3.17.89 \(\int \frac {17+60 x+(4+12 x) \log (4)}{20 e^4 x+4 e^4 x \log (4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {3 (4+x)-\frac {1}{4} \left (-4+\frac {3}{5+\log (4)}\right ) \log (x)}{e^4} \]

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6, 12, 186, 43} \begin {gather*} \frac {3 x}{e^4}+\frac {(17+\log (256)) \log (x)}{4 e^4 (5+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(17 + 60*x + (4 + 12*x)*Log[4])/(20*E^4*x + 4*E^4*x*Log[4]),x]

[Out]

(3*x)/E^4 + ((17 + Log[256])*Log[x])/(4*E^4*(5 + Log[4]))

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {17+60 x+(4+12 x) \log (4)}{e^4 x (20+4 \log (4))} \, dx\\ &=\frac {\int \frac {17+60 x+(4+12 x) \log (4)}{x} \, dx}{4 e^4 (5+\log (4))}\\ &=\frac {\int \frac {17+12 x (5+\log (4))+\log (256)}{x} \, dx}{4 e^4 (5+\log (4))}\\ &=\frac {\int \left (12 (5+\log (4))+\frac {17+\log (256)}{x}\right ) \, dx}{4 e^4 (5+\log (4))}\\ &=\frac {3 x}{e^4}+\frac {(17+\log (256)) \log (x)}{4 e^4 (5+\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.08 \begin {gather*} \frac {12 x (5+\log (4))+(17+\log (256)) \log (x)}{4 e^4 (5+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(17 + 60*x + (4 + 12*x)*Log[4])/(20*E^4*x + 4*E^4*x*Log[4]),x]

[Out]

(12*x*(5 + Log[4]) + (17 + Log[256])*Log[x])/(4*E^4*(5 + Log[4]))

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fricas [A]  time = 0.72, size = 33, normalized size = 1.27 \begin {gather*} \frac {24 \, x \log \relax (2) + {\left (8 \, \log \relax (2) + 17\right )} \log \relax (x) + 60 \, x}{4 \, {\left (2 \, e^{4} \log \relax (2) + 5 \, e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(12*x+4)*log(2)+60*x+17)/(8*x*exp(4)*log(2)+20*x*exp(4)),x, algorithm="fricas")

[Out]

1/4*(24*x*log(2) + (8*log(2) + 17)*log(x) + 60*x)/(2*e^4*log(2) + 5*e^4)

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giac [B]  time = 0.17, size = 49, normalized size = 1.88 \begin {gather*} \frac {{\left (8 \, \log \relax (2) + 17\right )} \log \left ({\left | x \right |}\right )}{4 \, {\left (2 \, e^{4} \log \relax (2) + 5 \, e^{4}\right )}} + \frac {3 \, {\left (2 \, x \log \relax (2) + 5 \, x\right )}}{2 \, e^{4} \log \relax (2) + 5 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(12*x+4)*log(2)+60*x+17)/(8*x*exp(4)*log(2)+20*x*exp(4)),x, algorithm="giac")

[Out]

1/4*(8*log(2) + 17)*log(abs(x))/(2*e^4*log(2) + 5*e^4) + 3*(2*x*log(2) + 5*x)/(2*e^4*log(2) + 5*e^4)

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maple [A]  time = 0.04, size = 31, normalized size = 1.19

method	result	size
norman	\(3 x \,{\mathrm e}^{-4}+\frac {\left (8 \ln \relax (2)+17\right ) {\mathrm e}^{-4} \ln \relax (x )}{8 \ln \relax (2)+20}\)	\(31\)
default	\(\frac {{\mathrm e}^{-4} \left (24 x \ln \relax (2)+60 x +\left (8 \ln \relax (2)+17\right ) \ln \relax (x )\right )}{8 \ln \relax (2)+20}\)	\(33\)
risch	\(3 x \,{\mathrm e}^{-4}+\frac {2 \,{\mathrm e}^{-4} \ln \relax (x ) \ln \relax (2)}{2 \ln \relax (2)+5}+\frac {17 \,{\mathrm e}^{-4} \ln \relax (x )}{4 \left (2 \ln \relax (2)+5\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(12*x+4)*ln(2)+60*x+17)/(8*x*exp(4)*ln(2)+20*x*exp(4)),x,method=_RETURNVERBOSE)

[Out]

3*x/exp(4)+1/4*(8*ln(2)+17)/exp(4)/(2*ln(2)+5)*ln(x)

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maxima [A]  time = 0.44, size = 29, normalized size = 1.12 \begin {gather*} 3 \, x e^{\left (-4\right )} + \frac {{\left (8 \, \log \relax (2) + 17\right )} \log \relax (x)}{4 \, {\left (2 \, e^{4} \log \relax (2) + 5 \, e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(12*x+4)*log(2)+60*x+17)/(8*x*exp(4)*log(2)+20*x*exp(4)),x, algorithm="maxima")

[Out]

3*x*e^(-4) + 1/4*(8*log(2) + 17)*log(x)/(2*e^4*log(2) + 5*e^4)

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mupad [B]  time = 1.12, size = 42, normalized size = 1.62 \begin {gather*} \frac {x\,\left (24\,\ln \relax (2)+60\right )}{20\,{\mathrm {e}}^4+8\,{\mathrm {e}}^4\,\ln \relax (2)}+\frac {\ln \relax (x)\,\left (\ln \left (256\right )+17\right )}{20\,{\mathrm {e}}^4+8\,{\mathrm {e}}^4\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x + 2*log(2)*(12*x + 4) + 17)/(20*x*exp(4) + 8*x*exp(4)*log(2)),x)

[Out]

(x*(24*log(2) + 60))/(20*exp(4) + 8*exp(4)*log(2)) + (log(x)*(log(256) + 17))/(20*exp(4) + 8*exp(4)*log(2))

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sympy [A]  time = 0.12, size = 31, normalized size = 1.19 \begin {gather*} \frac {x \left (24 \log {\relax (2 )} + 60\right ) + \left (8 \log {\relax (2 )} + 17\right ) \log {\relax (x )}}{8 e^{4} \log {\relax (2 )} + 20 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(12*x+4)*ln(2)+60*x+17)/(8*x*exp(4)*ln(2)+20*x*exp(4)),x)

[Out]

(x*(24*log(2) + 60) + (8*log(2) + 17)*log(x))/(8*exp(4)*log(2) + 20*exp(4))

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