3.39.61 \(\int \frac {-189-198 x+e^4 (22 x+23 x^2)+(-9-9 x+e^4 (x+x^2)) \log (\frac {9 x+9 x^2+e^4 (-x^2-x^3)}{e^4})}{-180-180 x+e^4 (20 x+20 x^2)} \, dx\)

Optimal. Leaf size=24 \[ 3+x+\frac {1}{20} x \log \left (\left (\frac {9}{e^4}-x\right ) \left (x+x^2\right )\right ) \]

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Rubi [B]  time = 0.35, antiderivative size = 86, normalized size of antiderivative = 3.58, number of steps used = 12, number of rules used = 7, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.084, Rules used = {6741, 6728, 893, 2523, 1657, 632, 31} \begin {gather*} \frac {1}{20} x \log \left (x \left (-e^4 x^2+\left (9-e^4\right ) x+9\right )\right )+\frac {4 x}{5}-\frac {1}{20} \log (x+1)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}-\frac {9 \log \left (e^4 x-9\right )}{20 e^4}+\frac {1}{20} \log \left (e^4 x+e^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-189 - 198*x + E^4*(22*x + 23*x^2) + (-9 - 9*x + E^4*(x + x^2))*Log[(9*x + 9*x^2 + E^4*(-x^2 - x^3))/E^4]
)/(-180 - 180*x + E^4*(20*x + 20*x^2)),x]

[Out]

(4*x)/5 - Log[1 + x]/20 + (9*Log[9 - E^4*x])/(20*E^4) - (9*Log[-9 + E^4*x])/(20*E^4) + Log[E^4 + E^4*x]/20 + (
x*Log[x*(9 + (9 - E^4)*x - E^4*x^2)])/20

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 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 {189+198 x-e^4 \left (22 x+23 x^2\right )-\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{180+20 \left (9-e^4\right ) x-20 e^4 x^2} \, dx\\ &=\int \left (\frac {189+22 \left (9-e^4\right ) x-23 e^4 x^2}{20 (1+x) \left (9-e^4 x\right )}+\frac {1}{20} \left (-4+\log \left (-x \left (-9+\left (-9+e^4\right ) x+e^4 x^2\right )\right )\right )\right ) \, dx\\ &=\frac {1}{20} \int \frac {189+22 \left (9-e^4\right ) x-23 e^4 x^2}{(1+x) \left (9-e^4 x\right )} \, dx+\frac {1}{20} \int \left (-4+\log \left (-x \left (-9+\left (-9+e^4\right ) x+e^4 x^2\right )\right )\right ) \, dx\\ &=-\frac {x}{5}+\frac {1}{20} \int \left (23+\frac {1}{-1-x}+\frac {9}{-9+e^4 x}\right ) \, dx+\frac {1}{20} \int \log \left (-x \left (-9+\left (-9+e^4\right ) x+e^4 x^2\right )\right ) \, dx\\ &=\frac {19 x}{20}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}+\frac {1}{20} x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )-\frac {1}{20} \int \frac {9+2 \left (9-e^4\right ) x-3 e^4 x^2}{9+\left (9-e^4\right ) x-e^4 x^2} \, dx\\ &=\frac {19 x}{20}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}+\frac {1}{20} x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )-\frac {1}{20} \int \left (3-\frac {18+\left (9-e^4\right ) x}{9+\left (9-e^4\right ) x-e^4 x^2}\right ) \, dx\\ &=\frac {4 x}{5}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}+\frac {1}{20} x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )+\frac {1}{20} \int \frac {18+\left (9-e^4\right ) x}{9+\left (9-e^4\right ) x-e^4 x^2} \, dx\\ &=\frac {4 x}{5}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}+\frac {1}{20} x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )+\frac {9}{20} \int \frac {1}{\frac {1}{2} \left (9-e^4\right )+\frac {1}{2} \left (9+e^4\right )-e^4 x} \, dx-\frac {1}{20} e^4 \int \frac {1}{\frac {1}{2} \left (-9-e^4\right )+\frac {1}{2} \left (9-e^4\right )-e^4 x} \, dx\\ &=\frac {4 x}{5}-\frac {1}{20} \log (1+x)+\frac {9 \log \left (9-e^4 x\right )}{20 e^4}-\frac {9 \log \left (-9+e^4 x\right )}{20 e^4}+\frac {1}{20} \log \left (e^4+e^4 x\right )+\frac {1}{20} x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 32, normalized size = 1.33 \begin {gather*} \frac {1}{20} \left (16 x+x \log \left (x \left (9+\left (9-e^4\right ) x-e^4 x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-189 - 198*x + E^4*(22*x + 23*x^2) + (-9 - 9*x + E^4*(x + x^2))*Log[(9*x + 9*x^2 + E^4*(-x^2 - x^3)
)/E^4])/(-180 - 180*x + E^4*(20*x + 20*x^2)),x]

[Out]

