Integrand size = 127, antiderivative size = 23 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=\frac {x-\log \left (3+x+\left (7+e^x\right )^2 x^2\right )}{x} \] Output:
(x-ln(x+(7+exp(x))^2*x^2+3))/x
Time = 0.45 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-\frac {\log \left (3+x+\left (7+e^x\right )^2 x^2\right )}{x} \] Input:
Integrate[(-x - 98*x^2 + E^x*(-28*x^2 - 14*x^3) + E^(2*x)*(-2*x^2 - 2*x^3) + (3 + x + 49*x^2 + 14*E^x*x^2 + E^(2*x)*x^2)*Log[3 + x + 49*x^2 + 14*E^x *x^2 + E^(2*x)*x^2])/(3*x^2 + x^3 + 49*x^4 + 14*E^x*x^4 + E^(2*x)*x^4),x]
Output:
-(Log[3 + x + (7 + E^x)^2*x^2]/x)
Time = 3.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {7239, 2010, 2009}
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.
\(\displaystyle \int \frac {-98 x^2+\left (14 e^x x^2+e^{2 x} x^2+49 x^2+x+3\right ) \log \left (14 e^x x^2+e^{2 x} x^2+49 x^2+x+3\right )+e^x \left (-14 x^3-28 x^2\right )+e^{2 x} \left (-2 x^3-2 x^2\right )-x}{14 e^x x^4+e^{2 x} x^4+49 x^4+x^3+3 x^2} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log \left (\left (e^x+7\right )^2 x^2+x+3\right )-\frac {x \left (2 e^x x^2 \left (e^x+7\right )+2 x \left (e^x+7\right )^2+1\right )}{\left (e^x+7\right )^2 x^2+x+3}}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {\log \left (\left (e^x+7\right )^2 x^2+x+3\right )-2 x-2}{x^2}+\frac {14 e^x x^3+98 x^3+2 x^2+7 x+6}{x^2 \left (14 e^x x^2+e^{2 x} x^2+49 x^2+x+3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (\left (e^x+7\right )^2 x^2+x+3\right )}{x}\) |
Input:
Int[(-x - 98*x^2 + E^x*(-28*x^2 - 14*x^3) + E^(2*x)*(-2*x^2 - 2*x^3) + (3 + x + 49*x^2 + 14*E^x*x^2 + E^(2*x)*x^2)*Log[3 + x + 49*x^2 + 14*E^x*x^2 + E^(2*x)*x^2])/(3*x^2 + x^3 + 49*x^4 + 14*E^x*x^4 + E^(2*x)*x^4),x]
Output:
-(Log[3 + x + (7 + E^x)^2*x^2]/x)
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.73 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30
method | result | size |
risch | \(-\frac {\ln \left ({\mathrm e}^{2 x} x^{2}+14 \,{\mathrm e}^{x} x^{2}+49 x^{2}+x +3\right )}{x}\) | \(30\) |
parallelrisch | \(-\frac {\ln \left ({\mathrm e}^{2 x} x^{2}+14 \,{\mathrm e}^{x} x^{2}+49 x^{2}+x +3\right )}{x}\) | \(30\) |
Input:
int(((exp(x)^2*x^2+14*exp(x)*x^2+49*x^2+x+3)*ln(exp(x)^2*x^2+14*exp(x)*x^2 +49*x^2+x+3)+(-2*x^3-2*x^2)*exp(x)^2+(-14*x^3-28*x^2)*exp(x)-98*x^2-x)/(ex p(x)^2*x^4+14*exp(x)*x^4+49*x^4+x^3+3*x^2),x,method=_RETURNVERBOSE)
Output:
-1/x*ln(exp(2*x)*x^2+14*exp(x)*x^2+49*x^2+x+3)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-\frac {\log \left (x^{2} e^{\left (2 \, x\right )} + 14 \, x^{2} e^{x} + 49 \, x^{2} + x + 3\right )}{x} \] Input:
integrate(((exp(x)^2*x^2+14*exp(x)*x^2+49*x^2+x+3)*log(exp(x)^2*x^2+14*exp (x)*x^2+49*x^2+x+3)+(-2*x^3-2*x^2)*exp(x)^2+(-14*x^3-28*x^2)*exp(x)-98*x^2 -x)/(exp(x)^2*x^4+14*exp(x)*x^4+49*x^4+x^3+3*x^2),x, algorithm="fricas")
Output:
-log(x^2*e^(2*x) + 14*x^2*e^x + 49*x^2 + x + 3)/x
Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=- \frac {\log {\left (x^{2} e^{2 x} + 14 x^{2} e^{x} + 49 x^{2} + x + 3 \right )}}{x} \] Input:
integrate(((exp(x)**2*x**2+14*exp(x)*x**2+49*x**2+x+3)*ln(exp(x)**2*x**2+1 4*exp(x)*x**2+49*x**2+x+3)+(-2*x**3-2*x**2)*exp(x)**2+(-14*x**3-28*x**2)*e xp(x)-98*x**2-x)/(exp(x)**2*x**4+14*exp(x)*x**4+49*x**4+x**3+3*x**2),x)
Output:
-log(x**2*exp(2*x) + 14*x**2*exp(x) + 49*x**2 + x + 3)/x
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-\frac {\log \left (x^{2} e^{\left (2 \, x\right )} + 14 \, x^{2} e^{x} + 49 \, x^{2} + x + 3\right )}{x} \] Input:
integrate(((exp(x)^2*x^2+14*exp(x)*x^2+49*x^2+x+3)*log(exp(x)^2*x^2+14*exp (x)*x^2+49*x^2+x+3)+(-2*x^3-2*x^2)*exp(x)^2+(-14*x^3-28*x^2)*exp(x)-98*x^2 -x)/(exp(x)^2*x^4+14*exp(x)*x^4+49*x^4+x^3+3*x^2),x, algorithm="maxima")
Output:
-log(x^2*e^(2*x) + 14*x^2*e^x + 49*x^2 + x + 3)/x
