3.20.77 \(\int \frac {1-8 x-4 e^5 x+24 x^2-32 x^3+16 x^4+e^{2 x} (-2 x+12 x^2-24 x^3+16 x^4)}{-x+6 x^2-12 x^3+8 x^4} \, dx\)

Optimal. Leaf size=25 \[ 4+e^{2 x}+\frac {e^5}{(1-2 x)^2}+2 x-\log (x) \]

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Rubi [B]  time = 0.71, antiderivative size = 61, normalized size of antiderivative = 2.44, number of steps used = 12, number of rules used = 7, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6741, 6742, 2194, 44, 37, 43} \begin {gather*} -\frac {12 x^2}{(1-2 x)^2}+2 x+e^{2 x}-\frac {5}{1-2 x}+\frac {1}{2 x-1}+\frac {2+e^5}{(1-2 x)^2}+\frac {1}{(1-2 x)^2}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 8*x - 4*E^5*x + 24*x^2 - 32*x^3 + 16*x^4 + E^(2*x)*(-2*x + 12*x^2 - 24*x^3 + 16*x^4))/(-x + 6*x^2 - 1
2*x^3 + 8*x^4),x]

[Out]

E^(2*x) + (1 - 2*x)^(-2) + (2 + E^5)/(1 - 2*x)^2 - 5/(1 - 2*x) + 2*x - (12*x^2)/(1 - 2*x)^2 + (-1 + 2*x)^(-1)
- Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 6741

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

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 \frac {1+\left (-8-4 e^5\right ) x+24 x^2-32 x^3+16 x^4+e^{2 x} \left (-2 x+12 x^2-24 x^3+16 x^4\right )}{-x+6 x^2-12 x^3+8 x^4} \, dx\\ &=\int \frac {-1-\left (-8-4 e^5\right ) x-24 x^2+32 x^3-16 x^4-e^{2 x} \left (-2 x+12 x^2-24 x^3+16 x^4\right )}{x \left (1-6 x+12 x^2-8 x^3\right )} \, dx\\ &=\int \left (2 e^{2 x}-\frac {4 \left (2+e^5\right )}{(-1+2 x)^3}+\frac {1}{x (-1+2 x)^3}+\frac {24 x}{(-1+2 x)^3}-\frac {32 x^2}{(-1+2 x)^3}+\frac {16 x^3}{(-1+2 x)^3}\right ) \, dx\\ &=\frac {2+e^5}{(1-2 x)^2}+2 \int e^{2 x} \, dx+16 \int \frac {x^3}{(-1+2 x)^3} \, dx+24 \int \frac {x}{(-1+2 x)^3} \, dx-32 \int \frac {x^2}{(-1+2 x)^3} \, dx+\int \frac {1}{x (-1+2 x)^3} \, dx\\ &=e^{2 x}+\frac {2+e^5}{(1-2 x)^2}-\frac {12 x^2}{(1-2 x)^2}+16 \int \left (\frac {1}{8}+\frac {1}{8 (-1+2 x)^3}+\frac {3}{8 (-1+2 x)^2}+\frac {3}{8 (-1+2 x)}\right ) \, dx-32 \int \left (\frac {1}{4 (-1+2 x)^3}+\frac {1}{2 (-1+2 x)^2}+\frac {1}{4 (-1+2 x)}\right ) \, dx+\int \left (-\frac {1}{x}+\frac {2}{(-1+2 x)^3}-\frac {2}{(-1+2 x)^2}+\frac {2}{-1+2 x}\right ) \, dx\\ &=e^{2 x}+\frac {1}{(1-2 x)^2}+\frac {2+e^5}{(1-2 x)^2}-\frac {5}{1-2 x}+2 x-\frac {12 x^2}{(1-2 x)^2}+\frac {1}{-1+2 x}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 25, normalized size = 1.00 \begin {gather*} -3+e^{2 x}+\frac {e^5}{(1-2 x)^2}+2 x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 8*x - 4*E^5*x + 24*x^2 - 32*x^3 + 16*x^4 + E^(2*x)*(-2*x + 12*x^2 - 24*x^3 + 16*x^4))/(-x + 6*x
^2 - 12*x^3 + 8*x^4),x]

[Out]

-3 + E^(2*x) + E^5/(1 - 2*x)^2 + 2*x - Log[x]

