Integrand size = 75, antiderivative size = 29 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=8-e^{\frac {5+\frac {1}{x}-x}{\frac {3}{4 x^2}+x}} x^2 \]
Time = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=-e^{-1+\frac {3+4 x+20 x^2}{3+4 x^3}} x^2 \]
Integrate[(E^((4*x + 20*x^2 - 4*x^3)/(3 + 4*x^3))*(-18*x - 12*x^2 - 120*x^ 3 - 12*x^4 + 32*x^5 + 80*x^6 - 32*x^7))/(9 + 24*x^3 + 16*x^6),x]
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.
\(\displaystyle \int \frac {e^{\frac {-4 x^3+20 x^2+4 x}{4 x^3+3}} \left (-32 x^7+80 x^6+32 x^5-12 x^4-120 x^3-12 x^2-18 x\right )}{16 x^6+24 x^3+9} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 16 \int -\frac {e^{\frac {4 \left (-x^3+5 x^2+x\right )}{4 x^3+3}} \left (16 x^7-40 x^6-16 x^5+6 x^4+60 x^3+6 x^2+9 x\right )}{8 \left (4 x^3+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 \int \frac {e^{\frac {4 \left (-x^3+5 x^2+x\right )}{4 x^3+3}} \left (16 x^7-40 x^6-16 x^5+6 x^4+60 x^3+6 x^2+9 x\right )}{\left (4 x^3+3\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -2 \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} x \left (16 x^6-40 x^5-16 x^4+6 x^3+60 x^2+6 x+9\right )}{\left (4 x^3+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} x-\frac {5}{2} e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}+\frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} \left (-8 x^2-9 x+60\right )}{2 \left (4 x^3+3\right )}+\frac {9 e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} \left (4 x^2+3 x-15\right )}{2 \left (4 x^3+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (-\frac {5}{2} \int e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}dx+\int e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} xdx-\frac {5 \sqrt [3]{2} \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{2 x+\sqrt [3]{6}}dx}{3^{2/3}}+\frac {1}{12} \left (4 (-2)^{2/3}-40 \sqrt [3]{3}+3 \sqrt [3]{-2} 3^{2/3}\right ) \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{-(-2)^{2/3} x-\sqrt [3]{3}}dx-\frac {5 \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{\left (2 \sqrt [3]{-1} x-\sqrt [3]{6}\right )^2}dx}{\sqrt [3]{6}}+\frac {15 i \sqrt [3]{-2} 3^{5/6} \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{2 \sqrt [3]{-1} x-\sqrt [3]{6}}dx}{\left (1+\sqrt [3]{-1}\right )^5}-\frac {5 \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{\left (2 (-1)^{2/3} x+\sqrt [3]{6}\right )^2}dx}{\sqrt [3]{6}}-\frac {1}{12} \left (40 \sqrt [3]{3}-\sqrt [3]{2} \left (4 \sqrt [3]{2}-3\ 3^{2/3}\right )\right ) \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{-2^{2/3} x-\sqrt [3]{3}}dx-\frac {1}{12} \left (40 \sqrt [3]{3}+\sqrt [3]{2} \left (3 (-3)^{2/3}+4 \sqrt [3]{-2}\right )\right ) \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{\sqrt [3]{-1} 2^{2/3} x-\sqrt [3]{3}}dx-\frac {5 \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{\left (2 \sqrt [6]{3} x+\sqrt [3]{2} \sqrt {3}\right )^2}dx}{\sqrt [3]{2}}-\frac {5 \sqrt [3]{2} \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}}}{\sqrt [3]{6}-\left (1-i \sqrt {3}\right ) x}dx}{3^{2/3}}+\frac {27}{2} \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} x}{\left (4 x^3+3\right )^2}dx+18 \int \frac {e^{\frac {4 x \left (-x^2+5 x+1\right )}{4 x^3+3}} x^2}{\left (4 x^3+3\right )^2}dx\right )\) |
Int[(E^((4*x + 20*x^2 - 4*x^3)/(3 + 4*x^3))*(-18*x - 12*x^2 - 120*x^3 - 12 *x^4 + 32*x^5 + 80*x^6 - 32*x^7))/(9 + 24*x^3 + 16*x^6),x]
3.25.66.3.1 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[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(-x^{2} {\mathrm e}^{-\frac {4 x \left (x^{2}-5 x -1\right )}{4 x^{3}+3}}\) | \(27\) |
risch | \(-x^{2} {\mathrm e}^{-\frac {4 x \left (x^{2}-5 x -1\right )}{4 x^{3}+3}}\) | \(27\) |
parallelrisch | \(-x^{2} {\mathrm e}^{\frac {-4 x^{3}+20 x^{2}+4 x}{4 x^{3}+3}}\) | \(31\) |
norman | \(\frac {-3 x^{2} {\mathrm e}^{\frac {-4 x^{3}+20 x^{2}+4 x}{4 x^{3}+3}}-4 x^{5} {\mathrm e}^{\frac {-4 x^{3}+20 x^{2}+4 x}{4 x^{3}+3}}}{4 x^{3}+3}\) | \(72\) |
int((-32*x^7+80*x^6+32*x^5-12*x^4-120*x^3-12*x^2-18*x)*exp((-4*x^3+20*x^2+ 4*x)/(4*x^3+3))/(16*x^6+24*x^3+9),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=-x^{2} e^{\left (-\frac {4 \, {\left (x^{3} - 5 \, x^{2} - x\right )}}{4 \, x^{3} + 3}\right )} \]
integrate((-32*x^7+80*x^6+32*x^5-12*x^4-120*x^3-12*x^2-18*x)*exp((-4*x^3+2 0*x^2+4*x)/(4*x^3+3))/(16*x^6+24*x^3+9),x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=- x^{2} e^{\frac {- 4 x^{3} + 20 x^{2} + 4 x}{4 x^{3} + 3}} \]
integrate((-32*x**7+80*x**6+32*x**5-12*x**4-120*x**3-12*x**2-18*x)*exp((-4 *x**3+20*x**2+4*x)/(4*x**3+3))/(16*x**6+24*x**3+9),x)
Time = 0.39 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=-x^{2} e^{\left (\frac {20 \, x^{2}}{4 \, x^{3} + 3} + \frac {4 \, x}{4 \, x^{3} + 3} + \frac {3}{4 \, x^{3} + 3} - 1\right )} \]
integrate((-32*x^7+80*x^6+32*x^5-12*x^4-120*x^3-12*x^2-18*x)*exp((-4*x^3+2 0*x^2+4*x)/(4*x^3+3))/(16*x^6+24*x^3+9),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=-x^{2} e^{\left (-\frac {4 \, {\left (x^{3} - 5 \, x^{2} - x\right )}}{4 \, x^{3} + 3}\right )} \]
integrate((-32*x^7+80*x^6+32*x^5-12*x^4-120*x^3-12*x^2-18*x)*exp((-4*x^3+2 0*x^2+4*x)/(4*x^3+3))/(16*x^6+24*x^3+9),x, algorithm=\
Time = 11.96 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {4 x+20 x^2-4 x^3}{3+4 x^3}} \left (-18 x-12 x^2-120 x^3-12 x^4+32 x^5+80 x^6-32 x^7\right )}{9+24 x^3+16 x^6} \, dx=-x^2\,{\mathrm {e}}^{\frac {-4\,x^3+20\,x^2+4\,x}{4\,x^3+3}} \]