Integrand size = 80, antiderivative size = 24 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=3 x-3 \left (e^{5-x}+\frac {\log (5)}{-4+x}\right )^2 \] Output:
3*x-3*(exp(5-x)+ln(5)/(-4+x))^2
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=-3 e^{10-2 x}+3 x-\frac {6 e^{5-x} \log (5)}{-4+x}-\frac {3 \log ^2(5)}{(-4+x)^2} \] Input:
Integrate[(-192 + 144*x - 36*x^2 + 3*x^3 + E^(10 - 2*x)*(-384 + 288*x - 72 *x^2 + 6*x^3) + E^(5 - x)*(72 - 42*x + 6*x^2)*Log[5] + 6*Log[5]^2)/(-64 + 48*x - 12*x^2 + x^3),x]
Output:
-3*E^(10 - 2*x) + 3*x - (6*E^(5 - x)*Log[5])/(-4 + x) - (3*Log[5]^2)/(-4 + x)^2
Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(24)=48\).
Time = 0.94 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2007, 7292, 7293, 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.
\(\displaystyle \int \frac {3 x^3-36 x^2+e^{5-x} \left (6 x^2-42 x+72\right ) \log (5)+e^{10-2 x} \left (6 x^3-72 x^2+288 x-384\right )+144 x-192+6 \log ^2(5)}{x^3-12 x^2+48 x-64} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {3 x^3-36 x^2+e^{5-x} \left (6 x^2-42 x+72\right ) \log (5)+e^{10-2 x} \left (6 x^3-72 x^2+288 x-384\right )+144 x-192+6 \log ^2(5)}{(x-4)^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-3 x^3+36 x^2-e^{5-x} \left (6 x^2-42 x+72\right ) \log (5)-e^{10-2 x} \left (6 x^3-72 x^2+288 x-384\right )-144 x+192 \left (1-\frac {\log ^2(5)}{32}\right )}{(4-x)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x^3}{(x-4)^3}-\frac {36 x^2}{(x-4)^3}+\frac {144 x}{(x-4)^3}+6 e^{10-2 x}+\frac {6 \left (\log ^2(5)-32\right )}{(x-4)^3}+\frac {6 e^{5-x} (x-3) \log (5)}{(x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {18 x^2}{(4-x)^2}+3 x-3 e^{10-2 x}-\frac {144}{4-x}+\frac {192}{(4-x)^2}+\frac {3 \left (32-\log ^2(5)\right )}{(4-x)^2}+\frac {6 e^{5-x} \log (5)}{4-x}\) |
Input:
Int[(-192 + 144*x - 36*x^2 + 3*x^3 + E^(10 - 2*x)*(-384 + 288*x - 72*x^2 + 6*x^3) + E^(5 - x)*(72 - 42*x + 6*x^2)*Log[5] + 6*Log[5]^2)/(-64 + 48*x - 12*x^2 + x^3),x]
Output:
-3*E^(10 - 2*x) + 192/(4 - x)^2 - 144/(4 - x) + 3*x - (18*x^2)/(4 - x)^2 + (6*E^(5 - x)*Log[5])/(4 - x) + (3*(32 - Log[5]^2))/(4 - x)^2
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.51 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79
method | result | size |
parts | \(-3 \,{\mathrm e}^{-2 x +10}+3 x -\frac {3 \ln \left (5\right )^{2}}{\left (x -4\right )^{2}}+\frac {6 \ln \left (5\right ) {\mathrm e}^{5-x}}{-x +4}\) | \(43\) |
risch | \(3 x -\frac {3 \ln \left (5\right )^{2}}{x^{2}-8 x +16}-3 \,{\mathrm e}^{-2 x +10}-\frac {6 \ln \left (5\right ) {\mathrm e}^{5-x}}{x -4}\) | \(44\) |
norman | \(\frac {-144 x +3 x^{3}-48 \,{\mathrm e}^{-2 x +10}+24 \,{\mathrm e}^{-2 x +10} x -3 \,{\mathrm e}^{-2 x +10} x^{2}+24 \ln \left (5\right ) {\mathrm e}^{5-x}-6 \ln \left (5\right ) {\mathrm e}^{5-x} x +384-3 \ln \left (5\right )^{2}}{\left (x -4\right )^{2}}\) | \(78\) |
parallelrisch | \(-\frac {3 \,{\mathrm e}^{-2 x +10} x^{2}+6 \ln \left (5\right ) {\mathrm e}^{5-x} x -3 x^{3}-384-24 \,{\mathrm e}^{-2 x +10} x +3 \ln \left (5\right )^{2}-24 \ln \left (5\right ) {\mathrm e}^{5-x}+48 \,{\mathrm e}^{-2 x +10}+144 x}{x^{2}-8 x +16}\) | \(84\) |
