3.85.51 \(\int \frac {e^4 (-15 x-6 x^2)+e^4 (6+3 x) \log (e^{2 x} (2+x))}{2 x^2+x^3} \, dx\)

Optimal. Leaf size=20 \[ 1-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x} \]

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Rubi [A]  time = 0.20, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {1593, 6742, 72, 2551} \begin {gather*} -\frac {3 e^4 \log \left (e^{2 x} (x+2)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(-15*x - 6*x^2) + E^4*(6 + 3*x)*Log[E^(2*x)*(2 + x)])/(2*x^2 + x^3),x]

[Out]

(-3*E^4*Log[E^(2*x)*(2 + x)])/x

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

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 {e^4 \left (-15 x-6 x^2\right )+e^4 (6+3 x) \log \left (e^{2 x} (2+x)\right )}{x^2 (2+x)} \, dx\\ &=\int \left (-\frac {3 e^4 (5+2 x)}{x (2+x)}+\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x^2}\right ) \, dx\\ &=-\left (\left (3 e^4\right ) \int \frac {5+2 x}{x (2+x)} \, dx\right )+\left (3 e^4\right ) \int \frac {\log \left (e^{2 x} (2+x)\right )}{x^2} \, dx\\ &=-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}+\left (3 e^4\right ) \int \frac {5+2 x}{x (2+x)} \, dx-\left (3 e^4\right ) \int \left (\frac {5}{2 x}-\frac {1}{2 (2+x)}\right ) \, dx\\ &=-\frac {15}{2} e^4 \log (x)+\frac {3}{2} e^4 \log (2+x)-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}+\left (3 e^4\right ) \int \left (\frac {5}{2 x}-\frac {1}{2 (2+x)}\right ) \, dx\\ &=-\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 0.90 \begin {gather*} -\frac {3 e^4 \log \left (e^{2 x} (2+x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(-15*x - 6*x^2) + E^4*(6 + 3*x)*Log[E^(2*x)*(2 + x)])/(2*x^2 + x^3),x]

[Out]

(-3*E^4*Log[E^(2*x)*(2 + x)])/x

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fricas [A]  time = 0.89, size = 16, normalized size = 0.80 \begin {gather*} -\frac {3 \, e^{4} \log \left ({\left (x + 2\right )} e^{\left (2 \, x\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="fricas")

[Out]

-3*e^4*log((x + 2)*e^(2*x))/x

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giac [A]  time = 0.12, size = 11, normalized size = 0.55 \begin {gather*} -\frac {3 \, e^{4} \log \left (x + 2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="giac")

[Out]

-3*e^4*log(x + 2)/x

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maple [A]  time = 0.43, size = 19, normalized size = 0.95

method	result	size
norman	\(-\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2 x} \left (2+x \right )\right )}{x}\)	\(19\)
default	\(-\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{2 x} \left (2+x \right )\right )}{x}+6 \,{\mathrm e}^{4} \ln \left (2 x +4\right )+\frac {15 \,{\mathrm e}^{4} \ln \left (-2 x \right )}{2}-\frac {15 \,{\mathrm e}^{4} \ln \left (-2 x -4\right )}{2}-\frac {15 \,{\mathrm e}^{4} \ln \relax (x )}{2}+\frac {3 \,{\mathrm e}^{4} \ln \left (2+x \right )}{2}\)	\(62\)
risch	\(\frac {3 \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{-2 x}\right )}{x}-\frac {3 \,{\mathrm e}^{4} \left (-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )+i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{2}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (2+x \right )\right )^{3}+2 \ln \left (2+x \right )\right )}{2 x}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6+3*x)*exp(4)*ln((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x,method=_RETURNVERBOSE)

[Out]

-3/x*exp(4)*ln((2+x)/exp(-2*x))

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maxima [B]  time = 0.40, size = 46, normalized size = 2.30 \begin {gather*} \frac {15}{2} \, {\left (\log \left (x + 2\right ) - \log \relax (x)\right )} e^{4} - 6 \, e^{4} \log \left (x + 2\right ) + \frac {15}{2} \, e^{4} \log \relax (x) - \frac {3 \, {\left (x e^{4} + 2 \, e^{4}\right )} \log \left (x + 2\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="maxima")

[Out]

15/2*(log(x + 2) - log(x))*e^4 - 6*e^4*log(x + 2) + 15/2*e^4*log(x) - 3/2*(x*e^4 + 2*e^4)*log(x + 2)/x

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mupad [B]  time = 5.34, size = 11, normalized size = 0.55 \begin {gather*} -\frac {3\,\ln \left (x+2\right )\,{\mathrm {e}}^4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4)*(15*x + 6*x^2) - exp(4)*log(exp(2*x)*(x + 2))*(3*x + 6))/(2*x^2 + x^3),x)

[Out]

-(3*log(x + 2)*exp(4))/x

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sympy [A]  time = 0.21, size = 17, normalized size = 0.85 \begin {gather*} - \frac {3 e^{4} \log {\left (\left (x + 2\right ) e^{2 x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*ln((2+x)/exp(-2*x))+(-6*x**2-15*x)*exp(4))/(x**3+2*x**2),x)

[Out]

-3*exp(4)*log((x + 2)*exp(2*x))/x

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