Integrand size = 17, antiderivative size = 139 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\frac {(3-x) (1+x)}{8 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {5 (3-x)^2 (1+x)}{64 \left (9+3 x-5 x^2+x^3\right )^{3/2}}-\frac {15 (3-x)^3 (1+x)}{256 \left (9+3 x-5 x^2+x^3\right )^{3/2}}+\frac {15 (3-x)^3 (1+x)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+x}}{2}\right )}{512 \left (9+3 x-5 x^2+x^3\right )^{3/2}} \]
1/8*(3-x)*(1+x)/(x^3-5*x^2+3*x+9)^(3/2)+5/64*(3-x)^2*(1+x)/(x^3-5*x^2+3*x+ 9)^(3/2)-15/256*(3-x)^3*(1+x)/(x^3-5*x^2+3*x+9)^(3/2)+15/512*(3-x)^3*(1+x) ^(3/2)*arctanh(1/2*(1+x)^(1/2))/(x^3-5*x^2+3*x+9)^(3/2)
Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\frac {86-140 x+30 x^2-15 (-3+x)^2 \sqrt {1+x} \text {arctanh}\left (\frac {\sqrt {1+x}}{2}\right )}{512 (-3+x) \sqrt {(-3+x)^2 (1+x)}} \]
(86 - 140*x + 30*x^2 - 15*(-3 + x)^2*Sqrt[1 + x]*ArcTanh[Sqrt[1 + x]/2])/( 512*(-3 + x)*Sqrt[(-3 + x)^2*(1 + x)])
Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2480, 27, 52, 52, 61, 73, 219}
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.
| \(\displaystyle \int \frac {1}{\left (x^3-5 x^2+3 x+9\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2480 |
| \(\displaystyle -\frac {2097152 (3-x)^3 (x+1)^{3/2} \int -\frac {1}{2097152 (3-x)^3 (x+1)^{3/2}}dx}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \int \frac {1}{(3-x)^3 (x+1)^{3/2}}dx}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \left (\frac {5}{16} \int \frac {1}{(3-x)^2 (x+1)^{3/2}}dx+\frac {1}{8 (3-x)^2 \sqrt {x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \left (\frac {5}{16} \left (\frac {3}{8} \int \frac {1}{(3-x) (x+1)^{3/2}}dx+\frac {1}{4 (3-x) \sqrt {x+1}}\right )+\frac {1}{8 (3-x)^2 \sqrt {x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \left (\frac {5}{16} \left (\frac {3}{8} \left (\frac {1}{4} \int \frac {1}{(3-x) \sqrt {x+1}}dx-\frac {1}{2 \sqrt {x+1}}\right )+\frac {1}{4 (3-x) \sqrt {x+1}}\right )+\frac {1}{8 (3-x)^2 \sqrt {x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \left (\frac {5}{16} \left (\frac {3}{8} \left (\frac {1}{2} \int \frac {1}{3-x}d\sqrt {x+1}-\frac {1}{2 \sqrt {x+1}}\right )+\frac {1}{4 (3-x) \sqrt {x+1}}\right )+\frac {1}{8 (3-x)^2 \sqrt {x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(3-x)^3 (x+1)^{3/2} \left (\frac {5}{16} \left (\frac {3}{8} \left (\frac {1}{4} \text {arctanh}\left (\frac {\sqrt {x+1}}{2}\right )-\frac {1}{2 \sqrt {x+1}}\right )+\frac {1}{4 (3-x) \sqrt {x+1}}\right )+\frac {1}{8 (3-x)^2 \sqrt {x+1}}\right )}{\left (x^3-5 x^2+3 x+9\right )^{3/2}}\) |
((3 - x)^3*(1 + x)^(3/2)*(1/(8*(3 - x)^2*Sqrt[1 + x]) + (5*(1/(4*(3 - x)*S qrt[1 + x]) + (3*(-1/2*1/Sqrt[1 + x] + ArcTanh[Sqrt[1 + x]/2]/4))/8))/16)) /(9 + 3*x - 5*x^2 + x^3)^(3/2)
3.3.31.3.1 Defintions of rubi rules used
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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*(( 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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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 = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 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 | \(\frac {\left (-3+x \right )^{3} \left (1+x \right ) \left (15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}-2\right )-15 \left (1+x \right )^{\frac {5}{2}} \ln \left (\sqrt {1+x}+2\right )-120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}-2\right )+120 \left (1+x \right )^{\frac {3}{2}} \ln \left (\sqrt {1+x}+2\right )+240 \ln \left (\sqrt {1+x}-2\right ) \sqrt {1+x}-240 \ln \left (\sqrt {1+x}+2\right ) \sqrt {1+x}+60 x^{2}-280 x +172\right )}{1024 \left (\sqrt {1+x}+2\right )^{2} \left (\sqrt {1+x}-2\right )^{2} \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {3}{2}}}\) | \(144\) |
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)
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=-\frac {15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (\frac {2 \, x + \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 15 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )} \log \left (-\frac {2 \, x - \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) - 4 \, \sqrt {x^{3} - 5 \, x^{2} + 3 \, x + 9} {\left (15 \, x^{2} - 70 \, x + 43\right )}}{1024 \, {\left (x^{4} - 8 \, x^{3} + 18 \, x^{2} - 27\right )}} \]
-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)
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=-\frac {15 \, \log \left (\sqrt {x + 1} + 2\right )}{1024 \, \mathrm {sgn}\left (x - 3\right )} + \frac {15 \, \log \left ({\left | \sqrt {x + 1} - 2 \right |}\right )}{1024 \, \mathrm {sgn}\left (x - 3\right )} + \frac {1}{32 \, \sqrt {x + 1} \mathrm {sgn}\left (x - 3\right )} + \frac {7 \, {\left (x + 1\right )}^{\frac {3}{2}} - 36 \, \sqrt {x + 1}}{256 \, {\left (x - 3\right )}^{2} \mathrm {sgn}\left (x - 3\right )} \]
-15/1024*log(sqrt(x + 1) + 2)/sgn(x - 3) + 15/1024*log(abs(sqrt(x + 1) - 2 ))/sgn(x - 3) + 1/32/(sqrt(x + 1)*sgn(x - 3)) + 1/256*(7*(x + 1)^(3/2) - 3 6*sqrt(x + 1))/((x - 3)^2*sgn(x - 3))
Timed out. \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{3/2}} \, dx=\int \frac {1}{{\left (x^3-5\,x^2+3\,x+9\right )}^{3/2}} \,d x \]