3.52.32 \(\int -\frac {54 e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x (180-12 e^4)} \, dx\)

Optimal. Leaf size=22 \[ e^{3+\frac {3}{5-\frac {e^4}{3}+2 e^x}} \]

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Rubi [A]  time = 0.33, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {12, 2282, 2230, 2209} \begin {gather*} e^{\frac {9}{6 e^x+15-e^4}+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-54*E^((54 - 3*E^4 + 18*E^x)/(15 - E^4 + 6*E^x) + x))/(225 - 30*E^4 + E^8 + 36*E^(2*x) + E^x*(180 - 12*E^
4)),x]

[Out]

E^(3 + 9/(15 - E^4 + 6*E^x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2209

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

Rule 2230

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (54 \int \frac {e^{\frac {54-3 e^4+18 e^x}{15-e^4+6 e^x}+x}}{225-30 e^4+e^8+36 e^{2 x}+e^x \left (180-12 e^4\right )} \, dx\right )\\ &=-\left (54 \operatorname {Subst}\left (\int \frac {e^{\frac {3 \left (-18+e^4-6 x\right )}{-15+e^4-6 x}}}{\left (15-e^4+6 x\right )^2} \, dx,x,e^x\right )\right )\\ &=-\left (54 \operatorname {Subst}\left (\int \frac {\exp \left (3+\frac {6 \left (-18+e^4\right )-6 \left (-15+e^4\right )}{2 \left (-15+e^4-6 x\right )}\right )}{\left (15-e^4+6 x\right )^2} \, dx,x,e^x\right )\right )\\ &=e^{3+\frac {9}{15-e^4+6 e^x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 20, normalized size = 0.91 \begin {gather*} e^{3+\frac {9}{15-e^4+6 e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-54*E^((54 - 3*E^4 + 18*E^x)/(15 - E^4 + 6*E^x) + x))/(225 - 30*E^4 + E^8 + 36*E^(2*x) + E^x*(180 -
12*E^4)),x]

[Out]

E^(3 + 9/(15 - E^4 + 6*E^x))

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fricas [B]  time = 0.53, size = 34, normalized size = 1.55 \begin {gather*} e^{\left (-x + \frac {{\left (x + 3\right )} e^{4} - 6 \, {\left (x + 3\right )} e^{x} - 15 \, x - 54}{e^{4} - 6 \, e^{x} - 15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-54*exp(x)*exp((18*exp(x)-3*exp(2)^2+54)/(6*exp(x)-exp(2)^2+15))/(36*exp(x)^2+(-12*exp(2)^2+180)*exp
(x)+exp(2)^4-30*exp(2)^2+225),x, algorithm="fricas")

[Out]

e^(-x + ((x + 3)*e^4 - 6*(x + 3)*e^x - 15*x - 54)/(e^4 - 6*e^x - 15))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {54 \, e^{\left (x + \frac {3 \, {\left (e^{4} - 6 \, e^{x} - 18\right )}}{e^{4} - 6 \, e^{x} - 15}\right )}}{12 \, {\left (e^{4} - 15\right )} e^{x} - e^{8} + 30 \, e^{4} - 36 \, e^{\left (2 \, x\right )} - 225}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-54*exp(x)*exp((18*exp(x)-3*exp(2)^2+54)/(6*exp(x)-exp(2)^2+15))/(36*exp(x)^2+(-12*exp(2)^2+180)*exp
(x)+exp(2)^4-30*exp(2)^2+225),x, algorithm="giac")

[Out]

integrate(54*e^(x + 3*(e^4 - 6*e^x - 18)/(e^4 - 6*e^x - 15))/(12*(e^4 - 15)*e^x - e^8 + 30*e^4 - 36*e^(2*x) -
225), x)

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maple [A]  time = 0.19, size = 22, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {-18 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{4}-54}{-6 \,{\mathrm e}^{x}+{\mathrm e}^{4}-15}}\)	\(22\)
norman	\(\frac {\left ({\mathrm e}^{4}-15\right ) {\mathrm e}^{\frac {18 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{4}+54}{6 \,{\mathrm e}^{x}-{\mathrm e}^{4}+15}}-6 \,{\mathrm e}^{x} {\mathrm e}^{\frac {18 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{4}+54}{6 \,{\mathrm e}^{x}-{\mathrm e}^{4}+15}}}{-6 \,{\mathrm e}^{x}+{\mathrm e}^{4}-15}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-54*exp(x)*exp((18*exp(x)-3*exp(2)^2+54)/(6*exp(x)-exp(2)^2+15))/(36*exp(x)^2+(-12*exp(2)^2+180)*exp(x)+ex
p(2)^4-30*exp(2)^2+225),x,method=_RETURNVERBOSE)

[Out]

exp(3*(-6*exp(x)+exp(4)-18)/(-6*exp(x)+exp(4)-15))

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maxima [B]  time = 0.43, size = 42, normalized size = 1.91 \begin {gather*} e^{\left (\frac {3 \, e^{4}}{e^{4} - 6 \, e^{x} - 15} - \frac {18 \, e^{x}}{e^{4} - 6 \, e^{x} - 15} - \frac {54}{e^{4} - 6 \, e^{x} - 15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-54*exp(x)*exp((18*exp(x)-3*exp(2)^2+54)/(6*exp(x)-exp(2)^2+15))/(36*exp(x)^2+(-12*exp(2)^2+180)*exp
(x)+exp(2)^4-30*exp(2)^2+225),x, algorithm="maxima")

[Out]

e^(3*e^4/(e^4 - 6*e^x - 15) - 18*e^x/(e^4 - 6*e^x - 15) - 54/(e^4 - 6*e^x - 15))

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mupad [B]  time = 3.73, size = 50, normalized size = 2.27 \begin {gather*} {\mathrm {e}}^{\frac {18\,{\mathrm {e}}^x}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}}\,{\mathrm {e}}^{\frac {54}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^4}{6\,{\mathrm {e}}^x-{\mathrm {e}}^4+15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(54*exp(x)*exp((18*exp(x) - 3*exp(4) + 54)/(6*exp(x) - exp(4) + 15)))/(36*exp(2*x) - 30*exp(4) + exp(8) -
exp(x)*(12*exp(4) - 180) + 225),x)

[Out]

exp((18*exp(x))/(6*exp(x) - exp(4) + 15))*exp(54/(6*exp(x) - exp(4) + 15))*exp(-(3*exp(4))/(6*exp(x) - exp(4)
+ 15))

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sympy [A]  time = 0.26, size = 22, normalized size = 1.00 \begin {gather*} e^{\frac {18 e^{x} - 3 e^{4} + 54}{6 e^{x} - e^{4} + 15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-54*exp(x)*exp((18*exp(x)-3*exp(2)**2+54)/(6*exp(x)-exp(2)**2+15))/(36*exp(x)**2+(-12*exp(2)**2+180)
*exp(x)+exp(2)**4-30*exp(2)**2+225),x)

[Out]

exp((18*exp(x) - 3*exp(4) + 54)/(6*exp(x) - exp(4) + 15))

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