3.6.50 \(\int \frac {3888+e^{50} (432-2592 x)-23328 x-2 e^{2 x} x+e^x (-36+e^{25} (12-156 x)+468 x)+e^{25} (-2592+15552 x)}{11664-7776 e^{25}+1296 e^{50}+e^{2 x}+e^x (-216+72 e^{25})} \, dx\)

Optimal. Leaf size=26 \[ \left (\frac {3}{9+\frac {e^x}{4 \left (-3+e^{25}\right )}}-x\right ) x \]

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Rubi [B]  time = 0.55, antiderivative size = 119, normalized size of antiderivative = 4.58, number of steps used = 17, number of rules used = 12, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6741, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} -\frac {5 x^2}{6}-\frac {1}{6} (1-x)^2-\frac {12 \left (3-e^{25}\right ) x}{e^x-36 \left (3-e^{25}\right )}-\frac {x}{3}-\frac {1}{3} (1-x) \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} x \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )+\frac {1}{3} \log \left (e^x-36 \left (3-e^{25}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3888 + E^50*(432 - 2592*x) - 23328*x - 2*E^(2*x)*x + E^x*(-36 + E^25*(12 - 156*x) + 468*x) + E^25*(-2592
+ 15552*x))/(11664 - 7776*E^25 + 1296*E^50 + E^(2*x) + E^x*(-216 + 72*E^25)),x]

[Out]

-1/6*(1 - x)^2 - x/3 - (12*(3 - E^25)*x)/(E^x - 36*(3 - E^25)) - (5*x^2)/6 - ((1 - x)*Log[1 - E^x/(36*(3 - E^2
5))])/3 - (x*Log[1 - E^x/(36*(3 - E^25))])/3 + Log[E^x - 36*(3 - E^25)]/3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
 (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
 - Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
 e, n}, x] && EqQ[c*d, 1]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3888+\left (1-\frac {6}{e^{25}}\right ) e^{50} (432-2592 x)-23328 x-2 e^{2 x} x+e^x \left (-36+e^{25} (12-156 x)+468 x\right )}{\left (e^x-108 \left (1-\frac {e^{25}}{3}\right )\right )^2} \, dx\\ &=\int \left (\frac {12 \left (3-e^{25}\right ) (-1+x)}{e^x-108 \left (1-\frac {e^{25}}{3}\right )}-2 x+\frac {432 \left (3-e^{25}\right )^2 x}{\left (e^x-108 \left (1-\frac {e^{25}}{3}\right )\right )^2}\right ) \, dx\\ &=-x^2+\left (12 \left (3-e^{25}\right )\right ) \int \frac {-1+x}{e^x-108 \left (1-\frac {e^{25}}{3}\right )} \, dx+\left (432 \left (3-e^{25}\right )^2\right ) \int \frac {x}{\left (e^x-108 \left (1-\frac {e^{25}}{3}\right )\right )^2} \, dx\\ &=-\frac {1}{6} (1-x)^2-x^2+\frac {1}{3} \int \frac {e^x (-1+x)}{e^x-108 \left (1-\frac {e^{25}}{3}\right )} \, dx+\left (12 \left (3-e^{25}\right )\right ) \int \frac {e^x x}{\left (e^x-108 \left (1-\frac {e^{25}}{3}\right )\right )^2} \, dx-\left (12 \left (3-e^{25}\right )\right ) \int \frac {x}{e^x-108 \left (1-\frac {e^{25}}{3}\right )} \, dx\\ &=-\frac {1}{6} (1-x)^2-\frac {12 \left (3-e^{25}\right ) x}{e^x-36 \left (3-e^{25}\right )}-\frac {5 x^2}{6}-\frac {1}{3} (1-x) \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} \int \frac {e^x x}{e^x-108 \left (1-\frac {e^{25}}{3}\right )} \, dx-\frac {1}{3} \int \log \left (1-\frac {e^x}{108 \left (1-\frac {e^{25}}{3}\right )}\right ) \, dx+\left (12 \left (3-e^{25}\right )\right ) \int \frac {1}{e^x-108 \left (1-\frac {e^{25}}{3}\right )} \, dx\\ &=-\frac {1}{6} (1-x)^2-\frac {12 \left (3-e^{25}\right ) x}{e^x-36 \left (3-e^{25}\right )}-\frac {5 x^2}{6}-\frac {1}{3} (1-x) \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} x \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )+\frac {1}{3} \int \log \left (1-\frac {e^x}{108 \left (1-\frac {e^{25}}{3}\right )}\right ) \, dx-\frac {1}{3} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{108 \left (1-\frac {e^{25}}{3}\right )}\right )}{x} \, dx,x,e^x\right )+\left (12 \left (3-e^{25}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-108+36 e^{25}+x\right )} \, dx,x,e^x\right )\\ &=-\frac {1}{6} (1-x)^2-\frac {12 \left (3-e^{25}\right ) x}{e^x-36 \left (3-e^{25}\right )}-\frac {5 x^2}{6}-\frac {1}{3} (1-x) \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} x \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )+\frac {1}{3} \text {Li}_2\left (\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-108+36 e^{25}+x} \, dx,x,e^x\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{108 \left (1-\frac {e^{25}}{3}\right )}\right )}{x} \, dx,x,e^x\right )\\ &=-\frac {1}{6} (1-x)^2-\frac {x}{3}-\frac {12 \left (3-e^{25}\right ) x}{e^x-36 \left (3-e^{25}\right )}-\frac {5 x^2}{6}-\frac {1}{3} (1-x) \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )-\frac {1}{3} x \log \left (1-\frac {e^x}{36 \left (3-e^{25}\right )}\right )+\frac {1}{3} \log \left (e^x-36 \left (3-e^{25}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 0.92 \begin {gather*} -x \left (-\frac {12 \left (-3+e^{25}\right )}{-108+36 e^{25}+e^x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3888 + E^50*(432 - 2592*x) - 23328*x - 2*E^(2*x)*x + E^x*(-36 + E^25*(12 - 156*x) + 468*x) + E^25*(
-2592 + 15552*x))/(11664 - 7776*E^25 + 1296*E^50 + E^(2*x) + E^x*(-216 + 72*E^25)),x]

