3.76.6 \(\int \frac {-9 x+36 x \log (\frac {1}{4} (16+e+4 x))+(32+2 e+8 x) \log ^3(\frac {1}{4} (16+e+4 x))}{(32 x+2 e x+8 x^2) \log ^3(\frac {1}{4} (16+e+4 x))} \, dx\)

Optimal. Leaf size=25 \[ \log (x)+\left (3-\frac {3}{4 \log \left (4+\frac {1}{4} (e+4 x)\right )}\right )^2 \]

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Rubi [A]  time = 0.47, antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 12, number of rules used = 7, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.101, Rules used = {6, 1593, 6688, 2390, 12, 2302, 30} \begin {gather*} \frac {9}{16 \log ^2\left (x+\frac {16+e}{4}\right )}+\log (x)-\frac {9}{2 \log \left (x+\frac {16+e}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*x + 36*x*Log[(16 + E + 4*x)/4] + (32 + 2*E + 8*x)*Log[(16 + E + 4*x)/4]^3)/((32*x + 2*E*x + 8*x^2)*Log
[(16 + E + 4*x)/4]^3),x]

[Out]

Log[x] + 9/(16*Log[(16 + E)/4 + x]^2) - 9/(2*Log[(16 + E)/4 + x])

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

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

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{\left ((32+2 e) x+8 x^2\right ) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx\\ &=\int \frac {-9 x+36 x \log \left (\frac {1}{4} (16+e+4 x)\right )+(32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )}{x (32+2 e+8 x) \log ^3\left (\frac {1}{4} (16+e+4 x)\right )} \, dx\\ &=\int \left (\frac {1}{x}-\frac {9}{2 (16+e+4 x) \log ^3\left (4+\frac {e}{4}+x\right )}+\frac {18}{(16+e+4 x) \log ^2\left (4+\frac {e}{4}+x\right )}\right ) \, dx\\ &=\log (x)-\frac {9}{2} \int \frac {1}{(16+e+4 x) \log ^3\left (4+\frac {e}{4}+x\right )} \, dx+18 \int \frac {1}{(16+e+4 x) \log ^2\left (4+\frac {e}{4}+x\right )} \, dx\\ &=\log (x)-\frac {9}{2} \operatorname {Subst}\left (\int \frac {4+\frac {e}{4}}{(16+e) x \log ^3(x)} \, dx,x,4+\frac {e}{4}+x\right )+18 \operatorname {Subst}\left (\int \frac {4+\frac {e}{4}}{(16+e) x \log ^2(x)} \, dx,x,4+\frac {e}{4}+x\right )\\ &=\log (x)-\frac {9}{8} \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,4+\frac {e}{4}+x\right )+\frac {9}{2} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,4+\frac {e}{4}+x\right )\\ &=\log (x)-\frac {9}{8} \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (4+\frac {e}{4}+x\right )\right )+\frac {9}{2} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (4+\frac {e}{4}+x\right )\right )\\ &=\log (x)+\frac {9}{16 \log ^2\left (\frac {16+e}{4}+x\right )}-\frac {9}{2 \log \left (\frac {16+e}{4}+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 33, normalized size = 1.32 \begin {gather*} \log (x)+\frac {9}{16 \log ^2\left (4+\frac {e}{4}+x\right )}-\frac {9}{2 \log \left (4+\frac {e}{4}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*x + 36*x*Log[(16 + E + 4*x)/4] + (32 + 2*E + 8*x)*Log[(16 + E + 4*x)/4]^3)/((32*x + 2*E*x + 8*x^
2)*Log[(16 + E + 4*x)/4]^3),x]

[Out]

Log[x] + 9/(16*Log[4 + E/4 + x]^2) - 9/(2*Log[4 + E/4 + x])

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fricas [A]  time = 0.59, size = 38, normalized size = 1.52 \begin {gather*} \frac {16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2} \log \relax (x) - 72 \, \log \left (x + \frac {1}{4} \, e + 4\right ) + 9}{16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4)-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1
/4*exp(1)+x+4)^3,x, algorithm="fricas")

[Out]

1/16*(16*log(x + 1/4*e + 4)^2*log(x) - 72*log(x + 1/4*e + 4) + 9)/log(x + 1/4*e + 4)^2

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giac [A]  time = 0.23, size = 38, normalized size = 1.52 \begin {gather*} \frac {16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2} \log \relax (x) - 72 \, \log \left (x + \frac {1}{4} \, e + 4\right ) + 9}{16 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4)-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1
/4*exp(1)+x+4)^3,x, algorithm="giac")

[Out]

1/16*(16*log(x + 1/4*e + 4)^2*log(x) - 72*log(x + 1/4*e + 4) + 9)/log(x + 1/4*e + 4)^2

