3.85.95 \(\int \frac {-144 x-80 x^2+4 e^5 x^2-72 \log (2)}{e^5 x^4+2 e^5 x^3 \log (2)+e^5 x^2 \log ^2(2)} \, dx\)

Optimal. Leaf size=29 \[ \frac {4 \left (-1+\frac {5 \left (4+\frac {3}{x}\right )+\frac {3}{x}}{e^5}\right )}{x+\log (2)} \]

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Rubi [A]  time = 0.06, antiderivative size = 40, normalized size of antiderivative = 1.38, number of steps used = 6, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6, 1594, 27, 12, 893} \begin {gather*} \frac {72}{e^5 x \log (2)}-\frac {4 \left (18-20 \log (2)+e^5 \log (2)\right )}{e^5 \log (2) (x+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-144*x - 80*x^2 + 4*E^5*x^2 - 72*Log[2])/(E^5*x^4 + 2*E^5*x^3*Log[2] + E^5*x^2*Log[2]^2),x]

[Out]

72/(E^5*x*Log[2]) - (4*(18 - 20*Log[2] + E^5*Log[2]))/(E^5*Log[2]*(x + Log[2]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-144 x+\left (-80+4 e^5\right ) x^2-72 \log (2)}{e^5 x^4+2 e^5 x^3 \log (2)+e^5 x^2 \log ^2(2)} \, dx\\ &=\int \frac {-144 x+\left (-80+4 e^5\right ) x^2-72 \log (2)}{x^2 \left (e^5 x^2+2 e^5 x \log (2)+e^5 \log ^2(2)\right )} \, dx\\ &=\int \frac {-144 x+\left (-80+4 e^5\right ) x^2-72 \log (2)}{e^5 x^2 (x+\log (2))^2} \, dx\\ &=\frac {\int \frac {-144 x+\left (-80+4 e^5\right ) x^2-72 \log (2)}{x^2 (x+\log (2))^2} \, dx}{e^5}\\ &=\frac {\int \left (-\frac {72}{x^2 \log (2)}+\frac {4 \left (18-20 \log (2)+e^5 \log (2)\right )}{\log (2) (x+\log (2))^2}\right ) \, dx}{e^5}\\ &=\frac {72}{e^5 x \log (2)}-\frac {4 \left (18-20 \log (2)+e^5 \log (2)\right )}{e^5 \log (2) (x+\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.90 \begin {gather*} \frac {4 \left (18+20 x-e^5 x\right )}{e^5 \left (x^2+x \log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-144*x - 80*x^2 + 4*E^5*x^2 - 72*Log[2])/(E^5*x^4 + 2*E^5*x^3*Log[2] + E^5*x^2*Log[2]^2),x]

[Out]

(4*(18 + 20*x - E^5*x))/(E^5*(x^2 + x*Log[2]))

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fricas [A]  time = 0.84, size = 26, normalized size = 0.90 \begin {gather*} -\frac {4 \, {\left (x e^{5} - 20 \, x - 18\right )}}{x^{2} e^{5} + x e^{5} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-72*log(2)+4*x^2*exp(5)-80*x^2-144*x)/(x^2*exp(5)*log(2)^2+2*x^3*exp(5)*log(2)+x^4*exp(5)),x, algor
ithm="fricas")

[Out]

-4*(x*e^5 - 20*x - 18)/(x^2*e^5 + x*e^5*log(2))

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giac [A]  time = 0.20, size = 23, normalized size = 0.79 \begin {gather*} -\frac {4 \, {\left (x e^{5} - 20 \, x - 18\right )} e^{\left (-5\right )}}{x^{2} + x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-72*log(2)+4*x^2*exp(5)-80*x^2-144*x)/(x^2*exp(5)*log(2)^2+2*x^3*exp(5)*log(2)+x^4*exp(5)),x, algor
ithm="giac")

[Out]

-4*(x*e^5 - 20*x - 18)*e^(-5)/(x^2 + x*log(2))

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maple [A]  time = 0.07, size = 23, normalized size = 0.79

method	result	size
risch	\(\frac {\left (\left (-4 \,{\mathrm e}^{5}+80\right ) x +72\right ) {\mathrm e}^{-5}}{x \left (\ln \relax (2)+x \right )}\)	\(23\)
gosper	\(-\frac {4 \left (x \,{\mathrm e}^{5}-20 x -18\right ) {\mathrm e}^{-5}}{x \left (\ln \relax (2)+x \right )}\)	\(25\)
norman	\(\frac {-4 \left ({\mathrm e}^{5}-20\right ) {\mathrm e}^{-5} x +72 \,{\mathrm e}^{-5}}{x \left (\ln \relax (2)+x \right )}\)	\(29\)
default	\(4 \,{\mathrm e}^{-5} \left (-\frac {{\mathrm e}^{5} \ln \relax (2)-20 \ln \relax (2)+18}{\ln \relax (2) \left (\ln \relax (2)+x \right )}+\frac {18}{x \ln \relax (2)}\right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-72*ln(2)+4*x^2*exp(5)-80*x^2-144*x)/(x^2*exp(5)*ln(2)^2+2*x^3*exp(5)*ln(2)+x^4*exp(5)),x,method=_RETURNV
ERBOSE)

[Out]

((-4*exp(5)+80)*x+72)/x/(ln(2)+x)*exp(-5)

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maxima [A]  time = 0.36, size = 25, normalized size = 0.86 \begin {gather*} -\frac {4 \, {\left (x {\left (e^{5} - 20\right )} - 18\right )}}{x^{2} e^{5} + x e^{5} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-72*log(2)+4*x^2*exp(5)-80*x^2-144*x)/(x^2*exp(5)*log(2)^2+2*x^3*exp(5)*log(2)+x^4*exp(5)),x, algor
ithm="maxima")

[Out]

-4*(x*(e^5 - 20) - 18)/(x^2*e^5 + x*e^5*log(2))

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mupad [B]  time = 5.96, size = 23, normalized size = 0.79 \begin {gather*} \frac {4\,{\mathrm {e}}^{-5}\,\left (20\,x-x\,{\mathrm {e}}^5+18\right )}{x\,\left (x+\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(144*x + 72*log(2) - 4*x^2*exp(5) + 80*x^2)/(x^4*exp(5) + 2*x^3*exp(5)*log(2) + x^2*exp(5)*log(2)^2),x)

[Out]

(4*exp(-5)*(20*x - x*exp(5) + 18))/(x*(x + log(2)))

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sympy [A]  time = 0.65, size = 26, normalized size = 0.90 \begin {gather*} - \frac {x \left (-80 + 4 e^{5}\right ) - 72}{x^{2} e^{5} + x e^{5} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-72*ln(2)+4*x**2*exp(5)-80*x**2-144*x)/(x**2*exp(5)*ln(2)**2+2*x**3*exp(5)*ln(2)+x**4*exp(5)),x)

[Out]

-(x*(-80 + 4*exp(5)) - 72)/(x**2*exp(5) + x*exp(5)*log(2))

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