derivativedivides | \(-15+3 x -\frac {3 \ln \left (5\right )^{2}}{\left (-x +4\right )^{2}}-3 \,{\mathrm e}^{-2 x +10}-12 \ln \left (5\right ) \left (\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}+\frac {{\mathrm e}^{5-x}}{-2 x +8}+\frac {{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}\right )+18 \ln \left (5\right ) \left (\frac {3 \,{\mathrm e}^{5-x}}{2 \left (-x +4\right )}+\frac {3 \,{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}+\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}\right )-6 \ln \left (5\right ) \left (\frac {5 \,{\mathrm e}^{5-x}}{2 \left (-x +4\right )}+\frac {7 \,{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}+\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}\right )\) | \(161\) |
default | \(-15+3 x -\frac {3 \ln \left (5\right )^{2}}{\left (-x +4\right )^{2}}-3 \,{\mathrm e}^{-2 x +10}-12 \ln \left (5\right ) \left (\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}+\frac {{\mathrm e}^{5-x}}{-2 x +8}+\frac {{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}\right )+18 \ln \left (5\right ) \left (\frac {3 \,{\mathrm e}^{5-x}}{2 \left (-x +4\right )}+\frac {3 \,{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}+\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}\right )-6 \ln \left (5\right ) \left (\frac {5 \,{\mathrm e}^{5-x}}{2 \left (-x +4\right )}+\frac {7 \,{\mathrm e} \,\operatorname {expIntegral}_{1}\left (x -4\right )}{2}+\frac {{\mathrm e}^{5-x}}{2 \left (-x +4\right )^{2}}\right )\) | \(161\) |
Input:
int(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*ln(5)*exp(5-x)+6* ln(5)^2+3*x^3-36*x^2+144*x-192)/(x^3-12*x^2+48*x-64),x,method=_RETURNVERBO SE)
Output:
-3*exp(5-x)^2+3*x-3*ln(5)^2/(x-4)^2+6*ln(5)*exp(5-x)/(-x+4)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=\frac {3 \, {\left (x^{3} - 2 \, {\left (x - 4\right )} e^{\left (-x + 5\right )} \log \left (5\right ) - 8 \, x^{2} - {\left (x^{2} - 8 \, x + 16\right )} e^{\left (-2 \, x + 10\right )} - \log \left (5\right )^{2} + 16 \, x\right )}}{x^{2} - 8 \, x + 16} \] Input:
integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp( 5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-192)/(x^3-12*x^2+48*x-64),x, algorithm= "fricas")
Output:
3*(x^3 - 2*(x - 4)*e^(-x + 5)*log(5) - 8*x^2 - (x^2 - 8*x + 16)*e^(-2*x + 10) - log(5)^2 + 16*x)/(x^2 - 8*x + 16)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=3 x - \frac {3 \log {\left (5 \right )}^{2}}{x^{2} - 8 x + 16} + \frac {\left (12 - 3 x\right ) e^{10 - 2 x} - 6 e^{5 - x} \log {\left (5 \right )}}{x - 4} \] Input:
integrate(((6*x**3-72*x**2+288*x-384)*exp(5-x)**2+(6*x**2-42*x+72)*ln(5)*e xp(5-x)+6*ln(5)**2+3*x**3-36*x**2+144*x-192)/(x**3-12*x**2+48*x-64),x)
Output:
3*x - 3*log(5)**2/(x**2 - 8*x + 16) + ((12 - 3*x)*exp(10 - 2*x) - 6*exp(5 - x)*log(5))/(x - 4)
