Integrand size = 87, antiderivative size = 25 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=e^{e^4}+\frac {625 e^x x^4}{16 \left (e+e^{2 x}\right )^4} \end {dmath*}
Time = 1.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625 e^x x^4}{16 \left (e+e^{2 x}\right )^4} \end {dmath*}
Integrate[(E^(3*x)*(2500*x^3 - 4375*x^4) + E^(1 + x)*(2500*x^3 + 625*x^4)) /(16*E^5 + 16*E^(10*x) + 80*E^(4 + 2*x) + 160*E^(3 + 4*x) + 160*E^(2 + 6*x ) + 80*E^(1 + 8*x)),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.
\(\displaystyle \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{x+1} \left (625 x^4+2500 x^3\right )}{16 e^{10 x}+80 e^{2 x+4}+160 e^{4 x+3}+160 e^{6 x+2}+80 e^{8 x+1}+16 e^5} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {625 e^x x^3 \left (e (x+4)-e^{2 x} (7 x-4)\right )}{16 \left (e^{2 x}+e\right )^5}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {625}{16} \int \frac {e^x x^3 \left (e^{2 x} (4-7 x)+e (x+4)\right )}{\left (e+e^{2 x}\right )^5}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {625}{16} \int \left (\frac {8 e^{x+1} x^4}{\left (e+e^{2 x}\right )^5}-\frac {e^x x^3 (7 x-4)}{\left (e+e^{2 x}\right )^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {625}{16} \left (8 \int \frac {e^{x+1} x^4}{\left (e+e^{2 x}\right )^5}dx+\frac {14}{3} \int \frac {e^{x-1} x^3}{\left (e+e^{2 x}\right )^3}dx+\frac {35}{6} \int \frac {e^{x-2} x^3}{\left (e+e^{2 x}\right )^2}dx+\frac {35}{4} \int \frac {e^{x-3} x^3}{e+e^{2 x}}dx-2 \int \frac {e^{x-1} x^2}{\left (e+e^{2 x}\right )^3}dx-\frac {5}{2} \int \frac {e^{x-2} x^2}{\left (e+e^{2 x}\right )^2}dx-\frac {15}{4} \int \frac {e^{x-3} x^2}{e+e^{2 x}}dx-\frac {35 x^4 \arctan \left (e^{x-\frac {1}{2}}\right )}{16 e^{7/2}}+\frac {5 x^3 \arctan \left (e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}+\frac {35 i x^3 \operatorname {PolyLog}\left (2,-i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}-\frac {35 i x^3 \operatorname {PolyLog}\left (2,i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}-\frac {15 i x^2 \operatorname {PolyLog}\left (2,-i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}+\frac {15 i x^2 \operatorname {PolyLog}\left (2,i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}-\frac {105 i x^2 \operatorname {PolyLog}\left (3,-i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}+\frac {105 i x^2 \operatorname {PolyLog}\left (3,i e^{x-\frac {1}{2}}\right )}{8 e^{7/2}}+\frac {15 i x \operatorname {PolyLog}\left (3,-i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}-\frac {15 i x \operatorname {PolyLog}\left (3,i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}+\frac {105 i x \operatorname {PolyLog}\left (4,-i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}-\frac {105 i x \operatorname {PolyLog}\left (4,i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}-\frac {15 i \operatorname {PolyLog}\left (4,-i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}+\frac {15 i \operatorname {PolyLog}\left (4,i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}-\frac {105 i \operatorname {PolyLog}\left (5,-i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}+\frac {105 i \operatorname {PolyLog}\left (5,i e^{x-\frac {1}{2}}\right )}{4 e^{7/2}}-\frac {35 e^{x-3} x^4}{16 \left (e^{2 x}+e\right )}-\frac {35 e^{x-2} x^4}{24 \left (e^{2 x}+e\right )^2}-\frac {7 e^{x-1} x^4}{6 \left (e^{2 x}+e\right )^3}+\frac {5 e^{x-3} x^3}{4 \left (e^{2 x}+e\right )}+\frac {5 e^{x-2} x^3}{6 \left (e^{2 x}+e\right )^2}+\frac {2 e^{x-1} x^3}{3 \left (e^{2 x}+e\right )^3}\right )\) |
