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

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

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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} $$\begin{array}{c}\frac{3x}{e^4}+\frac{(17+\log (256))\log (x)}{4e^4(5+\log (4))}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{17+60x+(4+12x)\log (4)}{e^{4}x(20+4\log (4))}dx\\ & =\frac{\int \frac{17+60x+(4+12x)\log (4)}{x}dx}{4e^4(5+\log (4))}\\ & =\frac{\int \frac{17+12x(5+\log (4))+\log (256)}{x}dx}{4e^4(5+\log (4))}\\ & =\frac{\int (12(5+\log (4))+\frac{17+\log (256)}{x})dx}{4e^4(5+\log (4))}\\ & =\frac{3x}{e^4}+\frac{(17+\log (256))\log (x)}{4e^4(5+\log (4))}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{24x\log (2)+(8\log (2)+17)\log (x)+60x}{4(2e^4\log (2)+5e^4)}\end{array}$$

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 $$\begin{array}{c}\frac{(8\log (2)+17)\log \left(\left|x\right|\right)}{4(2e^4\log (2)+5e^4)}+\frac{3(2x\log (2)+5x)}{2e^4\log (2)+5e^4}\end{array}$$

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	$3x{\mathrm{e}}^{-4}+\frac{(8\ln (2)+17){\mathrm{e}}^{-4}\ln (x)}{8\ln (2)+20}$	$31$
default	$\frac{{\mathrm{e}}^{-4}(24x\ln (2)+60x+(8\ln (2)+17)\ln (x))}{8\ln (2)+20}$	$33$
risch	$3x{\mathrm{e}}^{-4}+\frac{2{\mathrm{e}}^{-4}\ln (x)\ln (2)}{2\ln (2)+5}+\frac{17{\mathrm{e}}^{-4}\ln (x)}{4(2\ln (2)+5)}$	$37$

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 $$\begin{array}{c}3x{e}^{(-4)}+\frac{(8\log (2)+17)\log (x)}{4(2e^4\log (2)+5e^4)}\end{array}$$

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 $$\begin{array}{c}\frac{x(24\ln (2)+60)}{20{\mathrm{e}}^4+8{\mathrm{e}}^4\ln (2)}+\frac{\ln (x)(\ln \left(256\right)+17)}{20{\mathrm{e}}^4+8{\mathrm{e}}^4\ln (2)}\end{array}$$

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 $$\begin{array}{c}\frac{x(24\log (2)+60)+(8\log (2)+17)\log (x)}{8e^4\log (2)+20e^4}\end{array}$$

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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