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-\frac {\log \left (x^{2} e^{\left (2 \, x\right )} + 14 \, x^{2} e^{x} + 49 \, x^{2} + x + 3\right )}{x} \] Input:
integrate(((exp(x)^2*x^2+14*exp(x)*x^2+49*x^2+x+3)*log(exp(x)^2*x^2+14*exp (x)*x^2+49*x^2+x+3)+(-2*x^3-2*x^2)*exp(x)^2+(-14*x^3-28*x^2)*exp(x)-98*x^2 -x)/(exp(x)^2*x^4+14*exp(x)*x^4+49*x^4+x^3+3*x^2),x, algorithm="giac")
Output:
-log(x^2*e^(2*x) + 14*x^2*e^x + 49*x^2 + x + 3)/x
Time = 0.77 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-\frac {\ln \left (x+14\,x^2\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^{2\,x}+49\,x^2+3\right )}{x} \] Input:
int(-(x + exp(x)*(28*x^2 + 14*x^3) + exp(2*x)*(2*x^2 + 2*x^3) + 98*x^2 - l og(x + 14*x^2*exp(x) + x^2*exp(2*x) + 49*x^2 + 3)*(x + 14*x^2*exp(x) + x^2 *exp(2*x) + 49*x^2 + 3))/(14*x^4*exp(x) + x^4*exp(2*x) + 3*x^2 + x^3 + 49* x^4),x)
Output:
-log(x + 14*x^2*exp(x) + x^2*exp(2*x) + 49*x^2 + 3)/x
\[ \int \frac {-x-98 x^2+e^x \left (-28 x^2-14 x^3\right )+e^{2 x} \left (-2 x^2-2 x^3\right )+\left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right ) \log \left (3+x+49 x^2+14 e^x x^2+e^{2 x} x^2\right )}{3 x^2+x^3+49 x^4+14 e^x x^4+e^{2 x} x^4} \, dx=-2 \left (\int \frac {e^{2 x}}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )-28 \left (\int \frac {e^{x}}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )+3 \left (\int \frac {\mathrm {log}\left (e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3\right )}{e^{2 x} x^{4}+14 e^{x} x^{4}+49 x^{4}+x^{3}+3 x^{2}}d x \right )+\int \frac {\mathrm {log}\left (e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3\right )}{e^{2 x} x^{3}+14 e^{x} x^{3}+49 x^{3}+x^{2}+3 x}d x +49 \left (\int \frac {\mathrm {log}\left (e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3\right )}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )+\int \frac {e^{2 x} \mathrm {log}\left (e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3\right )}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x -2 \left (\int \frac {e^{2 x} x}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )+14 \left (\int \frac {e^{x} \mathrm {log}\left (e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3\right )}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )-14 \left (\int \frac {e^{x} x}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right )-\left (\int \frac {1}{e^{2 x} x^{3}+14 e^{x} x^{3}+49 x^{3}+x^{2}+3 x}d x \right )-98 \left (\int \frac {1}{e^{2 x} x^{2}+14 e^{x} x^{2}+49 x^{2}+x +3}d x \right ) \] Input:
int(((exp(x)^2*x^2+14*exp(x)*x^2+49*x^2+x+3)*log(exp(x)^2*x^2+14*exp(x)*x^ 2+49*x^2+x+3)+(-2*x^3-2*x^2)*exp(x)^2+(-14*x^3-28*x^2)*exp(x)-98*x^2-x)/(e xp(x)^2*x^4+14*exp(x)*x^4+49*x^4+x^3+3*x^2),x)
Output:
- 2*int(e**(2*x)/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) - 28 *int(e**x/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) + 3*int(log( e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3)/(e**(2*x)*x**4 + 14*e**x*x **4 + 49*x**4 + x**3 + 3*x**2),x) + int(log(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3)/(e**(2*x)*x**3 + 14*e**x*x**3 + 49*x**3 + x**2 + 3*x),x) + 49*int(log(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3)/(e**(2*x)*x* *2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) + int((e**(2*x)*log(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3))/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) - 2*int((e**(2*x)*x)/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) + 14*int((e**x*log(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3))/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) - 14*int((e**x*x )/(e**(2*x)*x**2 + 14*e**x*x**2 + 49*x**2 + x + 3),x) - int(1/(e**(2*x)*x* *3 + 14*e**x*x**3 + 49*x**3 + x**2 + 3*x),x) - 98*int(1/(e**(2*x)*x**2 + 1 4*e**x*x**2 + 49*x**2 + x + 3),x)