Int[(E^(3*x)*(2500*x^3 - 4375*x^4) + E^(1 + x)*(2500*x^3 + 625*x^4))/(16*E ^5 + 16*E^(10*x) + 80*E^(4 + 2*x) + 160*E^(3 + 4*x) + 160*E^(2 + 6*x) + 80 *E^(1 + 8*x)),x]
3.2.94.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {625 x^{4} {\mathrm e}^{x}}{16 \left ({\mathrm e}^{2 x}+{\mathrm e}\right )^{4}}\) | \(17\) |
parallelrisch | \(\frac {625 \,{\mathrm e}^{x} x^{4}}{16 \left ({\mathrm e}^{8 x}+4 \,{\mathrm e} \,{\mathrm e}^{6 x}+6 \,{\mathrm e}^{2} {\mathrm e}^{4 x}+4 \,{\mathrm e}^{3} {\mathrm e}^{2 x}+{\mathrm e}^{4}\right )}\) | \(47\) |
int(((-4375*x^4+2500*x^3)*exp(x)^3+(625*x^4+2500*x^3)*exp(1)*exp(x))/(16*e xp(x)^10+80*exp(1)*exp(x)^8+160*exp(1)^2*exp(x)^6+160*exp(1)^3*exp(x)^4+80 *exp(1)^4*exp(x)^2+16*exp(1)^5),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625 \, x^{4} e^{\left (x + 8\right )}}{16 \, {\left (e^{12} + e^{\left (8 \, x + 8\right )} + 4 \, e^{\left (6 \, x + 9\right )} + 6 \, e^{\left (4 \, x + 10\right )} + 4 \, e^{\left (2 \, x + 11\right )}\right )}} \end {dmath*}
integrate(((-4375*x^4+2500*x^3)*exp(x)^3+(625*x^4+2500*x^3)*exp(1)*exp(x)) /(16*exp(x)^10+80*exp(1)*exp(x)^8+160*exp(1)^2*exp(x)^6+160*exp(1)^3*exp(x )^4+80*exp(1)^4*exp(x)^2+16*exp(1)^5),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625 x^{4} e^{x}}{16 e^{8 x} + 64 e e^{6 x} + 96 e^{2} e^{4 x} + 64 e^{3} e^{2 x} + 16 e^{4}} \end {dmath*}
integrate(((-4375*x**4+2500*x**3)*exp(x)**3+(625*x**4+2500*x**3)*exp(1)*ex p(x))/(16*exp(x)**10+80*exp(1)*exp(x)**8+160*exp(1)**2*exp(x)**6+160*exp(1 )**3*exp(x)**4+80*exp(1)**4*exp(x)**2+16*exp(1)**5),x)
625*x**4*exp(x)/(16*exp(8*x) + 64*E*exp(6*x) + 96*exp(2)*exp(4*x) + 64*exp (3)*exp(2*x) + 16*exp(4))
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625 \, x^{4} e^{x}}{16 \, {\left (e^{4} + e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x + 1\right )} + 6 \, e^{\left (4 \, x + 2\right )} + 4 \, e^{\left (2 \, x + 3\right )}\right )}} \end {dmath*}
integrate(((-4375*x^4+2500*x^3)*exp(x)^3+(625*x^4+2500*x^3)*exp(1)*exp(x)) /(16*exp(x)^10+80*exp(1)*exp(x)^8+160*exp(1)^2*exp(x)^6+160*exp(1)^3*exp(x )^4+80*exp(1)^4*exp(x)^2+16*exp(1)^5),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.32 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625 \, {\left ({\left (x + 1\right )}^{4} e^{\left (x + 8\right )} - 4 \, {\left (x + 1\right )}^{3} e^{\left (x + 8\right )} + 6 \, {\left (x + 1\right )}^{2} e^{\left (x + 8\right )} - 4 \, {\left (x + 1\right )} e^{\left (x + 8\right )} + e^{\left (x + 8\right )}\right )}}{8 \, {\left (e^{12} + e^{\left (8 \, x + 8\right )} + 4 \, e^{\left (6 \, x + 9\right )} + 6 \, e^{\left (4 \, x + 10\right )} + 4 \, e^{\left (2 \, x + 11\right )}\right )}} \end {dmath*}
integrate(((-4375*x^4+2500*x^3)*exp(x)^3+(625*x^4+2500*x^3)*exp(1)*exp(x)) /(16*exp(x)^10+80*exp(1)*exp(x)^8+160*exp(1)^2*exp(x)^6+160*exp(1)^3*exp(x )^4+80*exp(1)^4*exp(x)^2+16*exp(1)^5),x, algorithm=\
625/8*((x + 1)^4*e^(x + 8) - 4*(x + 1)^3*e^(x + 8) + 6*(x + 1)^2*e^(x + 8) - 4*(x + 1)*e^(x + 8) + e^(x + 8))/(e^12 + e^(8*x + 8) + 4*e^(6*x + 9) + 6*e^(4*x + 10) + 4*e^(2*x + 11))
Time = 13.82 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \begin {dmath*} \int \frac {e^{3 x} \left (2500 x^3-4375 x^4\right )+e^{1+x} \left (2500 x^3+625 x^4\right )}{16 e^5+16 e^{10 x}+80 e^{4+2 x}+160 e^{3+4 x}+160 e^{2+6 x}+80 e^{1+8 x}} \, dx=\frac {625\,x^4\,{\mathrm {e}}^x}{16\,\left ({\mathrm {e}}^{8\,x}+{\mathrm {e}}^4+4\,{\mathrm {e}}^{2\,x+3}+6\,{\mathrm {e}}^{4\,x+2}+4\,{\mathrm {e}}^{6\,x+1}\right )} \end {dmath*}