3.74.27 \(\int \frac {-10-10 x+6 x^2+2 x^3+e^4 (2+2 x)+e^2 (-8 x-4 x^2)+(-5+e^4-2 e^2 x+x^2) \log (\frac {1}{16} (25 x^2+e^8 x^2-4 e^6 x^3-10 x^4+x^6+e^4 (-10 x^2+6 x^4)+e^2 (20 x^3-4 x^5)))}{-10+2 e^4-4 e^2 x+2 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{2} x \left (x+\log \left (\frac {1}{16} \left (-5+\left (e^2-x\right )^2\right )^2 x^2\right )\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 14, number of rules used = 6, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 1657, 632, 31, 2523, 12} \begin {gather*} \frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (-x^2+2 e^2 x-e^4+5\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 10*x + 6*x^2 + 2*x^3 + E^4*(2 + 2*x) + E^2*(-8*x - 4*x^2) + (-5 + E^4 - 2*E^2*x + x^2)*Log[(25*x^2
+ E^8*x^2 - 4*E^6*x^3 - 10*x^4 + x^6 + E^4*(-10*x^2 + 6*x^4) + E^2*(20*x^3 - 4*x^5))/16])/(-10 + 2*E^4 - 4*E^2
*x + 2*x^2),x]

[Out]

x^2/2 + (x*Log[(x^2*(5 - E^4 + 2*E^2*x - x^2)^2)/16])/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 31

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

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 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 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 \left (\frac {-5+e^4-\left (5+4 e^2-e^4\right ) x+\left (3-2 e^2\right ) x^2+x^3}{-5+e^4-2 e^2 x+x^2}+\frac {1}{2} \log \left (\frac {1}{16} x^2 \left (-5+e^4-2 e^2 x+x^2\right )^2\right )\right ) \, dx\\ &=\frac {1}{2} \int \log \left (\frac {1}{16} x^2 \left (-5+e^4-2 e^2 x+x^2\right )^2\right ) \, dx+\int \frac {-5+e^4-\left (5+4 e^2-e^4\right ) x+\left (3-2 e^2\right ) x^2+x^3}{-5+e^4-2 e^2 x+x^2} \, dx\\ &=\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )-\frac {1}{2} \int \frac {2 \left (5-e^4+4 e^2 x-3 x^2\right )}{5-e^4+2 e^2 x-x^2} \, dx+\int \left (3+x-\frac {2 \left (-5+e^4-e^2 x\right )}{-5+e^4-2 e^2 x+x^2}\right ) \, dx\\ &=3 x+\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )-2 \int \frac {-5+e^4-e^2 x}{-5+e^4-2 e^2 x+x^2} \, dx-\int \frac {5-e^4+4 e^2 x-3 x^2}{5-e^4+2 e^2 x-x^2} \, dx\\ &=3 x+\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )-\left (-\sqrt {5}-e^2\right ) \int \frac {1}{-\sqrt {5}-e^2+x} \, dx-\left (\sqrt {5}-e^2\right ) \int \frac {1}{\sqrt {5}-e^2+x} \, dx-\int \left (3+\frac {2 \left (-5+e^4-e^2 x\right )}{5-e^4+2 e^2 x-x^2}\right ) \, dx\\ &=\frac {x^2}{2}+\left (\sqrt {5}+e^2\right ) \log \left (\sqrt {5}+e^2-x\right )-\left (\sqrt {5}-e^2\right ) \log \left (\sqrt {5}-e^2+x\right )+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )-2 \int \frac {-5+e^4-e^2 x}{5-e^4+2 e^2 x-x^2} \, dx\\ &=\frac {x^2}{2}+\left (\sqrt {5}+e^2\right ) \log \left (\sqrt {5}+e^2-x\right )-\left (\sqrt {5}-e^2\right ) \log \left (\sqrt {5}-e^2+x\right )+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )-\left (-\sqrt {5}-e^2\right ) \int \frac {1}{\sqrt {5}+e^2-x} \, dx-\left (\sqrt {5}-e^2\right ) \int \frac {1}{-\sqrt {5}+e^2-x} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {1}{16} x^2 \left (5-e^4+2 e^2 x-x^2\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 31, normalized size = 1.11 \begin {gather*} \frac {1}{2} x \left (x+\log \left (\frac {1}{16} x^2 \left (-5+e^4-2 e^2 x+x^2\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 10*x + 6*x^2 + 2*x^3 + E^4*(2 + 2*x) + E^2*(-8*x - 4*x^2) + (-5 + E^4 - 2*E^2*x + x^2)*Log[(2
5*x^2 + E^8*x^2 - 4*E^6*x^3 - 10*x^4 + x^6 + E^4*(-10*x^2 + 6*x^4) + E^2*(20*x^3 - 4*x^5))/16])/(-10 + 2*E^4 -
4*E^2*x + 2*x^2),x]

[Out]

