3.27.2 \(\int \frac {e^{7 x+x^2+8 e^{7 x+x^2} x^3} (-96 x^2-224 x^3-64 x^4)}{e^{10}+e^{16 e^{7 x+x^2} x^3}+2 e^{5+8 e^{7 x+x^2} x^3}} \, dx\)

Optimal. Leaf size=22 \[ \frac {4}{e^5+e^{8 e^{x (7+x)} x^3}} \]

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Rubi [A]  time = 2.43, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {1594, 6688, 12, 6686} \begin {gather*} \frac {4}{e^{8 e^{x (x+7)} x^3}+e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(7*x + x^2 + 8*E^(7*x + x^2)*x^3)*(-96*x^2 - 224*x^3 - 64*x^4))/(E^10 + E^(16*E^(7*x + x^2)*x^3) + 2*E^
(5 + 8*E^(7*x + x^2)*x^3)),x]

[Out]

4/(E^5 + E^(8*E^(x*(7 + x))*x^3))

Rule 12

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

Rule 1594

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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{7 x+x^2+8 e^{7 x+x^2} x^3} x^2 \left (-96-224 x-64 x^2\right )}{e^{10}+e^{16 e^{7 x+x^2} x^3}+2 e^{5+8 e^{7 x+x^2} x^3}} \, dx\\ &=\int \frac {32 e^{7 x+x^2+8 e^{x (7+x)} x^3} x^2 \left (-3-7 x-2 x^2\right )}{\left (e^5+e^{8 e^{x (7+x)} x^3}\right )^2} \, dx\\ &=32 \int \frac {e^{7 x+x^2+8 e^{x (7+x)} x^3} x^2 \left (-3-7 x-2 x^2\right )}{\left (e^5+e^{8 e^{x (7+x)} x^3}\right )^2} \, dx\\ &=\frac {4}{e^5+e^{8 e^{x (7+x)} x^3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 22, normalized size = 1.00 \begin {gather*} \frac {4}{e^5+e^{8 e^{x (7+x)} x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(7*x + x^2 + 8*E^(7*x + x^2)*x^3)*(-96*x^2 - 224*x^3 - 64*x^4))/(E^10 + E^(16*E^(7*x + x^2)*x^3)
+ 2*E^(5 + 8*E^(7*x + x^2)*x^3)),x]

[Out]

4/(E^5 + E^(8*E^(x*(7 + x))*x^3))

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fricas [B]  time = 0.59, size = 43, normalized size = 1.95 \begin {gather*} \frac {4 \, e^{\left (x^{2} + 7 \, x\right )}}{e^{\left (8 \, x^{3} e^{\left (x^{2} + 7 \, x\right )} + x^{2} + 7 \, x\right )} + e^{\left (x^{2} + 7 \, x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*x^4-224*x^3-96*x^2)*exp(x^2+7*x)*exp(8*x^3*exp(x^2+7*x))/(exp(8*x^3*exp(x^2+7*x))^2+2*exp(5)*ex
p(8*x^3*exp(x^2+7*x))+exp(5)^2),x, algorithm="fricas")

[Out]

4*e^(x^2 + 7*x)/(e^(8*x^3*e^(x^2 + 7*x) + x^2 + 7*x) + e^(x^2 + 7*x + 5))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*x^4-224*x^3-96*x^2)*exp(x^2+7*x)*exp(8*x^3*exp(x^2+7*x))/(exp(8*x^3*exp(x^2+7*x))^2+2*exp(5)*ex
p(8*x^3*exp(x^2+7*x))+exp(5)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -2/sqrt(exp(10)-exp(5)^2)*(-i)*ln(abs(2*
exp(sageVARx^3*exp(sageVARx)^7*exp(sageVARx^2))^8+2*exp(5)-2*sqrt(exp(10)-exp(5)^2)*i)/abs(2*exp(sageVARx^3*ex
p(sageVARx)^7*exp(sag

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maple [A]  time = 0.21, size = 20, normalized size = 0.91

method	result	size
risch	\(\frac {4}{{\mathrm e}^{8 \,{\mathrm e}^{\left (x +7\right ) x} x^{3}}+{\mathrm e}^{5}}\)	\(20\)
norman	\(\frac {4}{{\mathrm e}^{8 x^{3} {\mathrm e}^{x^{2}+7 x}}+{\mathrm e}^{5}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-64*x^4-224*x^3-96*x^2)*exp(x^2+7*x)*exp(8*x^3*exp(x^2+7*x))/(exp(8*x^3*exp(x^2+7*x))^2+2*exp(5)*exp(8*x^
3*exp(x^2+7*x))+exp(5)^2),x,method=_RETURNVERBOSE)

[Out]

4/(exp(8*exp((x+7)*x)*x^3)+exp(5))

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maxima [A]  time = 0.61, size = 21, normalized size = 0.95 \begin {gather*} \frac {4}{e^{5} + e^{\left (8 \, x^{3} e^{\left (x^{2} + 7 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*x^4-224*x^3-96*x^2)*exp(x^2+7*x)*exp(8*x^3*exp(x^2+7*x))/(exp(8*x^3*exp(x^2+7*x))^2+2*exp(5)*ex
p(8*x^3*exp(x^2+7*x))+exp(5)^2),x, algorithm="maxima")

[Out]

4/(e^5 + e^(8*x^3*e^(x^2 + 7*x)))

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mupad [B]  time = 0.27, size = 21, normalized size = 0.95 \begin {gather*} \frac {4}{{\mathrm {e}}^{8\,x^3\,{\mathrm {e}}^{7\,x}\,{\mathrm {e}}^{x^2}}+{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8*x^3*exp(7*x + x^2))*exp(7*x + x^2)*(96*x^2 + 224*x^3 + 64*x^4))/(exp(16*x^3*exp(7*x + x^2)) + exp(
10) + 2*exp(8*x^3*exp(7*x + x^2))*exp(5)),x)

[Out]

4/(exp(8*x^3*exp(7*x)*exp(x^2)) + exp(5))

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sympy [A]  time = 0.31, size = 19, normalized size = 0.86 \begin {gather*} \frac {4}{e^{8 x^{3} e^{x^{2} + 7 x}} + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*x**4-224*x**3-96*x**2)*exp(x**2+7*x)*exp(8*x**3*exp(x**2+7*x))/(exp(8*x**3*exp(x**2+7*x))**2+2*
exp(5)*exp(8*x**3*exp(x**2+7*x))+exp(5)**2),x)

[Out]

4/(exp(8*x**3*exp(x**2 + 7*x)) + exp(5))

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