3.98.42 \(\int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+(8+8 x+2 x^2+e^4 (-2-x) \log (4)) \log (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x})}{-8 x^2-8 x^3-2 x^4+e^4 (2 x^2+x^3) \log (4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4-2 x+\log \left (-5+\frac {5 e^4 \log (4)}{4+2 x}\right )}{x} \]

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Rubi [B]  time = 1.35, antiderivative size = 305, normalized size of antiderivative = 11.73, number of steps used = 14, number of rules used = 10, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6741, 6728, 1628, 632, 31, 1586, 21, 2490, 36, 29} \begin {gather*} \frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {\left (e^8 \log ^2(4) \log (16)+4 e^4 \left (22 \log ^2(4)-22 \log (4) \log (16)+\log (16) \log (262144)\right )+64 \log (4)-4 \log (65536)\right ) \log (x+2)}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {\left (e^8 \log (4) \left (8 \log ^2(4)+\log (4) \log (16)-\log (16) \log (1024)\right )+e^4 \left (88 \log ^2(4)-\log (4) (88 \log (16)-\log (65536))+4 \log (16) \log (262144)\right )+64 \log (4)-4 \log (65536)\right ) \log \left (2 x+4-e^4 \log (4)\right )}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {e^4 \log (4) \log (x)}{2 \left (4-e^4 \log (4)\right )}+\frac {e^4 \log (4) \log \left (2 x+4-e^4 \log (4)\right )}{2 \left (4-e^4 \log (4)\right )}+\frac {(x+2) \log \left (-\frac {5 \left (2 x+4-e^4 \log (4)\right )}{2 (x+2)}\right )}{2 x}+\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(32 + 32*x + 8*x^2 + E^4*(-8 - 5*x)*Log[4] + (8 + 8*x + 2*x^2 + E^4*(-2 - x)*Log[4])*Log[(-20 - 10*x + 5*E
^4*Log[4])/(4 + 2*x)])/(-8*x^2 - 8*x^3 - 2*x^4 + E^4*(2*x^2 + x^3)*Log[4]),x]

[Out]

(8*(4 - E^4*Log[4]))/(x*(8 - E^4*Log[16])) - (E^4*Log[4]*Log[x])/(2*(4 - E^4*Log[4])) + (E^4*(E^4*(8*Log[4]^2
- Log[16]*Log[1024]) + Log[65536])*Log[x])/(8 - E^4*Log[16])^2 + ((64*Log[4] + E^8*Log[4]^2*Log[16] - 4*Log[65
536] + 4*E^4*(22*Log[4]^2 - 22*Log[4]*Log[16] + Log[16]*Log[262144]))*Log[2 + x])/(Log[4]*(8 - E^4*Log[16])^2)
+ (E^4*Log[4]*Log[4 + 2*x - E^4*Log[4]])/(2*(4 - E^4*Log[4])) - ((64*Log[4] + E^8*Log[4]*(8*Log[4]^2 + Log[4]
*Log[16] - Log[16]*Log[1024]) - 4*Log[65536] + E^4*(88*Log[4]^2 - Log[4]*(88*Log[16] - Log[65536]) + 4*Log[16]
*Log[262144]))*Log[4 + 2*x - E^4*Log[4]])/(Log[4]*(8 - E^4*Log[16])^2) + ((2 + x)*Log[(-5*(4 + 2*x - E^4*Log[4
]))/(2*(2 + x))])/(2*x)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2490

