3.54.24 \(\int \frac {-4+e^{7-2 x^2-x \log (9)} (784 x+448 x^2+64 x^3+(196+112 x+16 x^2) \log (9))}{49+28 x+4 x^2} \, dx\)

Optimal. Leaf size=25 \[ -4 e^{7-x (2 x+\log (9))}+\frac {2}{7+2 x} \]

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Rubi [A]  time = 0.21, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {27, 6688, 2287, 2236} \begin {gather*} \frac {2}{2 x+7}-4\ 9^{-x} e^{7-2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^(7 - 2*x^2 - x*Log[9])*(784*x + 448*x^2 + 64*x^3 + (196 + 112*x + 16*x^2)*Log[9]))/(49 + 28*x + 4*
x^2),x]

[Out]

(-4*E^(7 - 2*x^2))/9^x + 2/(7 + 2*x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

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 {-4+e^{7-2 x^2-x \log (9)} \left (784 x+448 x^2+64 x^3+\left (196+112 x+16 x^2\right ) \log (9)\right )}{(7+2 x)^2} \, dx\\ &=\int \left (-\frac {4}{(7+2 x)^2}+4\ 9^{-x} e^{7-2 x^2} (4 x+\log (9))\right ) \, dx\\ &=\frac {2}{7+2 x}+4 \int 9^{-x} e^{7-2 x^2} (4 x+\log (9)) \, dx\\ &=\frac {2}{7+2 x}+4 \int e^{7-2 x^2-x \log (9)} (4 x+\log (9)) \, dx\\ &=-4 9^{-x} e^{7-2 x^2}+\frac {2}{7+2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 26, normalized size = 1.04 \begin {gather*} -4 9^{-x} e^{7-2 x^2}+\frac {2}{7+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^(7 - 2*x^2 - x*Log[9])*(784*x + 448*x^2 + 64*x^3 + (196 + 112*x + 16*x^2)*Log[9]))/(49 + 28*
x + 4*x^2),x]

[Out]

(-4*E^(7 - 2*x^2))/9^x + 2/(7 + 2*x)

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fricas [A]  time = 1.10, size = 31, normalized size = 1.24 \begin {gather*} -\frac {2 \, {\left (2 \, {\left (2 \, x + 7\right )} e^{\left (-2 \, x^{2} - 2 \, x \log \relax (3) + 7\right )} - 1\right )}}{2 \, x + 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(16*x^2+112*x+196)*log(3)+64*x^3+448*x^2+784*x)*exp(-2*x*log(3)-2*x^2+7)-4)/(4*x^2+28*x+49),x, a
lgorithm="fricas")

[Out]

-2*(2*(2*x + 7)*e^(-2*x^2 - 2*x*log(3) + 7) - 1)/(2*x + 7)

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giac [A]  time = 0.20, size = 42, normalized size = 1.68 \begin {gather*} -\frac {2 \, {\left (4 \, x e^{\left (-2 \, x^{2} - 2 \, x \log \relax (3) + 7\right )} + 14 \, e^{\left (-2 \, x^{2} - 2 \, x \log \relax (3) + 7\right )} - 1\right )}}{2 \, x + 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(16*x^2+112*x+196)*log(3)+64*x^3+448*x^2+784*x)*exp(-2*x*log(3)-2*x^2+7)-4)/(4*x^2+28*x+49),x, a
lgorithm="giac")

[Out]

-2*(4*x*e^(-2*x^2 - 2*x*log(3) + 7) + 14*e^(-2*x^2 - 2*x*log(3) + 7) - 1)/(2*x + 7)

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maple [A]  time = 0.33, size = 20, normalized size = 0.80

method	result	size
risch	\(\frac {1}{x +\frac {7}{2}}-4 \left (\frac {1}{9}\right )^{x} {\mathrm e}^{-2 x^{2}+7}\)	\(20\)
default	\(-4 \,{\mathrm e}^{-2 x \ln \relax (3)-2 x^{2}+7}+\frac {2}{7+2 x}\)	\(26\)
norman	\(\frac {-8 \,{\mathrm e}^{-2 x \ln \relax (3)-2 x^{2}+7} x -28 \,{\mathrm e}^{-2 x \ln \relax (3)-2 x^{2}+7}+2}{7+2 x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(16*x^2+112*x+196)*ln(3)+64*x^3+448*x^2+784*x)*exp(-2*x*ln(3)-2*x^2+7)-4)/(4*x^2+28*x+49),x,method=_RE
TURNVERBOSE)

[Out]

1/(x+7/2)-4*(1/9)^x*exp(-2*x^2+7)

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maxima [A]  time = 0.58, size = 25, normalized size = 1.00 \begin {gather*} \frac {2}{2 \, x + 7} - 4 \, e^{\left (-2 \, x^{2} - 2 \, x \log \relax (3) + 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(16*x^2+112*x+196)*log(3)+64*x^3+448*x^2+784*x)*exp(-2*x*log(3)-2*x^2+7)-4)/(4*x^2+28*x+49),x, a
lgorithm="maxima")

[Out]

2/(2*x + 7) - 4*e^(-2*x^2 - 2*x*log(3) + 7)

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mupad [B]  time = 0.35, size = 60, normalized size = 2.40 \begin {gather*} -\frac {4\,x}{7\,\left (2\,x+7\right )}-\frac {28\,{\mathrm {e}}^7\,{\mathrm {e}}^{-2\,x^2}}{3^{2\,x}\,\left (2\,x+7\right )}-\frac {8\,x\,{\mathrm {e}}^7\,{\mathrm {e}}^{-2\,x^2}}{3^{2\,x}\,\left (2\,x+7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(7 - 2*x^2 - 2*x*log(3))*(784*x + 2*log(3)*(112*x + 16*x^2 + 196) + 448*x^2 + 64*x^3) - 4)/(28*x + 4*x
^2 + 49),x)

[Out]

- (4*x)/(7*(2*x + 7)) - (28*exp(7)*exp(-2*x^2))/(3^(2*x)*(2*x + 7)) - (8*x*exp(7)*exp(-2*x^2))/(3^(2*x)*(2*x +
7))

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sympy [A]  time = 0.19, size = 22, normalized size = 0.88 \begin {gather*} - 4 e^{- 2 x^{2} - 2 x \log {\relax (3 )} + 7} + \frac {4}{4 x + 14} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(16*x**2+112*x+196)*ln(3)+64*x**3+448*x**2+784*x)*exp(-2*x*ln(3)-2*x**2+7)-4)/(4*x**2+28*x+49),x
)

[Out]

-4*exp(-2*x**2 - 2*x*log(3) + 7) + 4/(4*x + 14)

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