(16*x + x*Log[x*(9 + (9 - E^4)*x - E^4*x^2)])/20

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fricas [A]  time = 0.56, size = 29, normalized size = 1.21 \begin {gather*} \frac {1}{20} \, x \log \left ({\left (9 \, x^{2} - {\left (x^{3} + x^{2}\right )} e^{4} + 9 \, x\right )} e^{\left (-4\right )}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x
-189)/((20*x^2+20*x)*exp(2)^2-180*x-180),x, algorithm="fricas")

[Out]

1/20*x*log((9*x^2 - (x^3 + x^2)*e^4 + 9*x)*e^(-4)) + x

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giac [A]  time = 0.82, size = 31, normalized size = 1.29 \begin {gather*} \frac {1}{20} \, x \log \left (-x^{3} e^{4} - x^{2} e^{4} + 9 \, x^{2} + 9 \, x\right ) + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x
-189)/((20*x^2+20*x)*exp(2)^2-180*x-180),x, algorithm="giac")

[Out]

1/20*x*log(-x^3*e^4 - x^2*e^4 + 9*x^2 + 9*x) + 4/5*x

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maple [A]  time = 0.20, size = 32, normalized size = 1.33

method	result	size
default	\(\frac {4 x}{5}+\frac {x \ln \left (x \left (-x^{2} {\mathrm e}^{4}-x \,{\mathrm e}^{4}+9 x +9\right )\right )}{20}\)	\(32\)
risch	\(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\)	\(33\)
norman	\(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+x)*exp(2)^2-9*x-9)*ln(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x-189)/(
(20*x^2+20*x)*exp(2)^2-180*x-180),x,method=_RETURNVERBOSE)

[Out]

4/5*x+1/20*x*ln(x*(-x^2*exp(2)^2-x*exp(2)^2+9*x+9))

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maxima [B]  time = 0.51, size = 162, normalized size = 6.75 \begin {gather*} \frac {23}{20} \, {\left (x e^{\left (-4\right )} + \frac {81 \, \log \left (x e^{4} - 9\right )}{e^{12} + 9 \, e^{8}} - \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {11}{10} \, {\left (\frac {9 \, \log \left (x e^{4} - 9\right )}{e^{8} + 9 \, e^{4}} + \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {1}{20} \, {\left (x e^{4} \log \relax (x) - 7 \, x e^{4} + {\left (x e^{4} - 9\right )} \log \left (-x e^{4} + 9\right ) + {\left (x e^{4} + e^{4}\right )} \log \left (x + 1\right )\right )} e^{\left (-4\right )} - \frac {891 \, \log \left (x e^{4} - 9\right )}{10 \, {\left (e^{8} + 9 \, e^{4}\right )}} - \frac {189 \, \log \left (x e^{4} - 9\right )}{20 \, {\left (e^{4} + 9\right )}} - \frac {9 \, \log \left (x + 1\right )}{20 \, {\left (e^{4} + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x
-189)/((20*x^2+20*x)*exp(2)^2-180*x-180),x, algorithm="maxima")

[Out]

23/20*(x*e^(-4) + 81*log(x*e^4 - 9)/(e^12 + 9*e^8) - log(x + 1)/(e^4 + 9))*e^4 + 11/10*(9*log(x*e^4 - 9)/(e^8
+ 9*e^4) + log(x + 1)/(e^4 + 9))*e^4 + 1/20*(x*e^4*log(x) - 7*x*e^4 + (x*e^4 - 9)*log(-x*e^4 + 9) + (x*e^4 + e
^4)*log(x + 1))*e^(-4) - 891/10*log(x*e^4 - 9)/(e^8 + 9*e^4) - 189/20*log(x*e^4 - 9)/(e^4 + 9) - 9/20*log(x +
1)/(e^4 + 9)

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mupad [B]  time = 3.05, size = 29, normalized size = 1.21 \begin {gather*} \frac {x\,\left (\ln \left ({\mathrm {e}}^{-4}\,\left (9\,x-{\mathrm {e}}^4\,\left (x^3+x^2\right )+9\,x^2\right )\right )+20\right )}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((198*x - exp(4)*(22*x + 23*x^2) + log(exp(-4)*(9*x - exp(4)*(x^2 + x^3) + 9*x^2))*(9*x - exp(4)*(x + x^2)
+ 9) + 189)/(180*x - exp(4)*(20*x + 20*x^2) + 180),x)

[Out]

(x*(log(exp(-4)*(9*x - exp(4)*(x^2 + x^3) + 9*x^2)) + 20))/20

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sympy [A]  time = 0.25, size = 29, normalized size = 1.21 \begin {gather*} \frac {x \log {\left (\frac {9 x^{2} + 9 x + \left (- x^{3} - x^{2}\right ) e^{4}}{e^{4}} \right )}}{20} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+x)*exp(2)**2-9*x-9)*ln(((-x**3-x**2)*exp(2)**2+9*x**2+9*x)/exp(2)**2)+(23*x**2+22*x)*exp(2)*
*2-198*x-189)/((20*x**2+20*x)*exp(2)**2-180*x-180),x)

[Out]

x*log((9*x**2 + 9*x + (-x**3 - x**2)*exp(4))*exp(-4))/20 + x

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