[Out]

-(x*((-12*(-3 + E^25))/(-108 + 36*E^25 + E^x) + x))

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fricas [A]  time = 0.72, size = 40, normalized size = 1.54 \begin {gather*} -\frac {x^{2} e^{x} - 108 \, x^{2} + 12 \, {\left (3 \, x^{2} - x\right )} e^{25} + 36 \, x}{36 \, e^{25} + e^{x} - 108} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2+((-156*x+12)*exp(25)+468*x-36)*exp(x)+(-2592*x+432)*exp(25)^2+(15552*x-2592)*exp(25)-
23328*x+3888)/(exp(x)^2+(72*exp(25)-216)*exp(x)+1296*exp(25)^2-7776*exp(25)+11664),x, algorithm="fricas")

[Out]

-(x^2*e^x - 108*x^2 + 12*(3*x^2 - x)*e^25 + 36*x)/(36*e^25 + e^x - 108)

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giac [B]  time = 0.87, size = 159, normalized size = 6.12 \begin {gather*} -\frac {108 \, x^{2} e^{50} - 648 \, x^{2} e^{25} + 3 \, x^{2} e^{\left (x + 25\right )} - 9 \, x^{2} e^{x} + 972 \, x^{2} - 36 \, x e^{50} + 216 \, x e^{25} + 468 \, e^{50} \log \left (36 \, e^{25} + e^{x} - 108\right ) - 1404 \, e^{25} \log \left (36 \, e^{25} + e^{x} - 108\right ) + 13 \, e^{\left (x + 25\right )} \log \left (36 \, e^{25} + e^{x} - 108\right ) - 468 \, e^{50} \log \left (-36 \, e^{25} - e^{x} + 108\right ) + 1404 \, e^{25} \log \left (-36 \, e^{25} - e^{x} + 108\right ) - 13 \, e^{\left (x + 25\right )} \log \left (-36 \, e^{25} - e^{x} + 108\right ) - 324 \, x}{3 \, {\left (36 \, e^{50} - 216 \, e^{25} + e^{\left (x + 25\right )} - 3 \, e^{x} + 324\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2+((-156*x+12)*exp(25)+468*x-36)*exp(x)+(-2592*x+432)*exp(25)^2+(15552*x-2592)*exp(25)-
23328*x+3888)/(exp(x)^2+(72*exp(25)-216)*exp(x)+1296*exp(25)^2-7776*exp(25)+11664),x, algorithm="giac")

[Out]

-1/3*(108*x^2*e^50 - 648*x^2*e^25 + 3*x^2*e^(x + 25) - 9*x^2*e^x + 972*x^2 - 36*x*e^50 + 216*x*e^25 + 468*e^50
*log(36*e^25 + e^x - 108) - 1404*e^25*log(36*e^25 + e^x - 108) + 13*e^(x + 25)*log(36*e^25 + e^x - 108) - 468*
e^50*log(-36*e^25 - e^x + 108) + 1404*e^25*log(-36*e^25 - e^x + 108) - 13*e^(x + 25)*log(-36*e^25 - e^x + 108)
 - 324*x)/(36*e^50 - 216*e^25 + e^(x + 25) - 3*e^x + 324)

