3.86.35 \(\int \frac {-192+144 x-36 x^2+3 x^3+e^{10-2 x} (-384+288 x-72 x^2+6 x^3)+e^{5-x} (72-42 x+6 x^2) \log (5)+6 \log ^2(5)}{-64+48 x-12 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ 3 x-3 \left (e^{5-x}+\frac {\log (5)}{-4+x}\right )^2 \]

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Rubi [A]  time = 0.28, antiderivative size = 44, normalized size of antiderivative = 1.83, number of steps used = 6, number of rules used = 4, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6688, 2194, 2197, 1850} \begin {gather*} 3 x-3 e^{10-2 x}-\frac {3 \log ^2(5)}{(4-x)^2}+\frac {6 e^{5-x} \log (5)}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-192 + 144*x - 36*x^2 + 3*x^3 + E^(10 - 2*x)*(-384 + 288*x - 72*x^2 + 6*x^3) + E^(5 - x)*(72 - 42*x + 6*x
^2)*Log[5] + 6*Log[5]^2)/(-64 + 48*x - 12*x^2 + x^3),x]

[Out]

-3*E^(10 - 2*x) + 3*x + (6*E^(5 - x)*Log[5])/(4 - x) - (3*Log[5]^2)/(4 - x)^2

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2194

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

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

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 (6 e^{10-2 x}+\frac {6 e^{5-x} (-3+x) \log (5)}{(-4+x)^2}+\frac {3 \left (48 x-12 x^2+x^3-2 \left (32-\log ^2(5)\right )\right )}{(-4+x)^3}\right ) \, dx\\ &=3 \int \frac {48 x-12 x^2+x^3-2 \left (32-\log ^2(5)\right )}{(-4+x)^3} \, dx+6 \int e^{10-2 x} \, dx+(6 \log (5)) \int \frac {e^{5-x} (-3+x)}{(-4+x)^2} \, dx\\ &=-3 e^{10-2 x}+\frac {6 e^{5-x} \log (5)}{4-x}+3 \int \left (1+\frac {2 \log ^2(5)}{(-4+x)^3}\right ) \, dx\\ &=-3 e^{10-2 x}+3 x+\frac {6 e^{5-x} \log (5)}{4-x}-\frac {3 \log ^2(5)}{(4-x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 40, normalized size = 1.67 \begin {gather*} -3 e^{10-2 x}+3 x-\frac {6 e^{5-x} \log (5)}{-4+x}-\frac {3 \log ^2(5)}{(-4+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192 + 144*x - 36*x^2 + 3*x^3 + E^(10 - 2*x)*(-384 + 288*x - 72*x^2 + 6*x^3) + E^(5 - x)*(72 - 42*x
+ 6*x^2)*Log[5] + 6*Log[5]^2)/(-64 + 48*x - 12*x^2 + x^3),x]

[Out]

-3*E^(10 - 2*x) + 3*x - (6*E^(5 - x)*Log[5])/(-4 + x) - (3*Log[5]^2)/(-4 + x)^2

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fricas [B]  time = 0.54, size = 59, normalized size = 2.46 \begin {gather*} \frac {3 \, {\left (x^{3} - 2 \, {\left (x - 4\right )} e^{\left (-x + 5\right )} \log \relax (5) - 8 \, x^{2} - {\left (x^{2} - 8 \, x + 16\right )} e^{\left (-2 \, x + 10\right )} - \log \relax (5)^{2} + 16 \, x\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp(5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-1
92)/(x^3-12*x^2+48*x-64),x, algorithm="fricas")

[Out]

3*(x^3 - 2*(x - 4)*e^(-x + 5)*log(5) - 8*x^2 - (x^2 - 8*x + 16)*e^(-2*x + 10) - log(5)^2 + 16*x)/(x^2 - 8*x +
16)

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giac [B]  time = 0.16, size = 79, normalized size = 3.29 \begin {gather*} \frac {3 \, {\left (x^{3} - x^{2} e^{\left (-2 \, x + 10\right )} - 2 \, x e^{\left (-x + 5\right )} \log \relax (5) - 8 \, x^{2} + 8 \, x e^{\left (-2 \, x + 10\right )} + 8 \, e^{\left (-x + 5\right )} \log \relax (5) - \log \relax (5)^{2} + 16 \, x - 16 \, e^{\left (-2 \, x + 10\right )}\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp(5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-1
92)/(x^3-12*x^2+48*x-64),x, algorithm="giac")

[Out]

3*(x^3 - x^2*e^(-2*x + 10) - 2*x*e^(-x + 5)*log(5) - 8*x^2 + 8*x*e^(-2*x + 10) + 8*e^(-x + 5)*log(5) - log(5)^
2 + 16*x - 16*e^(-2*x + 10))/(x^2 - 8*x + 16)

