3.98.53 \(\int \frac {x+(16+x) \log (-16-x)}{(16+x) \log ^2(\frac {1}{25} e^{\frac {8}{\log (3)}})} \, dx\)

Optimal. Leaf size=23 \[ \frac {x \log (-16-x)}{\log ^2\left (\frac {1}{25} e^{\frac {8}{\log (3)}}\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.96, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6742, 43, 2389, 2295} \begin {gather*} \frac {(x+16) \log ^2(3) \log (-x-16)}{(8-\log (3) \log (25))^2}-\frac {16 \log ^2(3) \log (x+16)}{(8-\log (3) \log (25))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + (16 + x)*Log[-16 - x])/((16 + x)*Log[E^(8/Log[3])/25]^2),x]

[Out]

((16 + x)*Log[3]^2*Log[-16 - x])/(8 - Log[3]*Log[25])^2 - (16*Log[3]^2*Log[16 + x])/(8 - Log[3]*Log[25])^2

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log ^2(3) \int \frac {x+(16+x) \log (-16-x)}{16+x} \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \left (\frac {x}{16+x}+\log (-16-x)\right ) \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \frac {x}{16+x} \, dx}{(8-\log (3) \log (25))^2}+\frac {\log ^2(3) \int \log (-16-x) \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \left (1-\frac {16}{16+x}\right ) \, dx}{(8-\log (3) \log (25))^2}-\frac {\log ^2(3) \operatorname {Subst}(\int \log (x) \, dx,x,-16-x)}{(8-\log (3) \log (25))^2}\\ &=\frac {(16+x) \log ^2(3) \log (-16-x)}{(8-\log (3) \log (25))^2}-\frac {16 \log ^2(3) \log (16+x)}{(8-\log (3) \log (25))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.26 \begin {gather*} \frac {(16+x) \log (-16-x)-16 \log (16+x)}{\left (-\frac {8}{\log (3)}+\log (25)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (16 + x)*Log[-16 - x])/((16 + x)*Log[E^(8/Log[3])/25]^2),x]

[Out]

((16 + x)*Log[-16 - x] - 16*Log[16 + x])/(-8/Log[3] + Log[25])^2

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fricas [A]  time = 0.58, size = 32, normalized size = 1.39 \begin {gather*} \frac {x \log \relax (3)^{2} \log \left (-x - 16\right )}{4 \, {\left (\log \relax (5)^{2} \log \relax (3)^{2} - 8 \, \log \relax (5) \log \relax (3) + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*log(-x-16)+x)/(x+16)/log(1/25*exp(4/log(3))^2)^2,x, algorithm="fricas")

[Out]

1/4*x*log(3)^2*log(-x - 16)/(log(5)^2*log(3)^2 - 8*log(5)*log(3) + 16)

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giac [A]  time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} \frac {x \log \left (-x - 16\right )}{\log \left (\frac {1}{25} \, e^{\frac {8}{\log \relax (3)}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*log(-x-16)+x)/(x+16)/log(1/25*exp(4/log(3))^2)^2,x, algorithm="giac")

[Out]

x*log(-x - 16)/log(1/25*e^(8/log(3)))^2

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maple [A]  time = 0.09, size = 22, normalized size = 0.96

method	result	size
risch	\(\frac {\ln \left (-x -16\right ) x}{\left (-2 \ln \relax (5)+\frac {8}{\ln \relax (3)}\right )^{2}}\)	\(22\)
norman	\(\frac {\ln \relax (3)^{2} x \ln \left (-x -16\right )}{4 \left (\ln \relax (3) \ln \relax (5)-4\right )^{2}}\)	\(23\)
derivativedivides	\(\frac {-\left (-x -16\right ) \ln \left (-x -16\right )-16 \ln \left (-x -16\right )}{\ln \left (\frac {{\mathrm e}^{\frac {8}{\ln \relax (3)}}}{25}\right )^{2}}\)	\(38\)
default	\(\frac {-\left (-x -16\right ) \ln \left (-x -16\right )-16 \ln \left (-x -16\right )}{\ln \left (\frac {{\mathrm e}^{\frac {8}{\ln \relax (3)}}}{25}\right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+16)*ln(-x-16)+x)/(x+16)/ln(1/25*exp(4/ln(3))^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/(-2*ln(5)+8/ln(3))^2*ln(-x-16)*x

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maxima [B]  time = 0.35, size = 47, normalized size = 2.04 \begin {gather*} \frac {8 \, \log \left (x + 16\right )^{2} + {\left (x - 16 \, \log \left (x + 16\right )\right )} \log \left (-x - 16\right ) + 8 \, \log \left (-x - 16\right )^{2}}{\log \left (\frac {1}{25} \, e^{\frac {8}{\log \relax (3)}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*log(-x-16)+x)/(x+16)/log(1/25*exp(4/log(3))^2)^2,x, algorithm="maxima")

[Out]

(8*log(x + 16)^2 + (x - 16*log(x + 16))*log(-x - 16) + 8*log(-x - 16)^2)/log(1/25*e^(8/log(3)))^2

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mupad [B]  time = 0.21, size = 32, normalized size = 1.39 \begin {gather*} \frac {x\,{\ln \relax (3)}^2\,\ln \left (-x-16\right )}{4\,{\ln \relax (3)}^2\,{\ln \relax (5)}^2-32\,\ln \relax (3)\,\ln \relax (5)+64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(- x - 16)*(x + 16))/(log(exp(8/log(3))/25)^2*(x + 16)),x)

[Out]

(x*log(3)^2*log(- x - 16))/(4*log(3)^2*log(5)^2 - 32*log(3)*log(5) + 64)

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sympy [A]  time = 0.20, size = 34, normalized size = 1.48 \begin {gather*} \frac {x \log {\relax (3 )}^{2} \log {\left (- x - 16 \right )}}{- 32 \log {\relax (3 )} \log {\relax (5 )} + 4 \log {\relax (3 )}^{2} \log {\relax (5 )}^{2} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*ln(-x-16)+x)/(x+16)/ln(1/25*exp(4/ln(3))**2)**2,x)

[Out]

x*log(3)**2*log(-x - 16)/(-32*log(3)*log(5) + 4*log(3)**2*log(5)**2 + 64)

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