3.102.50 \(\int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5)}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} (2 x-17 x^2-x^3+12 x^4+4 x^5)} \, dx\)

Optimal. Leaf size=33 \[ \log \left (\frac {5 \left (4 x^2-\frac {x}{2+x}\right )}{\left (16+e^{4 x}\right ) (1-x)}\right ) \]

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Rubi [A]  time = 1.21, antiderivative size = 38, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 7, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6741, 6742, 2282, 36, 29, 31, 628} \begin {gather*} \log \left (-4 x^2-8 x+1\right )-\log \left (e^{4 x}+16\right )-\log (1-x)+\log (x)-\log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 512*x - 240*x^2 + 128*x^3 + 64*x^4 + E^(4*x)*(2 - 40*x + 53*x^2 + 12*x^3 - 44*x^4 - 16*x^5))/(32*x -
272*x^2 - 16*x^3 + 192*x^4 + 64*x^5 + E^(4*x)*(2*x - 17*x^2 - x^3 + 12*x^4 + 4*x^5)),x]

[Out]

-Log[16 + E^(4*x)] - Log[1 - x] + Log[x] - Log[2 + x] + Log[1 - 8*x - 4*x^2]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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 {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{\left (16+e^{4 x}\right ) x \left (2-17 x-x^2+12 x^3+4 x^4\right )} \, dx\\ &=\int \left (\frac {64}{16+e^{4 x}}+\frac {2-40 x+53 x^2+12 x^3-44 x^4-16 x^5}{x \left (2-17 x-x^2+12 x^3+4 x^4\right )}\right ) \, dx\\ &=64 \int \frac {1}{16+e^{4 x}} \, dx+\int \frac {2-40 x+53 x^2+12 x^3-44 x^4-16 x^5}{x \left (2-17 x-x^2+12 x^3+4 x^4\right )} \, dx\\ &=16 \operatorname {Subst}\left (\int \frac {1}{x (16+x)} \, dx,x,e^{4 x}\right )+\int \left (-4+\frac {1}{-2-x}+\frac {1}{1-x}+\frac {1}{x}+\frac {8 (1+x)}{-1+8 x+4 x^2}\right ) \, dx\\ &=-4 x-\log (1-x)+\log (x)-\log (2+x)+8 \int \frac {1+x}{-1+8 x+4 x^2} \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{4 x}\right )-\operatorname {Subst}\left (\int \frac {1}{16+x} \, dx,x,e^{4 x}\right )\\ &=-\log \left (16+e^{4 x}\right )-\log (1-x)+\log (x)-\log (2+x)+\log \left (1-8 x-4 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 37, normalized size = 1.12 \begin {gather*} -\log \left (16+e^{4 x}\right )+\log (x)+\log \left (1-8 x-4 x^2\right )-\log \left (2-x-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 512*x - 240*x^2 + 128*x^3 + 64*x^4 + E^(4*x)*(2 - 40*x + 53*x^2 + 12*x^3 - 44*x^4 - 16*x^5))/(
32*x - 272*x^2 - 16*x^3 + 192*x^4 + 64*x^5 + E^(4*x)*(2*x - 17*x^2 - x^3 + 12*x^4 + 4*x^5)),x]

[Out]

-Log[16 + E^(4*x)] + Log[x] + Log[1 - 8*x - 4*x^2] - Log[2 - x - x^2]

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fricas [A]  time = 0.59, size = 34, normalized size = 1.03 \begin {gather*} \log \left (4 \, x^{3} + 8 \, x^{2} - x\right ) - \log \left (x^{2} + x - 2\right ) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2-512*x+32)/((4*x^5+12*x^4-x^3-
17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-272*x^2+32*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 32, normalized size = 0.97 \begin {gather*} \log \left (4 \, x^{2} + 8 \, x - 1\right ) - \log \left (x^{2} + x - 2\right ) + \log \relax (x) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2-512*x+32)/((4*x^5+12*x^4-x^3-
17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-272*x^2+32*x),x, algorithm="giac")

[Out]

log(4*x^2 + 8*x - 1) - log(x^2 + x - 2) + log(x) - log(e^(4*x) + 16)

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maple [A]  time = 0.23, size = 35, normalized size = 1.06

method	result	size
risch	\(-\ln \left (x^{2}+x -2\right )+\ln \left (4 x^{3}+8 x^{2}-x \right )-\ln \left ({\mathrm e}^{4 x}+16\right )\)	\(35\)
norman	\(-\ln \left (x -1\right )-\ln \left (2+x \right )-\ln \left ({\mathrm e}^{4 x}+16\right )+\ln \relax (x )+\ln \left (4 x^{2}+8 x -1\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2-512*x+32)/((4*x^5+12*x^4-x^3-17*x^2
+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-272*x^2+32*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x^2+x-2)+ln(4*x^3+8*x^2-x)-ln(exp(4*x)+16)

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maxima [A]  time = 0.50, size = 35, normalized size = 1.06 \begin {gather*} \log \left (4 \, x^{2} + 8 \, x - 1\right ) - \log \left (x + 2\right ) - \log \left (x - 1\right ) + \log \relax (x) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2-512*x+32)/((4*x^5+12*x^4-x^3-
17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-272*x^2+32*x),x, algorithm="maxima")

[Out]

log(4*x^2 + 8*x - 1) - log(x + 2) - log(x - 1) + log(x) - log(e^(4*x) + 16)

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mupad [B]  time = 0.42, size = 32, normalized size = 0.97 \begin {gather*} \ln \left (x\,\left (4\,x^2+8\,x-1\right )\right )-\ln \left (x^2+x-2\right )-\ln \left ({\mathrm {e}}^{4\,x}+16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(512*x + exp(4*x)*(40*x - 53*x^2 - 12*x^3 + 44*x^4 + 16*x^5 - 2) + 240*x^2 - 128*x^3 - 64*x^4 - 32)/(32*x
+ exp(4*x)*(2*x - 17*x^2 - x^3 + 12*x^4 + 4*x^5) - 272*x^2 - 16*x^3 + 192*x^4 + 64*x^5),x)

[Out]

log(x*(8*x + 4*x^2 - 1)) - log(x + x^2 - 2) - log(exp(4*x) + 16)

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sympy [A]  time = 0.23, size = 29, normalized size = 0.88 \begin {gather*} - \log {\left (e^{4 x} + 16 \right )} - \log {\left (x^{2} + x - 2 \right )} + \log {\left (4 x^{3} + 8 x^{2} - x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**5-44*x**4+12*x**3+53*x**2-40*x+2)*exp(4*x)+64*x**4+128*x**3-240*x**2-512*x+32)/((4*x**5+12*
x**4-x**3-17*x**2+2*x)*exp(4*x)+64*x**5+192*x**4-16*x**3-272*x**2+32*x),x)

[Out]

-log(exp(4*x) + 16) - log(x**2 + x - 2) + log(4*x**3 + 8*x**2 - x)

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