3.29.44 \(\int \frac {1200 x-800 \log (4)}{9 e^{10} x^5-18 e^{10} x^4 \log (4)+9 e^{10} x^3 \log ^2(4)} \, dx\)

Optimal. Leaf size=18 \[ \frac {400}{9 e^{10} x^2 (-x+\log (4))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1594, 27, 12, 74} \begin {gather*} -\frac {400}{9 e^{10} x^2 (x-\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1200*x - 800*Log[4])/(9*E^10*x^5 - 18*E^10*x^4*Log[4] + 9*E^10*x^3*Log[4]^2),x]

[Out]

-400/(9*E^10*x^2*(x - Log[4]))

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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 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 {1200 x-800 \log (4)}{x^3 \left (9 e^{10} x^2-18 e^{10} x \log (4)+9 e^{10} \log ^2(4)\right )} \, dx\\ &=\int \frac {1200 x-800 \log (4)}{9 e^{10} x^3 (x-\log (4))^2} \, dx\\ &=\frac {\int \frac {1200 x-800 \log (4)}{x^3 (x-\log (4))^2} \, dx}{9 e^{10}}\\ &=-\frac {400}{9 e^{10} x^2 (x-\log (4))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} -\frac {400}{9 e^{10} x^2 (x-\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1200*x - 800*Log[4])/(9*E^10*x^5 - 18*E^10*x^4*Log[4] + 9*E^10*x^3*Log[4]^2),x]

[Out]

-400/(9*E^10*x^2*(x - Log[4]))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 20, normalized size = 1.11 \begin {gather*} -\frac {400}{9 \, {\left (x^{3} e^{10} - 2 \, x^{2} e^{10} \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1600*log(2)+1200*x)/(36*x^3*exp(5)^2*log(2)^2-36*x^4*exp(5)^2*log(2)+9*x^5*exp(5)^2),x, algorithm=
"fricas")

[Out]

-400/9/(x^3*e^10 - 2*x^2*e^10*log(2))

________________________________________________________________________________________

giac [B]  time = 0.45, size = 34, normalized size = 1.89 \begin {gather*} -\frac {100 \, e^{\left (-10\right )}}{9 \, {\left (x - 2 \, \log \relax (2)\right )} \log \relax (2)^{2}} + \frac {100 \, {\left (x + 2 \, \log \relax (2)\right )} e^{\left (-10\right )}}{9 \, x^{2} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1600*log(2)+1200*x)/(36*x^3*exp(5)^2*log(2)^2-36*x^4*exp(5)^2*log(2)+9*x^5*exp(5)^2),x, algorithm=
"giac")

[Out]

-100/9*e^(-10)/((x - 2*log(2))*log(2)^2) + 100/9*(x + 2*log(2))*e^(-10)/(x^2*log(2)^2)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 18, normalized size = 1.00

method	result	size
risch	\(\frac {400 \,{\mathrm e}^{-10}}{9 \left (2 \ln \relax (2)-x \right ) x^{2}}\)	\(18\)
gosper	\(\frac {400 \,{\mathrm e}^{-10}}{9 \left (2 \ln \relax (2)-x \right ) x^{2}}\)	\(20\)
norman	\(\frac {400 \,{\mathrm e}^{-10}}{9 \left (2 \ln \relax (2)-x \right ) x^{2}}\)	\(20\)
default	\(\frac {400 \,{\mathrm e}^{-10} \left (\frac {1}{4 \ln \relax (2)^{2} x}+\frac {1}{2 \ln \relax (2) x^{2}}-\frac {1}{4 \ln \relax (2)^{2} \left (x -2 \ln \relax (2)\right )}\right )}{9}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1600*ln(2)+1200*x)/(36*x^3*exp(5)^2*ln(2)^2-36*x^4*exp(5)^2*ln(2)+9*x^5*exp(5)^2),x,method=_RETURNVERBOS
E)

[Out]

400/9*exp(-10)/(2*ln(2)-x)/x^2

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 20, normalized size = 1.11 \begin {gather*} -\frac {400}{9 \, {\left (x^{3} e^{10} - 2 \, x^{2} e^{10} \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1600*log(2)+1200*x)/(36*x^3*exp(5)^2*log(2)^2-36*x^4*exp(5)^2*log(2)+9*x^5*exp(5)^2),x, algorithm=
"maxima")

[Out]

-400/9/(x^3*e^10 - 2*x^2*e^10*log(2))

________________________________________________________________________________________

mupad [B]  time = 1.86, size = 19, normalized size = 1.06 \begin {gather*} \frac {400\,{\mathrm {e}}^{-10}}{9\,\left (2\,x^2\,\ln \relax (2)-x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1200*x - 1600*log(2))/(9*x^5*exp(10) - 36*x^4*exp(10)*log(2) + 36*x^3*exp(10)*log(2)^2),x)

[Out]

(400*exp(-10))/(9*(2*x^2*log(2) - x^3))

________________________________________________________________________________________

sympy [A]  time = 0.20, size = 22, normalized size = 1.22 \begin {gather*} - \frac {400}{9 x^{3} e^{10} - 18 x^{2} e^{10} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1600*ln(2)+1200*x)/(36*x**3*exp(5)**2*ln(2)**2-36*x**4*exp(5)**2*ln(2)+9*x**5*exp(5)**2),x)

[Out]

-400/(9*x**3*exp(10) - 18*x**2*exp(10)*log(2))

________________________________________________________________________________________