3.85.41 \(\int \frac {15-45 x+45 x^2-15 x^3+(600 x+210 x^2-162 x^3-6 x^4+6 x^5+e^{2 x} (96 x-216 x^2+48 x^3+192 x^4-144 x^5+24 x^6)) \log ^2(-4+x)}{(4-13 x+15 x^2-7 x^3+x^4) \log ^2(-4+x)} \, dx\)

Optimal. Leaf size=35 \[ 3 \left (4 e^{2 x} x^2+\frac {x^2 (5+x)^2}{(-1+x)^2}+\frac {5}{\log (-4+x)}\right ) \]

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Rubi [A]  time = 0.71, antiderivative size = 57, normalized size of antiderivative = 1.63, number of steps used = 21, number of rules used = 10, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6688, 6742, 37, 43, 2196, 2176, 2194, 2390, 2302, 30} \begin {gather*} 12 e^{2 x} x^2+\frac {75 x^2}{(1-x)^2}+3 x^2+36 x-\frac {102}{1-x}+\frac {33}{(1-x)^2}+\frac {15}{\log (x-4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 - 45*x + 45*x^2 - 15*x^3 + (600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + E^(2*x)*(96*x - 216*x^2 + 48*x
^3 + 192*x^4 - 144*x^5 + 24*x^6))*Log[-4 + x]^2)/((4 - 13*x + 15*x^2 - 7*x^3 + x^4)*Log[-4 + x]^2),x]

[Out]

33/(1 - x)^2 - 102/(1 - x) + 36*x + 3*x^2 + 12*E^(2*x)*x^2 + (75*x^2)/(1 - x)^2 + 15/Log[-4 + x]

Rule 30

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

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]]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {6 x \left (-25-15 x+3 x^2+x^3+4 e^{2 x} (-1+x)^3 (1+x)\right )}{(-1+x)^3}-\frac {15}{(-4+x) \log ^2(-4+x)}\right ) \, dx\\ &=6 \int \frac {x \left (-25-15 x+3 x^2+x^3+4 e^{2 x} (-1+x)^3 (1+x)\right )}{(-1+x)^3} \, dx-15 \int \frac {1}{(-4+x) \log ^2(-4+x)} \, dx\\ &=6 \int \left (-\frac {25 x}{(-1+x)^3}-\frac {15 x^2}{(-1+x)^3}+\frac {3 x^3}{(-1+x)^3}+\frac {x^4}{(-1+x)^3}+4 e^{2 x} x (1+x)\right ) \, dx-15 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-4+x\right )\\ &=6 \int \frac {x^4}{(-1+x)^3} \, dx-15 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-4+x)\right )+18 \int \frac {x^3}{(-1+x)^3} \, dx+24 \int e^{2 x} x (1+x) \, dx-90 \int \frac {x^2}{(-1+x)^3} \, dx-150 \int \frac {x}{(-1+x)^3} \, dx\\ &=\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)}+6 \int \left (3+\frac {1}{(-1+x)^3}+\frac {4}{(-1+x)^2}+\frac {6}{-1+x}+x\right ) \, dx+18 \int \left (1+\frac {1}{(-1+x)^3}+\frac {3}{(-1+x)^2}+\frac {3}{-1+x}\right ) \, dx+24 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx-90 \int \left (\frac {1}{(-1+x)^3}+\frac {2}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx\\ &=\frac {33}{(1-x)^2}-\frac {102}{1-x}+36 x+3 x^2+\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)}+24 \int e^{2 x} x \, dx+24 \int e^{2 x} x^2 \, dx\\ &=\frac {33}{(1-x)^2}-\frac {102}{1-x}+36 x+12 e^{2 x} x+3 x^2+12 e^{2 x} x^2+\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)}-12 \int e^{2 x} \, dx-24 \int e^{2 x} x \, dx\\ &=-6 e^{2 x}+\frac {33}{(1-x)^2}-\frac {102}{1-x}+36 x+3 x^2+12 e^{2 x} x^2+\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)}+12 \int e^{2 x} \, dx\\ &=\frac {33}{(1-x)^2}-\frac {102}{1-x}+36 x+3 x^2+12 e^{2 x} x^2+\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 57, normalized size = 1.63 \begin {gather*} \frac {33}{(1-x)^2}-\frac {102}{1-x}+36 x+3 x^2+12 e^{2 x} x^2+\frac {75 x^2}{(1-x)^2}+\frac {15}{\log (-4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 - 45*x + 45*x^2 - 15*x^3 + (600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + E^(2*x)*(96*x - 216*x^2
+ 48*x^3 + 192*x^4 - 144*x^5 + 24*x^6))*Log[-4 + x]^2)/((4 - 13*x + 15*x^2 - 7*x^3 + x^4)*Log[-4 + x]^2),x]

[Out]

33/(1 - x)^2 - 102/(1 - x) + 36*x + 3*x^2 + 12*E^(2*x)*x^2 + (75*x^2)/(1 - x)^2 + 15/Log[-4 + x]