(x*(x + Log[(x^2*(-5 + E^4 - 2*E^2*x + x^2)^2)/16]))/2

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fricas [B]  time = 1.28, size = 68, normalized size = 2.43 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {1}{2} \, x \log \left (\frac {1}{16} \, x^{6} - \frac {5}{8} \, x^{4} - \frac {1}{4} \, x^{3} e^{6} + \frac {1}{16} \, x^{2} e^{8} + \frac {25}{16} \, x^{2} + \frac {1}{8} \, {\left (3 \, x^{4} - 5 \, x^{2}\right )} e^{4} - \frac {1}{4} \, {\left (x^{5} - 5 \, x^{3}\right )} e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/1
6*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/16*x^2)+(2*x+2)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)
/(2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="fricas")

[Out]

1/2*x^2 + 1/2*x*log(1/16*x^6 - 5/8*x^4 - 1/4*x^3*e^6 + 1/16*x^2*e^8 + 25/16*x^2 + 1/8*(3*x^4 - 5*x^2)*e^4 - 1/
4*(x^5 - 5*x^3)*e^2)

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giac [B]  time = 0.61, size = 68, normalized size = 2.43 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {1}{2} \, x \log \left (\frac {1}{16} \, x^{6} - \frac {1}{4} \, x^{5} e^{2} + \frac {3}{8} \, x^{4} e^{4} - \frac {5}{8} \, x^{4} - \frac {1}{4} \, x^{3} e^{6} + \frac {5}{4} \, x^{3} e^{2} + \frac {1}{16} \, x^{2} e^{8} - \frac {5}{8} \, x^{2} e^{4} + \frac {25}{16} \, x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/1
6*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/16*x^2)+(2*x+2)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)
/(2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="giac")

[Out]

1/2*x^2 + 1/2*x*log(1/16*x^6 - 1/4*x^5*e^2 + 3/8*x^4*e^4 - 5/8*x^4 - 1/4*x^3*e^6 + 5/4*x^3*e^2 + 1/16*x^2*e^8
- 5/8*x^2*e^4 + 25/16*x^2)

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maple [B]  time = 0.27, size = 71, normalized size = 2.54

method	result	size
risch	\(\frac {x^{2}}{2}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{8}}{16}-\frac {x^{3} {\mathrm e}^{6}}{4}+\frac {\left (6 x^{4}-10 x^{2}\right ) {\mathrm e}^{4}}{16}+\frac {\left (-4 x^{5}+20 x^{3}\right ) {\mathrm e}^{2}}{16}+\frac {x^{6}}{16}-\frac {5 x^{4}}{8}+\frac {25 x^{2}}{16}\right )}{2}\)	\(71\)
norman	\(\frac {x^{2}}{2}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{8}}{16}-\frac {x^{3} {\mathrm e}^{6}}{4}+\frac {\left (6 x^{4}-10 x^{2}\right ) {\mathrm e}^{4}}{16}+\frac {\left (-4 x^{5}+20 x^{3}\right ) {\mathrm e}^{2}}{16}+\frac {x^{6}}{16}-\frac {5 x^{4}}{8}+\frac {25 x^{2}}{16}\right )}{2}\)	\(77\)
default	\(\frac {x^{2}}{2}-2 x \ln \relax (2)+\frac {x \ln \left (x^{2} {\mathrm e}^{8}-4 x^{3} {\mathrm e}^{6}+6 x^{4} {\mathrm e}^{4}-10 x^{2} {\mathrm e}^{4}-4 \,{\mathrm e}^{2} x^{5}+20 x^{3} {\mathrm e}^{2}+x^{6}-10 x^{4}+25 x^{2}\right )}{2}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)^2-2*exp(2)*x+x^2-5)*ln(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/16*(-4*x
^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/16*x^2)+(2*x+2)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)/(2*exp
(2)^2-4*exp(2)*x+2*x^2-10),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2+1/2*x*ln(1/16*x^2*exp(8)-1/4*x^3*exp(6)+1/16*(6*x^4-10*x^2)*exp(4)+1/16*(-4*x^5+20*x^3)*exp(2)+1/16*x^
6-5/8*x^4+25/16*x^2)

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maxima [B]  time = 0.55, size = 417, normalized size = 14.89 \begin {gather*} \frac {1}{10} \, \sqrt {5} {\left (e^{6} + 15 \, e^{2}\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {3}{10} \, \sqrt {5} {\left (e^{4} + 5\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {1}{10} \, \sqrt {5} e^{4} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) - \frac {1}{2} \, \sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + \frac {1}{2} \, x^{2} - x {\left (2 \, \log \relax (2) + 3\right )} + \frac {1}{10} \, {\left (\sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 5 \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right )\right )} e^{4} - \frac {1}{5} \, {\left (\sqrt {5} {\left (e^{4} + 5\right )} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 10 \, e^{2} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + 10 \, x\right )} e^{2} - \frac {2}{5} \, {\left (\sqrt {5} e^{2} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 5 \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right )\right )} e^{2} + 2 \, x e^{2} + {\left (x - e^{2}\right )} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + \frac {1}{2} \, {\left (3 \, e^{4} + 5\right )} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + 3 \, e^{2} \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) + x \log \relax (x) - \frac {3}{2} \, \sqrt {5} \log \left (\frac {x - \sqrt {5} - e^{2}}{x + \sqrt {5} - e^{2}}\right ) + 3 \, x - \frac {5}{2} \, \log \left (x^{2} - 2 \, x e^{2} + e^{4} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)^2-2*exp(2)*x+x^2-5)*log(1/16*x^2*exp(2)^4-1/4*x^3*exp(2)^3+1/16*(6*x^4-10*x^2)*exp(2)^2+1/1
6*(-4*x^5+20*x^3)*exp(2)+1/16*x^6-5/8*x^4+25/16*x^2)+(2*x+2)*exp(2)^2+(-4*x^2-8*x)*exp(2)+2*x^3+6*x^2-10*x-10)
/(2*exp(2)^2-4*exp(2)*x+2*x^2-10),x, algorithm="maxima")

