3.13.40 \(\int \frac {e^{2 e^5-2 x} (-3 x^2-2 x^3+(-1-2 x) \log (4))}{40 x^6+80 x^4 \log (4)+40 x^2 \log ^2(4)} \, dx\) [1240]
Optimal. Leaf size=26 \[ \frac {e^{2 e^5-2 x}}{40 x \left (x^2+\log (4)\right )} \]
[Out]
1/40/(x^2+2*ln(2))/x/exp(-exp(5)+x)^2
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Rubi [A]
time = 0.58, antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps
used = 4, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {1608, 28, 6873,
2326} \begin {gather*} \frac {e^{2 e^5-2 x} \left (x^3+x \log (4)\right )}{40 x^2 \left (x^2+\log (4)\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(E^(2*E^5 - 2*x)*(-3*x^2 - 2*x^3 + (-1 - 2*x)*Log[4]))/(40*x^6 + 80*x^4*Log[4] + 40*x^2*Log[4]^2),x]
[Out]
(E^(2*E^5 - 2*x)*(x^3 + x*Log[4]))/(40*x^2*(x^2 + Log[4])^2)
Rule 28
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Rule 1608
Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]
Rule 2326
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]
Rule 6873
Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3+(-1-2 x) \log (4)\right )}{x^2 \left (40 x^4+80 x^2 \log (4)+40 \log ^2(4)\right )} \, dx\\ &=40 \int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3+(-1-2 x) \log (4)\right )}{x^2 \left (40 x^2+40 \log (4)\right )^2} \, dx\\ &=40 \int \frac {e^{2 e^5-2 x} \left (-3 x^2-2 x^3-\log (4)-2 x \log (4)\right )}{x^2 \left (40 x^2+40 \log (4)\right )^2} \, dx\\ &=\frac {e^{2 e^5-2 x} \left (x^3+x \log (4)\right )}{40 x^2 \left (x^2+\log (4)\right )^2}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.04, size = 36, normalized size = 1.38 \begin {gather*} \frac {e^{2 e^5-2 x} \left (2 x^3+x \log (16)\right )}{80 x^2 \left (x^2+\log (4)\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^(2*E^5 - 2*x)*(-3*x^2 - 2*x^3 + (-1 - 2*x)*Log[4]))/(40*x^6 + 80*x^4*Log[4] + 40*x^2*Log[4]^2),x]
[Out]
(E^(2*E^5 - 2*x)*(2*x^3 + x*Log[16]))/(80*x^2*(x^2 + Log[4])^2)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.30, size = 2581, normalized size = 99.27
| | |
| method |
result |
size |
| | |
| gosper |
\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \left (2\right )\right ) x}\) |
\(25\) |
| norman |
\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \left (2\right )\right ) x}\) |
\(25\) |
| risch |
\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{5}-2 x}}{40 \left (x^{2}+2 \ln \left (2\right )\right ) x}\) |
\(25\) |
| derivativedivides |
\(\text {Expression too large to display}\) |
\(2581\) |
| default |
\(\text {Expression too large to display}\) |
\(2581\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*(-2*x-1)*ln(2)-2*x^3-3*x^2)/(160*x^2*ln(2)^2+160*x^4*ln(2)+40*x^6)/exp(-exp(5)+x)^2,x,method=_RETURNVER
BOSE)
[Out]
3/160*exp(5)^2/ln(2)^2*sum((6*exp(5)*_R1^2+12*exp(5)*ln(2)+3*exp(5)*_R1+4*ln(2)*_R1+12*exp(10)*_R1+4*ln(2)+6*e
xp(15)+3*exp(10))/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z
^3+3*exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-1/40*exp(5)*exp(2*exp(5)-2*x)/(exp(5)^3+3*exp
(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))-1/80/ln(2)*exp(5)*s
um((12*exp(5)*_R1+6*_R1^2+3*exp(5)+8*ln(2)+6*exp(10)+3*_R1)/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R
1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z^3+3*exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-1/4
0*exp(2*exp(5)-2*x)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(
2)*(-exp(5)+x))*(-exp(5)+x)+3/320*exp(2*exp(5)-2*x)*(3*exp(5)^4+6*exp(5)^3*(-exp(5)+x)+3*exp(5)^2*(-exp(5)+x)^
2+6*exp(5)^2*ln(2)-2*ln(2)*(-exp(5)+x)^2)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-ex
p(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))+3/160*exp(2*exp(5)-2*x)/ln(2)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3
*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*(-exp(5)+x)^2+3/160*exp(5)^5*exp(2*exp
(5)-2*x)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*
(-exp(5)+x))+1/40*exp(5)^3*exp(2*exp(5)-2*x)/ln(2)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-e
xp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))-9/320*exp(5)^4*exp(2*exp(5)-2*x)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(
-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))-9/160*exp(5)^2*exp(2*exp(5
)-2*x)/ln(2)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-ex
p(5)+x))-1/160*exp(2*exp(5)-2*x)*(3*exp(5)^5+6*exp(5)^4*(-exp(5)+x)+3*exp(5)^3*(-exp(5)+x)^2+4*exp(5)^3*ln(2)-
6*exp(5)^2*ln(2)*(-exp(5)+x)-6*exp(5)*ln(2)*(-exp(5)+x)^2-4*exp(5)*ln(2)^2-4*ln(2)^2*(-exp(5)+x))/ln(2)^2/(exp
(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))+3/160*ex