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_))^
2, x_Symbol] :> Simp[((a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/((b*g - a*h)*(g + h*x)), x] - Dist[(p*
r*s*(b*c - a*d))/(b*g - a*h), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/((c + d*x)*(g + h*x)), x], x] /
; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0] && NeQ[b*g - a*h, 0] &&
IGtQ[s, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32-32 x-8 x^2-e^4 (-8-5 x) \log (4)-\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{x^2 \left (8+2 x^2-x \left (-8+e^4 \log (4)\right )-e^4 \log (16)\right )} \, dx\\ &=\int \left (\frac {-8 x^2-x \left (32-5 e^4 \log (4)\right )-8 \left (4-e^4 \log (4)\right )}{x^2 \left (8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)\right )}+\frac {(2+x) \left (-4-2 x+e^4 \log (4)\right ) \log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x^2 \left (8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)\right )}\right ) \, dx\\ &=\int \frac {-8 x^2-x \left (32-5 e^4 \log (4)\right )-8 \left (4-e^4 \log (4)\right )}{x^2 \left (8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)\right )} \, dx+\int \frac {(2+x) \left (-4-2 x+e^4 \log (4)\right ) \log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x^2 \left (8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)\right )} \, dx\\ &=\int \left (-\frac {8 \left (-4+e^4 \log (4)\right )}{x^2 \left (-8+e^4 \log (16)\right )}+\frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right )}{x \left (8-e^4 \log (16)\right )^2}+\frac {e^4 \left (-64 \log (4)-e^8 \log ^2(4) \log (16)-8 e^4 \left (15 \log ^2(4)-11 \log (4) \log (16)+\log ^2(16)\right )-2 x \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right )\right )}{\left (8-e^4 \log (16)\right )^2 \left (8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)\right )}\right ) \, dx+\int \frac {\left (-4-2 x+e^4 \log (4)\right ) \log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x^2 \left (4+2 x-e^4 \log (4)\right )} \, dx\\ &=\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )}+\frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {e^4 \int \frac {-64 \log (4)-e^8 \log ^2(4) \log (16)-8 e^4 \left (15 \log ^2(4)-11 \log (4) \log (16)+\log ^2(16)\right )-2 x \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right )}{8+2 x^2+x \left (8-e^4 \log (4)\right )-e^4 \log (16)} \, dx}{\left (8-e^4 \log (16)\right )^2}-\int \frac {\log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x^2} \, dx\\ &=\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )}+\frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {(2+x) \log \left (-\frac {5 \left (4+2 x-e^4 \log (4)\right )}{2 (2+x)}\right )}{2 x}+\frac {1}{2} \left (e^4 \log (4)\right ) \int \frac {1}{x \left (-4-2 x+e^4 \log (4)\right )} \, dx+\frac {\left (2 \left (64 \log (4)+e^8 \log ^2(4) \log (16)-4 \log (65536)+4 e^4 \left (22 \log ^2(4)-22 \log (4) \log (16)+\log (16) \log (262144)\right )\right )\right ) \int \frac {1}{2 x+\frac {1}{2} e^4 \log (4)+\frac {1}{2} \left (8-e^4 \log (4)\right )} \, dx}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {\left (2 \left (64 \log (4)+e^8 \log (4) \left (8 \log ^2(4)+\log (4) \log (16)-\log (16) \log (1024)\right )-4 \log (65536)+e^4 \left (88 \log ^2(4)-\log (4) (88 \log (16)-\log (65536))+4 \log (16) \log (262144)\right )\right )\right ) \int \frac {1}{2 x-\frac {1}{2} e^4 \log (4)+\frac {1}{2} \left (8-e^4 \log (4)\right )} \, dx}{\log (4) \left (8-e^4 \log (16)\right )^2}\\ &=\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )}+\frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {\left (64 \log (4)+e^8 \log ^2(4) \log (16)-4 \log (65536)+4 e^4 \left (22 \log ^2(4)-22 \log (4) \log (16)+\log (16) \log (262144)\right )\right ) \log (2+x)}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {\left (64 \log (4)+e^8 \log (4) \left (8 \log ^2(4)+\log (4) \log (16)-\log (16) \log (1024)\right )-4 \log (65536)+e^4 \left (88 \log ^2(4)-\log (4) (88 \log (16)-\log (65536))+4 \log (16) \log (262144)\right )\right ) \log \left (4+2 x-e^4 \log (4)\right )}{\log (4) \left (8-e^4 \log (16)\right )^2}+\frac {(2+x) \log \left (-\frac {5 \left (4+2 x-e^4 \log (4)\right )}{2 (2+x)}\right )}{2 x}-\frac {\left (e^4 \log (4)\right ) \int \frac {1}{-4-2 x+e^4 \log (4)} \, dx}{4-e^4 \log (4)}+\frac {\left (e^4 \log (4)\right ) \int \frac {1}{x} \, dx}{2 \left (-4+e^4 \log (4)\right )}\\ &=\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )}-\frac {e^4 \log (4) \log (x)}{2 \left (4-e^4 \log (4)\right )}+\frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {\left (64 \log (4)+e^8 \log ^2(4) \log (16)-4 \log (65536)+4 e^4 \left (22 \log ^2(4)-22 \log (4) \log (16)+\log (16) \log (262144)\right )\right ) \log (2+x)}{\log (4) \left (8-e^4 \log (16)\right )^2}+\frac {e^4 \log (4) \log \left (4+2 x-e^4 \log (4)\right )}{2 \left (4-e^4 \log (4)\right )}-\frac {\left (64 \log (4)+e^8 \log (4) \left (8 \log ^2(4)+\log (4) \log (16)-\log (16) \log (1024)\right )-4 \log (65536)+e^4 \left (88 \log ^2(4)-\log (4) (88 \log (16)-\log (65536))+4 \log (16) \log (262144)\right )\right ) \log \left (4+2 x-e^4 \log (4)\right )}{\log (4) \left (8-e^4 \log (16)\right )^2}+\frac {(2+x) \log \left (-\frac {5 \left (4+2 x-e^4 \log (4)\right )}{2 (2+x)}\right )}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 31, normalized size = 1.19 \begin {gather*} \frac {4}{x}+\frac {\log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 + 32*x + 8*x^2 + E^4*(-8 - 5*x)*Log[4] + (8 + 8*x + 2*x^2 + E^4*(-2 - x)*Log[4])*Log[(-20 - 10*x
+ 5*E^4*Log[4])/(4 + 2*x)])/(-8*x^2 - 8*x^3 - 2*x^4 + E^4*(2*x^2 + x^3)*Log[4]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 24, normalized size = 0.92 \begin {gather*} \frac {\log \left (\frac {5 \, {\left (e^{4} \log \relax (2) - x - 2\right )}}{x + 2}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x-2)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x-20)/(2*x+4))+2*(-5*x-8)*exp(4)*log(
2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4)*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="fricas")