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maple [A]  time = 0.18, size = 27, normalized size = 1.08

method	result	size
norman	\(\frac {\frac {9}{16}-\frac {9 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}{2}}{\ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}+\ln \relax (x )\)	\(27\)
risch	\(\ln \relax (x )-\frac {9 \left (-1+8 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )\right )}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}\)	\(28\)
derivativedivides	\(\ln \left (-4 x \right )+\frac {9}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}-\frac {9}{2 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}\)	\(30\)
default	\(\ln \left (-4 x \right )+\frac {9}{16 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )^{2}}-\frac {9}{2 \ln \left (\frac {{\mathrm e}}{4}+x +4\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(1)+8*x+32)*ln(1/4*exp(1)+x+4)^3+36*x*ln(1/4*exp(1)+x+4)-9*x)/(2*x*exp(1)+8*x^2+32*x)/ln(1/4*exp(1)
+x+4)^3,x,method=_RETURNVERBOSE)

[Out]

(9/16-9/2*ln(1/4*exp(1)+x+4))/ln(1/4*exp(1)+x+4)^2+ln(x)

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maxima [B]  time = 0.56, size = 287, normalized size = 11.48 \begin {gather*} -{\left (\frac {\log \left (4 \, x + e + 16\right )}{e + 16} - \frac {\log \relax (x)}{e + 16}\right )} e - \frac {\log \left (x + \frac {1}{4} \, e + 4\right )^{3}}{2 \, {\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + 3 \, {\left (2 \, \log \relax (2) - \log \left (4 \, x + e + 16\right )\right )} \log \left (-2 \, \log \relax (2) + \log \left (4 \, x + e + 16\right )\right ) + 3 \, \log \left (x + \frac {1}{4} \, e + 4\right ) \log \left (-2 \, \log \relax (2) + \log \left (4 \, x + e + 16\right )\right ) + \frac {3 \, \log \left (x + \frac {1}{4} \, e + 4\right )^{2}}{2 \, {\left (2 \, \log \relax (2) - \log \left (4 \, x + e + 16\right )\right )}} - \frac {16 \, \log \left (4 \, x + e + 16\right )}{e + 16} - \frac {9 \, \log \left (x + \frac {1}{4} \, e + 4\right )}{4 \, {\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + \frac {16 \, \log \relax (x)}{e + 16} + \frac {9}{16 \, {\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (4 \, x + e + 16\right ) + \log \left (4 \, x + e + 16\right )^{2}\right )}} + \frac {9}{4 \, {\left (2 \, \log \relax (2) - \log \left (4 \, x + e + 16\right )\right )}} + 3 \, \log \left (4 \, x + e + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(1)+8*x+32)*log(1/4*exp(1)+x+4)^3+36*x*log(1/4*exp(1)+x+4)-9*x)/(2*x*exp(1)+8*x^2+32*x)/log(1
/4*exp(1)+x+4)^3,x, algorithm="maxima")

[Out]

-(log(4*x + e + 16)/(e + 16) - log(x)/(e + 16))*e - 1/2*log(x + 1/4*e + 4)^3/(4*log(2)^2 - 4*log(2)*log(4*x +
e + 16) + log(4*x + e + 16)^2) + 3*(2*log(2) - log(4*x + e + 16))*log(-2*log(2) + log(4*x + e + 16)) + 3*log(x
+ 1/4*e + 4)*log(-2*log(2) + log(4*x + e + 16)) + 3/2*log(x + 1/4*e + 4)^2/(2*log(2) - log(4*x + e + 16)) - 1
6*log(4*x + e + 16)/(e + 16) - 9/4*log(x + 1/4*e + 4)/(4*log(2)^2 - 4*log(2)*log(4*x + e + 16) + log(4*x + e +
16)^2) + 16*log(x)/(e + 16) + 9/16/(4*log(2)^2 - 4*log(2)*log(4*x + e + 16) + log(4*x + e + 16)^2) + 9/4/(2*l
og(2) - log(4*x + e + 16)) + 3*log(4*x + e + 16)

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mupad [B]  time = 5.00, size = 38, normalized size = 1.52 \begin {gather*} \frac {16\,\ln \relax (x)\,{\ln \left (x+\frac {\mathrm {e}}{4}+4\right )}^2-72\,\ln \left (x+\frac {\mathrm {e}}{4}+4\right )+9}{16\,{\ln \left (x+\frac {\mathrm {e}}{4}+4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + exp(1)/4 + 4)^3*(8*x + 2*exp(1) + 32) - 9*x + 36*x*log(x + exp(1)/4 + 4))/(log(x + exp(1)/4 + 4)^
3*(32*x + 2*x*exp(1) + 8*x^2)),x)

[Out]

(16*log(x + exp(1)/4 + 4)^2*log(x) - 72*log(x + exp(1)/4 + 4) + 9)/(16*log(x + exp(1)/4 + 4)^2)

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sympy [A]  time = 0.15, size = 29, normalized size = 1.16 \begin {gather*} \frac {9 - 72 \log {\left (x + \frac {e}{4} + 4 \right )}}{16 \log {\left (x + \frac {e}{4} + 4 \right )}^{2}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(1)+8*x+32)*ln(1/4*exp(1)+x+4)**3+36*x*ln(1/4*exp(1)+x+4)-9*x)/(2*x*exp(1)+8*x**2+32*x)/ln(1/
4*exp(1)+x+4)**3,x)

[Out]

(9 - 72*log(x + E/4 + 4))/(16*log(x + E/4 + 4)**2) + log(x)

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