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fricas [B]  time = 0.64, size = 69, normalized size = 1.97 \begin {gather*} \frac {3 \, {\left (5 \, x^{2} + {\left (x^{4} + 10 \, x^{3} - 23 \, x^{2} + 4 \, {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + 96 \, x - 48\right )} \log \left (x - 4\right ) - 10 \, x + 5\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="fricas")

[Out]

3*(5*x^2 + (x^4 + 10*x^3 - 23*x^2 + 4*(x^4 - 2*x^3 + x^2)*e^(2*x) + 96*x - 48)*log(x - 4) - 10*x + 5)/((x^2 -
2*x + 1)*log(x - 4))

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giac [B]  time = 0.26, size = 112, normalized size = 3.20 \begin {gather*} \frac {3 \, {\left (4 \, x^{4} e^{\left (2 \, x\right )} \log \left (x - 4\right ) + x^{4} \log \left (x - 4\right ) - 8 \, x^{3} e^{\left (2 \, x\right )} \log \left (x - 4\right ) + 10 \, x^{3} \log \left (x - 4\right ) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \left (x - 4\right ) - 23 \, x^{2} \log \left (x - 4\right ) + 5 \, x^{2} + 96 \, x \log \left (x - 4\right ) - 10 \, x - 48 \, \log \left (x - 4\right ) + 5\right )}}{x^{2} \log \left (x - 4\right ) - 2 \, x \log \left (x - 4\right ) + \log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="giac")

[Out]

3*(4*x^4*e^(2*x)*log(x - 4) + x^4*log(x - 4) - 8*x^3*e^(2*x)*log(x - 4) + 10*x^3*log(x - 4) + 4*x^2*e^(2*x)*lo
g(x - 4) - 23*x^2*log(x - 4) + 5*x^2 + 96*x*log(x - 4) - 10*x - 48*log(x - 4) + 5)/(x^2*log(x - 4) - 2*x*log(x
- 4) + log(x - 4))

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maple [A]  time = 0.04, size = 41, normalized size = 1.17

method	result	size
default	\(\frac {15}{\ln \left (x -4\right )}+12 \,{\mathrm e}^{2 x} x^{2}+3 x^{2}+36 x +\frac {108}{\left (x -1\right )^{2}}+\frac {252}{x -1}\)	\(41\)
risch	\(\frac {12 \,{\mathrm e}^{2 x} x^{4}-24 \,{\mathrm e}^{2 x} x^{3}+3 x^{4}+12 \,{\mathrm e}^{2 x} x^{2}+30 x^{3}-69 x^{2}+288 x -144}{x^{2}-2 x +1}+\frac {15}{\ln \left (x -4\right )}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*ln(x-4)^2-15*x^
3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/ln(x-4)^2,x,method=_RETURNVERBOSE)

[Out]

15/ln(x-4)+12*exp(x)^2*x^2+3*x^2+36*x+108/(x-1)^2+252/(x-1)

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maxima [B]  time = 0.46, size = 69, normalized size = 1.97 \begin {gather*} \frac {3 \, {\left (5 \, x^{2} + {\left (x^{4} + 10 \, x^{3} - 23 \, x^{2} + 4 \, {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + 96 \, x - 48\right )} \log \left (x - 4\right ) - 10 \, x + 5\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="maxima")

[Out]

3*(5*x^2 + (x^4 + 10*x^3 - 23*x^2 + 4*(x^4 - 2*x^3 + x^2)*e^(2*x) + 96*x - 48)*log(x - 4) - 10*x + 5)/((x^2 -
2*x + 1)*log(x - 4))

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mupad [B]  time = 0.22, size = 42, normalized size = 1.20 \begin {gather*} 36\,x+\frac {252\,x-144}{x^2-2\,x+1}+12\,x^2\,{\mathrm {e}}^{2\,x}+\frac {15}{\ln \left (x-4\right )}+3\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*x^2 - 45*x - 15*x^3 + log(x - 4)^2*(600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + exp(2*x)*(96*x - 216*x
^2 + 48*x^3 + 192*x^4 - 144*x^5 + 24*x^6)) + 15)/(log(x - 4)^2*(15*x^2 - 13*x - 7*x^3 + x^4 + 4)),x)

[Out]

36*x + (252*x - 144)/(x^2 - 2*x + 1) + 12*x^2*exp(2*x) + 15/log(x - 4) + 3*x^2

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sympy [A]  time = 0.47, size = 37, normalized size = 1.06 \begin {gather*} 12 x^{2} e^{2 x} + 3 x^{2} + 36 x + \frac {252 x - 144}{x^{2} - 2 x + 1} + \frac {15}{\log {\left (x - 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x**6-144*x**5+192*x**4+48*x**3-216*x**2+96*x)*exp(x)**2+6*x**5-6*x**4-162*x**3+210*x**2+600*x)
*ln(x-4)**2-15*x**3+45*x**2-45*x+15)/(x**4-7*x**3+15*x**2-13*x+4)/ln(x-4)**2,x)

[Out]

12*x**2*exp(2*x) + 3*x**2 + 36*x + (252*x - 144)/(x**2 - 2*x + 1) + 15/log(x - 4)

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