[Out]

1/10*sqrt(5)*(e^6 + 15*e^2)*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 3/10*sqrt(5)*(e^4 + 5)*log((x - sqr
t(5) - e^2)/(x + sqrt(5) - e^2)) + 1/10*sqrt(5)*e^4*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) - 1/2*sqrt(5)
*e^2*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 1/2*x^2 - x*(2*log(2) + 3) + 1/10*(sqrt(5)*e^2*log((x - sq
rt(5) - e^2)/(x + sqrt(5) - e^2)) + 5*log(x^2 - 2*x*e^2 + e^4 - 5))*e^4 - 1/5*(sqrt(5)*(e^4 + 5)*log((x - sqrt
(5) - e^2)/(x + sqrt(5) - e^2)) + 10*e^2*log(x^2 - 2*x*e^2 + e^4 - 5) + 10*x)*e^2 - 2/5*(sqrt(5)*e^2*log((x -
sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 5*log(x^2 - 2*x*e^2 + e^4 - 5))*e^2 + 2*x*e^2 + (x - e^2)*log(x^2 - 2*x*
e^2 + e^4 - 5) + 1/2*(3*e^4 + 5)*log(x^2 - 2*x*e^2 + e^4 - 5) + 3*e^2*log(x^2 - 2*x*e^2 + e^4 - 5) + x*log(x)
- 3/2*sqrt(5)*log((x - sqrt(5) - e^2)/(x + sqrt(5) - e^2)) + 3*x - 5/2*log(x^2 - 2*x*e^2 + e^4 - 5)

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mupad [B]  time = 5.23, size = 66, normalized size = 2.36 \begin {gather*} \frac {x\,\left (x+\ln \left (\frac {{\mathrm {e}}^2\,\left (20\,x^3-4\,x^5\right )}{16}-\frac {{\mathrm {e}}^4\,\left (10\,x^2-6\,x^4\right )}{16}-\frac {x^3\,{\mathrm {e}}^6}{4}+\frac {x^2\,{\mathrm {e}}^8}{16}+\frac {25\,x^2}{16}-\frac {5\,x^4}{8}+\frac {x^6}{16}\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((exp(2)*(20*x^3 - 4*x^5))/16 - (exp(4)*(10*x^2 - 6*x^4))/16 - (x^3*exp(6))/4 + (x^2*exp(8))/16 + (25*
x^2)/16 - (5*x^4)/8 + x^6/16)*(exp(4) - 2*x*exp(2) + x^2 - 5) - 10*x - exp(2)*(8*x + 4*x^2) + 6*x^2 + 2*x^3 +
exp(4)*(2*x + 2) - 10)/(2*exp(4) - 4*x*exp(2) + 2*x^2 - 10),x)

[Out]

(x*(x + log((exp(2)*(20*x^3 - 4*x^5))/16 - (exp(4)*(10*x^2 - 6*x^4))/16 - (x^3*exp(6))/4 + (x^2*exp(8))/16 + (
25*x^2)/16 - (5*x^4)/8 + x^6/16)))/2

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sympy [B]  time = 0.31, size = 76, normalized size = 2.71 \begin {gather*} \frac {x^{2}}{2} + \frac {x \log {\left (\frac {x^{6}}{16} - \frac {5 x^{4}}{8} - \frac {x^{3} e^{6}}{4} + \frac {25 x^{2}}{16} + \frac {x^{2} e^{8}}{16} + \left (\frac {3 x^{4}}{8} - \frac {5 x^{2}}{8}\right ) e^{4} + \left (- \frac {x^{5}}{4} + \frac {5 x^{3}}{4}\right ) e^{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)**2-2*exp(2)*x+x**2-5)*ln(1/16*x**2*exp(2)**4-1/4*x**3*exp(2)**3+1/16*(6*x**4-10*x**2)*exp(2
)**2+1/16*(-4*x**5+20*x**3)*exp(2)+1/16*x**6-5/8*x**4+25/16*x**2)+(2*x+2)*exp(2)**2+(-4*x**2-8*x)*exp(2)+2*x**
3+6*x**2-10*x-10)/(2*exp(2)**2-4*exp(2)*x+2*x**2-10),x)

[Out]

x**2/2 + x*log(x**6/16 - 5*x**4/8 - x**3*exp(6)/4 + 25*x**2/16 + x**2*exp(8)/16 + (3*x**4/8 - 5*x**2/8)*exp(4)
+ (-x**5/4 + 5*x**3/4)*exp(2))/2

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