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maple [A]  time = 0.11, size = 44, normalized size = 1.83

method	result	size
risch	\(3 x -\frac {3 \ln \relax (5)^{2}}{x^{2}-8 x +16}-3 \,{\mathrm e}^{-2 x +10}-\frac {6 \ln \relax (5) {\mathrm e}^{5-x}}{x -4}\)	\(44\)
derivativedivides	\(-15+3 x -\frac {3 \ln \relax (5)^{2}}{\left (-x +4\right )^{2}}-3 \,{\mathrm e}^{-2 x +10}+\frac {6 \ln \relax (5) {\mathrm e}^{5-x}}{-x +4}\)	\(46\)
default	\(-15+3 x -\frac {3 \ln \relax (5)^{2}}{\left (-x +4\right )^{2}}-3 \,{\mathrm e}^{-2 x +10}+\frac {6 \ln \relax (5) {\mathrm e}^{5-x}}{-x +4}\)	\(46\)
norman	\(\frac {-144 x +3 x^{3}-48 \,{\mathrm e}^{-2 x +10}+24 x \,{\mathrm e}^{-2 x +10}-3 x^{2} {\mathrm e}^{-2 x +10}+24 \ln \relax (5) {\mathrm e}^{5-x}-6 \ln \relax (5) {\mathrm e}^{5-x} x +384-3 \ln \relax (5)^{2}}{\left (x -4\right )^{2}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*ln(5)*exp(5-x)+6*ln(5)^2+3*x^3-36*x^2+144*x-192)/(x^3
-12*x^2+48*x-64),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 3 \, x - \frac {3 \, \log \relax (5)^{2}}{x^{2} - 8 \, x + 16} - \frac {3 \, {\left (2 \, {\left (x^{2} e^{5} \log \relax (5) - 8 \, x e^{5} \log \relax (5) + 16 \, e^{5} \log \relax (5)\right )} e^{\left (-x\right )} + {\left (x^{3} e^{10} - 12 \, x^{2} e^{10} + 48 \, x e^{10}\right )} e^{\left (-2 \, x\right )}\right )}}{x^{3} - 12 \, x^{2} + 48 \, x - 64} - \frac {48 \, {\left (3 \, x - 10\right )}}{x^{2} - 8 \, x + 16} - \frac {144 \, {\left (x - 2\right )}}{x^{2} - 8 \, x + 16} + \frac {288 \, {\left (x - 3\right )}}{x^{2} - 8 \, x + 16} + \frac {384 \, e^{2} E_{3}\left (2 \, x - 8\right )}{{\left (x - 4\right )}^{2}} + \frac {96}{x^{2} - 8 \, x + 16} - 576 \, \int \frac {e^{\left (-2 \, x + 10\right )}}{x^{4} - 16 \, x^{3} + 96 \, x^{2} - 256 \, x + 256}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-72*x^2+288*x-384)*exp(5-x)^2+(6*x^2-42*x+72)*log(5)*exp(5-x)+6*log(5)^2+3*x^3-36*x^2+144*x-1
92)/(x^3-12*x^2+48*x-64),x, algorithm="maxima")

[Out]

3*x - 3*log(5)^2/(x^2 - 8*x + 16) - 3*(2*(x^2*e^5*log(5) - 8*x*e^5*log(5) + 16*e^5*log(5))*e^(-x) + (x^3*e^10
- 12*x^2*e^10 + 48*x*e^10)*e^(-2*x))/(x^3 - 12*x^2 + 48*x - 64) - 48*(3*x - 10)/(x^2 - 8*x + 16) - 144*(x - 2)
/(x^2 - 8*x + 16) + 288*(x - 3)/(x^2 - 8*x + 16) + 384*e^2*exp_integral_e(3, 2*x - 8)/(x - 4)^2 + 96/(x^2 - 8*
x + 16) - 576*integrate(e^(-2*x + 10)/(x^4 - 16*x^3 + 96*x^2 - 256*x + 256), x)

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mupad [B]  time = 5.31, size = 52, normalized size = 2.17 \begin {gather*} 3\,x-3\,{\mathrm {e}}^{10-2\,x}-\frac {3\,{\ln \relax (5)}^2-24\,{\mathrm {e}}^{5-x}\,\ln \relax (5)+6\,x\,{\mathrm {e}}^{5-x}\,\ln \relax (5)}{x^2-8\,x+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((144*x + exp(10 - 2*x)*(288*x - 72*x^2 + 6*x^3 - 384) + 6*log(5)^2 - 36*x^2 + 3*x^3 + exp(5 - x)*log(5)*(6
*x^2 - 42*x + 72) - 192)/(48*x - 12*x^2 + x^3 - 64),x)

[Out]

3*x - 3*exp(10 - 2*x) - (3*log(5)^2 - 24*exp(5 - x)*log(5) + 6*x*exp(5 - x)*log(5))/(x^2 - 8*x + 16)

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sympy [B]  time = 0.27, size = 42, normalized size = 1.75 \begin {gather*} 3 x - \frac {3 \log {\relax (5 )}^{2}}{x^{2} - 8 x + 16} + \frac {\left (12 - 3 x\right ) e^{10 - 2 x} - 6 e^{5 - x} \log {\relax (5 )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**3-72*x**2+288*x-384)*exp(5-x)**2+(6*x**2-42*x+72)*ln(5)*exp(5-x)+6*ln(5)**2+3*x**3-36*x**2+14
4*x-192)/(x**3-12*x**2+48*x-64),x)

[Out]

3*x - 3*log(5)**2/(x**2 - 8*x + 16) + ((12 - 3*x)*exp(10 - 2*x) - 6*exp(5 - x)*log(5))/(x - 4)

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