[(Px_)^(p_), x_Symbol] :> With[{a = Coeff[Px, x, 0], b = Coeff[Px, x, 1] , c = Coeff[Px, x, 2], d = Coeff[Px, x, 3]}, Simp[Px^p/((c^3 - 4*b*c*d + 9* a*d^2 + d*(c^2 - 3*b*d)*x)^p*(b*c - 9*a*d + 2*(c^2 - 3*b*d)*x)^(2*p)) Int [(c^3 - 4*b*c*d + 9*a*d^2 + d*(c^2 - 3*b*d)*x)^p*(b*c - 9*a*d + 2*(c^2 - 3* b*d)*x)^(2*p), x], x] /; EqQ[b^2*c^2 - 4*a*c^3 - 4*b^3*d + 18*a*b*c*d - 27* a^2*d^2, 0] && NeQ[c^2 - 3*b*d, 0]] /; FreeQ[p, x] && PolyQ[Px, x, 3] && ! IntegerQ[p]
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.32
method | result | size |
risch | \(\frac {\frac {27}{320} x^{2}-\frac {63}{160} x +\frac {87}{320}}{\left (-3+x \right ) \left (\left (1+x \right ) \left (-3+x \right )^{2}\right )^{\frac {1}{3}}}\) | \(29\) |
gosper | \(\frac {3 \left (1+x \right ) \left (-3+x \right ) \left (9 x^{2}-42 x +29\right )}{320 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {4}{3}}}\) | \(34\) |
orering | \(\frac {3 \left (1+x \right ) \left (-3+x \right ) \left (9 x^{2}-42 x +29\right )}{320 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {4}{3}}}\) | \(34\) |
trager | \(\frac {3 \left (9 x^{2}-42 x +29\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}}{320 \left (-3+x \right )^{3} \left (1+x \right )}\) | \(38\) |
Input:
int(1/(x^3-5*x^2+3*x+9)^(4/3),x,method=_RETURNVERBOSE)
Output:
3/320*(9*x^2-42*x+29)/(-3+x)/((1+x)*(-3+x)^2)^(1/3)
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\frac {3 \, {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}} {\left (9 \, x^{2} - 42 \, x + 29\right )}}{320 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \] Input:
integrate(1/(x^3-5*x^2+3*x+9)^(4/3),x, algorithm="fricas")
Output:
3/320*(x^3 - 5*x^2 + 3*x + 9)^(2/3)*(9*x^2 - 42*x + 29)/(x^4 - 8*x^3 + 18* x^2 - 27)
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(1/(x**3-5*x**2+3*x+9)**(4/3),x)
Output:
Integral((x**3 - 5*x**2 + 3*x + 9)**(-4/3), x)
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(1/(x^3-5*x^2+3*x+9)^(4/3),x, algorithm="maxima")
Output:
integrate((x^3 - 5*x^2 + 3*x + 9)^(-4/3), x)
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(1/(x^3-5*x^2+3*x+9)^(4/3),x, algorithm="giac")
Output:
integrate((x^3 - 5*x^2 + 3*x + 9)^(-4/3), x)
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\frac {3\,\left (9\,x^2-42\,x+29\right )\,{\left (x^3-5\,x^2+3\,x+9\right )}^{2/3}}{320\,\left (x+1\right )\,{\left (x-3\right )}^3} \] Input:
int(1/(3*x - 5*x^2 + x^3 + 9)^(4/3),x)
Output:
(3*(9*x^2 - 42*x + 29)*(3*x - 5*x^2 + x^3 + 9)^(2/3))/(320*(x + 1)*(x - 3) ^3)
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{4/3}} \, dx=\int \frac {1}{\left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x^{3}-5 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x^{2}+3 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x +9 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}}d x \] Input:
int(1/(x^3-5*x^2+3*x+9)^(4/3),x)
Output:
int(1/((x**3 - 5*x**2 + 3*x + 9)**(1/3)*x**3 - 5*(x**3 - 5*x**2 + 3*x + 9) **(1/3)*x**2 + 3*(x**3 - 5*x**2 + 3*x + 9)**(1/3)*x + 9*(x**3 - 5*x**2 + 3 *x + 9)**(1/3)),x)