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fricas [B]  time = 0.87, size = 58, normalized size = 2.32 \begin {gather*} \frac {8 \, x^{3} - 8 \, x^{2} + {\left (4 \, x^{2} - 4 \, x + 1\right )} e^{\left (2 \, x\right )} - {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \relax (x) + 2 \, x + e^{5}}{4 \, x^{2} - 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-24*x^3+12*x^2-2*x)*exp(2*x)-4*x*exp(5)+16*x^4-32*x^3+24*x^2-8*x+1)/(8*x^4-12*x^3+6*x^2-x),x
, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 65, normalized size = 2.60 \begin {gather*} \frac {8 \, x^{3} + 4 \, x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} \log \relax (x) - 8 \, x^{2} - 4 \, x e^{\left (2 \, x\right )} + 4 \, x \log \relax (x) + 2 \, x + e^{5} + e^{\left (2 \, x\right )} - \log \relax (x)}{4 \, x^{2} - 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-24*x^3+12*x^2-2*x)*exp(2*x)-4*x*exp(5)+16*x^4-32*x^3+24*x^2-8*x+1)/(8*x^4-12*x^3+6*x^2-x),x
, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 25, normalized size = 1.00

method	result	size
derivativedivides	\(-\ln \left (2 x \right )+2 x +{\mathrm e}^{2 x}+\frac {{\mathrm e}^{5}}{\left (2 x -1\right )^{2}}\)	\(25\)
default	\(-\ln \left (2 x \right )+2 x +{\mathrm e}^{2 x}+\frac {{\mathrm e}^{5}}{\left (2 x -1\right )^{2}}\)	\(25\)
risch	\(2 x +\frac {{\mathrm e}^{5}}{4 x^{2}-4 x +1}-\ln \relax (x )+{\mathrm e}^{2 x}\)	\(27\)
norman	\(\frac {-6 x +8 x^{3}-4 x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{2 x} x^{2}+2+{\mathrm e}^{5}+{\mathrm e}^{2 x}}{\left (2 x -1\right )^{2}}-\ln \relax (x )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^4-24*x^3+12*x^2-2*x)*exp(2*x)-4*x*exp(5)+16*x^4-32*x^3+24*x^2-8*x+1)/(8*x^4-12*x^3+6*x^2-x),x,metho
d=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x + \frac {2 \, {\left (4 \, x^{3} - 6 \, x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )}}{8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1} - \frac {12 \, x - 5}{2 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {2 \, {\left (8 \, x - 3\right )}}{4 \, x^{2} - 4 \, x + 1} - \frac {3 \, {\left (4 \, x - 1\right )}}{4 \, x^{2} - 4 \, x + 1} + \frac {4 \, x - 3}{2 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {e^{5}}{4 \, x^{2} - 4 \, x + 1} + \frac {e E_{3}\left (-2 \, x + 1\right )}{{\left (2 \, x - 1\right )}^{2}} + \frac {2}{4 \, x^{2} - 4 \, x + 1} + 6 \, \int \frac {e^{\left (2 \, x\right )}}{16 \, x^{4} - 32 \, x^{3} + 24 \, x^{2} - 8 \, x + 1}\,{d x} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-24*x^3+12*x^2-2*x)*exp(2*x)-4*x*exp(5)+16*x^4-32*x^3+24*x^2-8*x+1)/(8*x^4-12*x^3+6*x^2-x),x
, algorithm="maxima")

[Out]

2*x + 2*(4*x^3 - 6*x^2 + 3*x)*e^(2*x)/(8*x^3 - 12*x^2 + 6*x - 1) - 1/2*(12*x - 5)/(4*x^2 - 4*x + 1) + 2*(8*x -
3)/(4*x^2 - 4*x + 1) - 3*(4*x - 1)/(4*x^2 - 4*x + 1) + 1/2*(4*x - 3)/(4*x^2 - 4*x + 1) + e^5/(4*x^2 - 4*x + 1
) + e*exp_integral_e(3, -2*x + 1)/(2*x - 1)^2 + 2/(4*x^2 - 4*x + 1) + 6*integrate(e^(2*x)/(16*x^4 - 32*x^3 + 2
4*x^2 - 8*x + 1), x) - log(x)

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mupad [B]  time = 1.17, size = 22, normalized size = 0.88 \begin {gather*} 2\,x+{\mathrm {e}}^{2\,x}-\ln \relax (x)+\frac {{\mathrm {e}}^5}{{\left (2\,x-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 4*x*exp(5) + exp(2*x)*(2*x - 12*x^2 + 24*x^3 - 16*x^4) - 24*x^2 + 32*x^3 - 16*x^4 - 1)/(x - 6*x^2 +
12*x^3 - 8*x^4),x)

[Out]

2*x + exp(2*x) - log(x) + exp(5)/(2*x - 1)^2

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sympy [A]  time = 0.27, size = 24, normalized size = 0.96 \begin {gather*} 2 x + e^{2 x} - \log {\relax (x )} + \frac {e^{5}}{4 x^{2} - 4 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**4-24*x**3+12*x**2-2*x)*exp(2*x)-4*x*exp(5)+16*x**4-32*x**3+24*x**2-8*x+1)/(8*x**4-12*x**3+6*
x**2-x),x)

[Out]

2*x + exp(2*x) - log(x) + exp(5)/(4*x**2 - 4*x + 1)

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