[Out]

(log(5*(e^4*log(2) - x - 2)/(x + 2)) + 4)/x

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giac [B]  time = 0.46, size = 120, normalized size = 4.62 \begin {gather*} \frac {{\left (2 \, e^{8} \log \relax (2)^{2} + \frac {{\left (e^{4} \log \relax (2) - x - 2\right )} e^{4} \log \relax (2) \log \left (\frac {5 \, {\left (e^{4} \log \relax (2) - x - 2\right )}}{x + 2}\right )}{x + 2} + e^{4} \log \relax (2) \log \left (\frac {5 \, {\left (e^{4} \log \relax (2) - x - 2\right )}}{x + 2}\right )\right )} {\left (\frac {{\left (e^{4} \log \relax (2) - 2\right )} e^{\left (-8\right )}}{\log \relax (2)^{2}} + \frac {2 \, e^{\left (-8\right )}}{\log \relax (2)^{2}}\right )}}{e^{4} \log \relax (2) - \frac {2 \, {\left (e^{4} \log \relax (2) - x - 2\right )}}{x + 2} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x-2)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x-20)/(2*x+4))+2*(-5*x-8)*exp(4)*log(
2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4)*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 27, normalized size = 1.04

method	result	size
norman	\(\frac {4+\ln \left (\frac {10 \,{\mathrm e}^{4} \ln \relax (2)-10 x -20}{2 x +4}\right )}{x}\)	\(27\)
risch	\(\frac {\ln \left (\frac {10 \,{\mathrm e}^{4} \ln \relax (2)-10 x -20}{2 x +4}\right )}{x}+\frac {4}{x}\)	\(31\)
derivativedivides	\(-\frac {5 \,{\mathrm e}^{4} \ln \relax (2) \left (-\frac {\ln \left (5 \,{\mathrm e}^{4} \ln \relax (2)-\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \relax (2)-2\right )}-\frac {2 \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right ) \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \relax (2)-2\right ) \left (5 \,{\mathrm e}^{4} \ln \relax (2)-\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}+\frac {4}{-5 \,{\mathrm e}^{4} \ln \relax (2)+\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}}+\frac {\ln \left (-5 \,{\mathrm e}^{4} \ln \relax (2)+\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \,{\mathrm e}^{4} \ln \relax (2)-10}-\frac {2 \,{\mathrm e}^{-4} \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \ln \relax (2) \left ({\mathrm e}^{4} \ln \relax (2)-2\right )}\right )}{2}\)	\(181\)
default	\(-\frac {5 \,{\mathrm e}^{4} \ln \relax (2) \left (-\frac {\ln \left (5 \,{\mathrm e}^{4} \ln \relax (2)-\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \relax (2)-2\right )}-\frac {2 \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right ) \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \relax (2)-2\right ) \left (5 \,{\mathrm e}^{4} \ln \relax (2)-\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}+\frac {4}{-5 \,{\mathrm e}^{4} \ln \relax (2)+\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}}+\frac {\ln \left (-5 \,{\mathrm e}^{4} \ln \relax (2)+\frac {10 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \,{\mathrm e}^{4} \ln \relax (2)-10}-\frac {2 \,{\mathrm e}^{-4} \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \relax (2)}{2+x}\right )}{5 \ln \relax (2) \left ({\mathrm e}^{4} \ln \relax (2)-2\right )}\right )}{2}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(-x-2)*exp(4)*ln(2)+2*x^2+8*x+8)*ln((10*exp(4)*ln(2)-10*x-20)/(2*x+4))+2*(-5*x-8)*exp(4)*ln(2)+8*x^2+3
2*x+32)/(2*(x^3+2*x^2)*exp(4)*ln(2)-2*x^4-8*x^3-8*x^2),x,method=_RETURNVERBOSE)

[Out]

(4+ln((10*exp(4)*ln(2)-10*x-20)/(2*x+4)))/x

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maxima [C]  time = 0.58, size = 391, normalized size = 15.04 \begin {gather*} -2 \, {\left (\frac {e^{\left (-4\right )} \log \left (x + 2\right )}{\log \relax (2)} - \frac {{\left (e^{4} \log \relax (2) - 4\right )} \log \relax (x)}{e^{8} \log \relax (2)^{2} - 4 \, e^{4} \log \relax (2) + 4} - \frac {4 \, \log \left (-e^{4} \log \relax (2) + x + 2\right )}{e^{12} \log \relax (2)^{3} - 4 \, e^{8} \log \relax (2)^{2} + 4 \, e^{4} \log \relax (2)} - \frac {2}{{\left (e^{4} \log \relax (2) - 2\right )} x}\right )} e^{4} \log \relax (2) + \frac {5}{2} \, {\left (\frac {e^{\left (-4\right )} \log \left (x + 2\right )}{\log \relax (2)} + \frac {2 \, \log \left (-e^{4} \log \relax (2) + x + 2\right )}{e^{8} \log \relax (2)^{2} - 2 \, e^{4} \log \relax (2)} - \frac {\log \relax (x)}{e^{4} \log \relax (2) - 2}\right )} e^{4} \log \relax (2) + \frac {e^{4} \log \relax (2) \log \relax (x)}{2 \, {\left (e^{4} \log \relax (2) - 2\right )}} - \frac {4 \, e^{\left (-4\right )} \log \left (-e^{4} \log \relax (2) + x + 2\right )}{\log \relax (2)} - \frac {4 \, {\left (e^{4} \log \relax (2) - 4\right )} \log \relax (x)}{e^{8} \log \relax (2)^{2} - 4 \, e^{4} \log \relax (2) + 4} - \frac {16 \, \log \left (-e^{4} \log \relax (2) + x + 2\right )}{e^{12} \log \relax (2)^{3} - 4 \, e^{8} \log \relax (2)^{2} + 4 \, e^{4} \log \relax (2)} - \frac {16 \, \log \left (-e^{4} \log \relax (2) + x + 2\right )}{e^{8} \log \relax (2)^{2} - 2 \, e^{4} \log \relax (2)} + \frac {8 \, \log \relax (x)}{e^{4} \log \relax (2) - 2} - \frac {4 i \, \pi - 2 \, {\left (i \, \pi \log \relax (2) + \log \relax (5) \log \relax (2)\right )} e^{4} - 2 \, {\left (e^{4} \log \relax (2) - x - 2\right )} \log \left (-e^{4} \log \relax (2) + x + 2\right ) + {\left ({\left (e^{4} \log \relax (2) - 2\right )} x + 2 \, e^{4} \log \relax (2) - 4\right )} \log \left (x + 2\right ) + 4 \, \log \relax (5)}{2 \, {\left (e^{4} \log \relax (2) - 2\right )} x} - \frac {8}{{\left (e^{4} \log \relax (2) - 2\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x-2)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x-20)/(2*x+4))+2*(-5*x-8)*exp(4)*log(
2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4)*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="maxima")

