3.57.1 \(\int \frac {(16+20 x) \log (\frac {1}{8} (8 x+5 x^2))}{8 x+5 x^2} \, dx\)

Optimal. Leaf size=12 \[ \log ^2\left (x+\frac {5 x^2}{8}\right ) \]

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Rubi [B]  time = 0.32, antiderivative size = 75, normalized size of antiderivative = 6.25, number of steps used = 18, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {1593, 2528, 2524, 2357, 2301, 2317, 2391, 2418, 2392, 2390} \begin {gather*} 2 \log \left (\frac {1}{8} \left (5 x^2+8 x\right )\right ) \log (x)+2 \log (5 x+8) \log \left (\frac {1}{8} \left (5 x^2+8 x\right )\right )-\log ^2(x)-\log ^2(5 x+8)-2 \log \left (\frac {5 x}{8}+1\right ) \log (x)-2 \log (8) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((16 + 20*x)*Log[(8*x + 5*x^2)/8])/(8*x + 5*x^2),x]

[Out]

-2*Log[8]*Log[x] - 2*Log[1 + (5*x)/8]*Log[x] - Log[x]^2 - Log[8 + 5*x]^2 + 2*Log[x]*Log[(8*x + 5*x^2)/8] + 2*L
og[8 + 5*x]*Log[(8*x + 5*x^2)/8]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(16+20 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )}{x (8+5 x)} \, dx\\ &=\int \left (\frac {2 \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )}{x}+\frac {10 \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )}{8+5 x}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )}{x} \, dx+10 \int \frac {\log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )}{8+5 x} \, dx\\ &=2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )-2 \int \frac {(8+10 x) \log (x)}{8 x+5 x^2} \, dx-2 \int \frac {(8+10 x) \log (8+5 x)}{8 x+5 x^2} \, dx\\ &=2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )-2 \int \frac {(8+10 x) \log (x)}{x (8+5 x)} \, dx-2 \int \frac {(8+10 x) \log (8+5 x)}{x (8+5 x)} \, dx\\ &=2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )-2 \int \left (\frac {\log (x)}{x}+\frac {5 \log (x)}{8+5 x}\right ) \, dx-2 \int \left (\frac {\log (8+5 x)}{x}+\frac {5 \log (8+5 x)}{8+5 x}\right ) \, dx\\ &=2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )-2 \int \frac {\log (x)}{x} \, dx-2 \int \frac {\log (8+5 x)}{x} \, dx-10 \int \frac {\log (x)}{8+5 x} \, dx-10 \int \frac {\log (8+5 x)}{8+5 x} \, dx\\ &=-2 \log (8) \log (x)-2 \log \left (1+\frac {5 x}{8}\right ) \log (x)-\log ^2(x)+2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )-2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,8+5 x\right )\\ &=-2 \log (8) \log (x)-2 \log \left (1+\frac {5 x}{8}\right ) \log (x)-\log ^2(x)-\log ^2(8+5 x)+2 \log (x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )+2 \log (8+5 x) \log \left (\frac {1}{8} \left (8 x+5 x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 83, normalized size = 6.92 \begin {gather*} 4 \left (-\frac {1}{2} \log (8) \log (x)-\frac {\log ^2(x)}{4}-\frac {1}{2} \log (x) \log \left (\frac {1}{8} (8+5 x)\right )-\frac {1}{4} \log ^2(8+5 x)+\frac {1}{2} \log (x) \log \left (x+\frac {5 x^2}{8}\right )+\frac {1}{2} \log (8+5 x) \log \left (x+\frac {5 x^2}{8}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16 + 20*x)*Log[(8*x + 5*x^2)/8])/(8*x + 5*x^2),x]

[Out]

4*(-1/2*(Log[8]*Log[x]) - Log[x]^2/4 - (Log[x]*Log[(8 + 5*x)/8])/2 - Log[8 + 5*x]^2/4 + (Log[x]*Log[x + (5*x^2
)/8])/2 + (Log[8 + 5*x]*Log[x + (5*x^2)/8])/2)

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fricas [A]  time = 0.67, size = 10, normalized size = 0.83 \begin {gather*} \log \left (\frac {5}{8} \, x^{2} + x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(5/8*x^2+x)/(5*x^2+8*x),x, algorithm="fricas")

[Out]

log(5/8*x^2 + x)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (5 \, x + 4\right )} \log \left (\frac {5}{8} \, x^{2} + x\right )}{5 \, x^{2} + 8 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(5/8*x^2+x)/(5*x^2+8*x),x, algorithm="giac")

[Out]

integrate(4*(5*x + 4)*log(5/8*x^2 + x)/(5*x^2 + 8*x), x)

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maple [A]  time = 0.24, size = 11, normalized size = 0.92

method	result	size
norman	\(\ln \left (\frac {5}{8} x^{2}+x \right )^{2}\)	\(11\)
risch	\(\ln \left (\frac {5}{8} x^{2}+x \right )^{2}\)	\(11\)
default	\(2 \ln \left (5 x +8\right ) \ln \left (5 x^{2}+8 x \right )-2 \left (\ln \left (5 x +8\right )-\ln \left (\frac {5 x}{8}+1\right )\right ) \ln \left (-\frac {5 x}{8}\right )-\ln \left (5 x +8\right )^{2}+2 \ln \relax (x ) \ln \left (5 x^{2}+8 x \right )-2 \ln \relax (x ) \ln \left (\frac {5 x}{8}+1\right )-\ln \relax (x )^{2}-6 \ln \relax (2) \ln \left (x \left (5 x +8\right )\right )\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x+16)*ln(5/8*x^2+x)/(5*x^2+8*x),x,method=_RETURNVERBOSE)

[Out]

ln(5/8*x^2+x)^2

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maxima [B]  time = 0.41, size = 47, normalized size = 3.92 \begin {gather*} 2 \, \log \left (5 \, x^{2} + 8 \, x\right ) \log \left (\frac {5}{8} \, x^{2} + x\right ) - \log \left (5 \, x + 8\right )^{2} - 2 \, \log \left (5 \, x + 8\right ) \log \relax (x) - \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*log(5/8*x^2+x)/(5*x^2+8*x),x, algorithm="maxima")

[Out]

2*log(5*x^2 + 8*x)*log(5/8*x^2 + x) - log(5*x + 8)^2 - 2*log(5*x + 8)*log(x) - log(x)^2

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mupad [B]  time = 3.80, size = 10, normalized size = 0.83 \begin {gather*} {\ln \left (\frac {5\,x^2}{8}+x\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + (5*x^2)/8)*(20*x + 16))/(8*x + 5*x^2),x)

[Out]

log(x + (5*x^2)/8)^2

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sympy [A]  time = 0.12, size = 10, normalized size = 0.83 \begin {gather*} \log {\left (\frac {5 x^{2}}{8} + x \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x+16)*ln(5/8*x**2+x)/(5*x**2+8*x),x)

[Out]

log(5*x**2/8 + x)**2

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