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maple [A]  time = 0.39, size = 24, normalized size = 0.92




method result size



risch \(-x^{2}+\frac {12 \left ({\mathrm e}^{25}-3\right ) x}{{\mathrm e}^{x}+36 \,{\mathrm e}^{25}-108}\) \(24\)
norman \(\frac {\left (-36 \,{\mathrm e}^{25}+108\right ) x^{2}+\left (12 \,{\mathrm e}^{25}-36\right ) x -{\mathrm e}^{x} x^{2}}{{\mathrm e}^{x}+36 \,{\mathrm e}^{25}-108}\) \(38\)
default \(\text {Expression too large to display}\) \(2970\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(x)^2+((-156*x+12)*exp(25)+468*x-36)*exp(x)+(-2592*x+432)*exp(25)^2+(15552*x-2592)*exp(25)-23328*
x+3888)/(exp(x)^2+(72*exp(25)-216)*exp(x)+1296*exp(25)^2-7776*exp(25)+11664),x,method=_RETURNVERBOSE)

[Out]

-x^2+12*(exp(25)-3)*x/(exp(x)+36*exp(25)-108)

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maxima [B]  time = 0.58, size = 222, normalized size = 8.54 \begin {gather*} \frac {1}{3} \, {\left (\frac {x}{e^{50} - 6 \, e^{25} + 9} - \frac {\log \left (36 \, e^{25} + e^{x} - 108\right )}{e^{50} - 6 \, e^{25} + 9} + \frac {36}{{\left (e^{25} - 3\right )} e^{x} + 36 \, e^{50} - 216 \, e^{25} + 324}\right )} e^{50} - 2 \, {\left (\frac {x}{e^{50} - 6 \, e^{25} + 9} - \frac {\log \left (36 \, e^{25} + e^{x} - 108\right )}{e^{50} - 6 \, e^{25} + 9} + \frac {36}{{\left (e^{25} - 3\right )} e^{x} + 36 \, e^{50} - 216 \, e^{25} + 324}\right )} e^{25} + \frac {3 \, x}{e^{50} - 6 \, e^{25} + 9} - \frac {108 \, x^{2} {\left (e^{25} - 3\right )} + {\left (3 \, x^{2} + x\right )} e^{x} + 36 \, e^{25} - 108}{3 \, {\left (36 \, e^{25} + e^{x} - 108\right )}} - \frac {3 \, \log \left (36 \, e^{25} + e^{x} - 108\right )}{e^{50} - 6 \, e^{25} + 9} + \frac {108}{{\left (e^{25} - 3\right )} e^{x} + 36 \, e^{50} - 216 \, e^{25} + 324} + \frac {1}{3} \, \log \left (36 \, e^{25} + e^{x} - 108\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2+((-156*x+12)*exp(25)+468*x-36)*exp(x)+(-2592*x+432)*exp(25)^2+(15552*x-2592)*exp(25)-
23328*x+3888)/(exp(x)^2+(72*exp(25)-216)*exp(x)+1296*exp(25)^2-7776*exp(25)+11664),x, algorithm="maxima")

[Out]

1/3*(x/(e^50 - 6*e^25 + 9) - log(36*e^25 + e^x - 108)/(e^50 - 6*e^25 + 9) + 36/((e^25 - 3)*e^x + 36*e^50 - 216
*e^25 + 324))*e^50 - 2*(x/(e^50 - 6*e^25 + 9) - log(36*e^25 + e^x - 108)/(e^50 - 6*e^25 + 9) + 36/((e^25 - 3)*
e^x + 36*e^50 - 216*e^25 + 324))*e^25 + 3*x/(e^50 - 6*e^25 + 9) - 1/3*(108*x^2*(e^25 - 3) + (3*x^2 + x)*e^x +
36*e^25 - 108)/(36*e^25 + e^x - 108) - 3*log(36*e^25 + e^x - 108)/(e^50 - 6*e^25 + 9) + 108/((e^25 - 3)*e^x +
36*e^50 - 216*e^25 + 324) + 1/3*log(36*e^25 + e^x - 108)

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mupad [B]  time = 0.19, size = 23, normalized size = 0.88 \begin {gather*} \frac {12\,x\,\left ({\mathrm {e}}^{25}-3\right )}{36\,{\mathrm {e}}^{25}+{\mathrm {e}}^x-108}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(23328*x + 2*x*exp(2*x) + exp(x)*(exp(25)*(156*x - 12) - 468*x + 36) + exp(50)*(2592*x - 432) - exp(25)*(
15552*x - 2592) - 3888)/(exp(2*x) - 7776*exp(25) + 1296*exp(50) + exp(x)*(72*exp(25) - 216) + 11664),x)

[Out]

(12*x*(exp(25) - 3))/(36*exp(25) + exp(x) - 108) - x^2

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sympy [A]  time = 0.14, size = 22, normalized size = 0.85 \begin {gather*} - x^{2} + \frac {- 36 x + 12 x e^{25}}{e^{x} - 108 + 36 e^{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)**2+((-156*x+12)*exp(25)+468*x-36)*exp(x)+(-2592*x+432)*exp(25)**2+(15552*x-2592)*exp(25
)-23328*x+3888)/(exp(x)**2+(72*exp(25)-216)*exp(x)+1296*exp(25)**2-7776*exp(25)+11664),x)

[Out]

-x**2 + (-36*x + 12*x*exp(25))/(exp(x) - 108 + 36*exp(25))

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