[Out]

-2*(e^(-4)*log(x + 2)/log(2) - (e^4*log(2) - 4)*log(x)/(e^8*log(2)^2 - 4*e^4*log(2) + 4) - 4*log(-e^4*log(2) +
x + 2)/(e^12*log(2)^3 - 4*e^8*log(2)^2 + 4*e^4*log(2)) - 2/((e^4*log(2) - 2)*x))*e^4*log(2) + 5/2*(e^(-4)*log
(x + 2)/log(2) + 2*log(-e^4*log(2) + x + 2)/(e^8*log(2)^2 - 2*e^4*log(2)) - log(x)/(e^4*log(2) - 2))*e^4*log(2
) + 1/2*e^4*log(2)*log(x)/(e^4*log(2) - 2) - 4*e^(-4)*log(-e^4*log(2) + x + 2)/log(2) - 4*(e^4*log(2) - 4)*log
(x)/(e^8*log(2)^2 - 4*e^4*log(2) + 4) - 16*log(-e^4*log(2) + x + 2)/(e^12*log(2)^3 - 4*e^8*log(2)^2 + 4*e^4*lo
g(2)) - 16*log(-e^4*log(2) + x + 2)/(e^8*log(2)^2 - 2*e^4*log(2)) + 8*log(x)/(e^4*log(2) - 2) - 1/2*(4*I*pi -
2*(I*pi*log(2) + log(5)*log(2))*e^4 - 2*(e^4*log(2) - x - 2)*log(-e^4*log(2) + x + 2) + ((e^4*log(2) - 2)*x +
2*e^4*log(2) - 4)*log(x + 2) + 4*log(5))/((e^4*log(2) - 2)*x) - 8/((e^4*log(2) - 2)*x)

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mupad [B]  time = 6.92, size = 23, normalized size = 0.88 \begin {gather*} \frac {\ln \left (-\frac {5\,\left (x-{\mathrm {e}}^4\,\ln \relax (2)+2\right )}{x+2}\right )+4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x + log(-(10*x - 10*exp(4)*log(2) + 20)/(2*x + 4))*(8*x + 2*x^2 - 2*exp(4)*log(2)*(x + 2) + 8) + 8*x^
2 - 2*exp(4)*log(2)*(5*x + 8) + 32)/(8*x^2 + 8*x^3 + 2*x^4 - 2*exp(4)*log(2)*(2*x^2 + x^3)),x)

[Out]

(log(-(5*(x - exp(4)*log(2) + 2))/(x + 2)) + 4)/x

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sympy [A]  time = 0.28, size = 24, normalized size = 0.92 \begin {gather*} \frac {\log {\left (\frac {- 10 x - 20 + 10 e^{4} \log {\relax (2 )}}{2 x + 4} \right )}}{x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x-2)*exp(4)*ln(2)+2*x**2+8*x+8)*ln((10*exp(4)*ln(2)-10*x-20)/(2*x+4))+2*(-5*x-8)*exp(4)*ln(2)+
8*x**2+32*x+32)/(2*(x**3+2*x**2)*exp(4)*ln(2)-2*x**4-8*x**3-8*x**2),x)

[Out]

log((-10*x - 20 + 10*exp(4)*log(2))/(2*x + 4))/x + 4/x

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