\[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=\int { \frac {3 \, {\left (x^{3} + 2 \, {\left (x^{2} - 7 \, x + 12\right )} e^{\left (-x + 5\right )} \log \left (5\right ) - 12 \, x^{2} + 2 \, {\left (x^{3} - 12 \, x^{2} + 48 \, x - 64\right )} e^{\left (-2 \, x + 10\right )} + 2 \, \log \left (5\right )^{2} + 48 \, x - 64\right )}}{x^{3} - 12 \, x^{2} + 48 \, x - 64} \,d x } \] Input:
integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp( 5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-192)/(x^3-12*x^2+48*x-64),x, algorithm= "maxima")
Output:
3*x - 3*log(5)^2/(x^2 - 8*x + 16) - 3*(2*(x^2*e^5*log(5) - 8*x*e^5*log(5) + 16*e^5*log(5))*e^(-x) + (x^3*e^10 - 12*x^2*e^10 + 48*x*e^10)*e^(-2*x))/( x^3 - 12*x^2 + 48*x - 64) - 48*(3*x - 10)/(x^2 - 8*x + 16) - 144*(x - 2)/( x^2 - 8*x + 16) + 288*(x - 3)/(x^2 - 8*x + 16) + 384*e^2*exp_integral_e(3, 2*x - 8)/(x - 4)^2 + 96/(x^2 - 8*x + 16) - 576*integrate(e^(-2*x + 10)/(x ^4 - 16*x^3 + 96*x^2 - 256*x + 256), x)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.29 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=\frac {3 \, {\left (x^{3} - x^{2} e^{\left (-2 \, x + 10\right )} - 2 \, x e^{\left (-x + 5\right )} \log \left (5\right ) - 8 \, x^{2} + 8 \, x e^{\left (-2 \, x + 10\right )} + 8 \, e^{\left (-x + 5\right )} \log \left (5\right ) - \log \left (5\right )^{2} + 16 \, x - 16 \, e^{\left (-2 \, x + 10\right )}\right )}}{x^{2} - 8 \, x + 16} \] Input:
integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp( 5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-192)/(x^3-12*x^2+48*x-64),x, algorithm= "giac")
Output:
3*(x^3 - x^2*e^(-2*x + 10) - 2*x*e^(-x + 5)*log(5) - 8*x^2 + 8*x*e^(-2*x + 10) + 8*e^(-x + 5)*log(5) - log(5)^2 + 16*x - 16*e^(-2*x + 10))/(x^2 - 8* x + 16)
Time = 3.56 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=3\,x-3\,{\mathrm {e}}^{10-2\,x}-\frac {3\,{\ln \left (5\right )}^2-24\,{\mathrm {e}}^{5-x}\,\ln \left (5\right )+6\,x\,{\mathrm {e}}^{5-x}\,\ln \left (5\right )}{x^2-8\,x+16} \] Input:
int((144*x + exp(10 - 2*x)*(288*x - 72*x^2 + 6*x^3 - 384) + 6*log(5)^2 - 3 6*x^2 + 3*x^3 + exp(5 - x)*log(5)*(6*x^2 - 42*x + 72) - 192)/(48*x - 12*x^ 2 + x^3 - 64),x)
Output:
3*x - 3*exp(10 - 2*x) - (3*log(5)^2 - 24*exp(5 - x)*log(5) + 6*x*exp(5 - x )*log(5))/(x^2 - 8*x + 16)
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.04 \[ \int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} \left (-384+288 x-72 x^2+6 x^3\right )+e^{5-x} \left (72-42 x+6 x^2\right ) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx=\frac {-3 e^{2 x} \mathrm {log}\left (5\right )^{2}+3 e^{2 x} x^{3}-18 e^{2 x} x^{2}+96 e^{2 x}-6 e^{x} \mathrm {log}\left (5\right ) e^{5} x +24 e^{x} \mathrm {log}\left (5\right ) e^{5}-3 e^{10} x^{2}+24 e^{10} x -48 e^{10}}{e^{2 x} \left (x^{2}-8 x +16\right )} \] Input:
int(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp(5-x)+6 *log(5)^2+3*x^3-36*x^2+144*x-192)/(x^3-12*x^2+48*x-64),x)
Output:
(3*( - e**(2*x)*log(5)**2 + e**(2*x)*x**3 - 6*e**(2*x)*x**2 + 32*e**(2*x) - 2*e**x*log(5)*e**5*x + 8*e**x*log(5)*e**5 - e**10*x**2 + 8*e**10*x - 16* e**10))/(e**(2*x)*(x**2 - 8*x + 16))