p(5)/ln(2)^2*sum((6*exp(5)*_R1^2+12*exp(5)*ln(2)+3*exp(5)*_R1+4*ln(2)*_R1+12*exp(10)*_R1+4*ln(2)+6*exp(15)+3*e
xp(10))/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z^3+3*exp(5
)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-3/160*exp(5)/ln(2)^2*sum((12*exp(5)^2*ln(2)-4*ln(2)*_R1
^2+6*_R1^2*exp(10)+6*exp(5)*ln(2)+6*exp(5)*exp(15)+3*exp(5)*exp(10)-2*ln(2)*_R1+12*_R1*exp(15)+3*exp(10)*_R1)/
(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z^3+3*exp(5)*_Z^2+(
3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-1/160*exp(5)^3/ln(2)^2*sum((12*exp(5)*_R1+6*_R1^2+3*exp(5)+8*ln
(2)+6*exp(10)+3*_R1)/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf
(_Z^3+3*exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-3/320*exp(5)^2/ln(2)^2*sum((12*exp(5)*_R1+
6*_R1^2+3*exp(5)+8*ln(2)+6*exp(10)+3*_R1)/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+
2*x-2*_R1),_R1=RootOf(_Z^3+3*exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))+3/80*exp(5)^4*exp(2*e
xp(5)-2*x)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2
)*(-exp(5)+x))*(-exp(5)+x)+3/160*exp(5)^3*exp(2*exp(5)-2*x)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*
(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*(-exp(5)+x)^2-3/80*exp(5)*exp(2*exp(5)-2*x)/ln
(2)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*
(-exp(5)+x)^2-3/80/ln(2)*exp(5)^2*exp(2*exp(5)-2*x)/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-
exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*(-exp(5)+x)-9/160*exp(5)^3*exp(2*exp(5)-2*x)/ln(2)^2/(exp(5)^3
+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*(-exp(5)+x)-9
/320*exp(5)^2*exp(2*exp(5)-2*x)/ln(2)^2/(exp(5)^3+3*exp(5)^2*(-exp(5)+x)+3*exp(5)*(-exp(5)+x)^2+(-exp(5)+x)^3+
2*exp(5)*ln(2)+2*ln(2)*(-exp(5)+x))*(-exp(5)+x)^2+1/160/ln(2)^2*sum((-12*exp(5)*ln(2)*_R1^2+6*exp(5)*_R1^2*exp
(10)+6*exp(5)^2*ln(2)-8*exp(5)*ln(2)^2+8*exp(5)*ln(2)*exp(10)-6*exp(5)*ln(2)*_R1-8*ln(2)^2*_R1-12*ln(2)*exp(10
)*_R1+6*exp(5)*exp(20)+3*exp(5)*exp(15)+12*exp(20)*_R1+3*_R1*exp(15))/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))
*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z^3+3*exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp
(15)))+1/80/ln(2)*sum((6*exp(5)*_R1^2+12*exp(5)*ln(2)+3*exp(5)*_R1+4*ln(2)*_R1+12*exp(10)*_R1+4*ln(2)+6*exp(15
)+3*exp(10))/(3*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_R1)*Ei(1,-2*exp(5)+2*x-2*_R1),_R1=RootOf(_Z^3+3*
exp(5)*_Z^2+(3*exp(10)+2*ln(2))*_Z+2*exp(5)*ln(2)+exp(15)))-3/320/ln(2)^2*sum((12*exp(5)^2*ln(2)-4*ln(2)*_R1^2
+6*_R1^2*exp(10)+6*exp(5)*ln(2)+6*exp(5)*exp(15)+3*exp(5)*exp(10)-2*ln(2)*_R1+12*_R1*exp(15)+3*exp(10)*_R1)/(3
*exp(10)+6*exp(5)*_R1+3*_R1^2+2*ln(2))*exp(-2*_...
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Maxima [A]
time = 0.52, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \left (2\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*(-1-2*x)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="maxima")
[Out]
1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))
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Fricas [A]
time = 0.36, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \left (2\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*(-1-2*x)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="fricas")
[Out]
1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))
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Sympy [A]
time = 0.06, size = 20, normalized size = 0.77 \begin {gather*} \frac {e^{- 2 x + 2 e^{5}}}{40 x^{3} + 80 x \log {\left (2 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*(-1-2*x)*ln(2)-2*x**3-3*x**2)/(160*x**2*ln(2)**2+160*x**4*ln(2)+40*x**6)/exp(-exp(5)+x)**2,x)
[Out]
exp(-2*x + 2*exp(5))/(40*x**3 + 80*x*log(2))
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Giac [A]
time = 0.38, size = 22, normalized size = 0.85 \begin {gather*} \frac {e^{\left (-2 \, x + 2 \, e^{5}\right )}}{40 \, {\left (x^{3} + 2 \, x \log \left (2\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*(-1-2*x)*log(2)-2*x^3-3*x^2)/(160*x^2*log(2)^2+160*x^4*log(2)+40*x^6)/exp(-exp(5)+x)^2,x, algorit
hm="giac")
[Out]
1/40*e^(-2*x + 2*e^5)/(x^3 + 2*x*log(2))
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Mupad [B]
time = 1.08, size = 23, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-2\,x}}{40\,\left (x^3+2\,\ln \left (2\right )\,x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-(exp(2*exp(5) - 2*x)*(2*log(2)*(2*x + 1) + 3*x^2 + 2*x^3))/(160*x^2*log(2)^2 + 160*x^4*log(2) + 40*x^6),x
)
[Out]
(exp(2*exp(5))*exp(-2*x))/(40*(2*x*log(2) + x^3))
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