3.64.26 \(\int \frac {-16-x-x^3+e (16 x^2+5 x^3)+x^3 \log (5)+(32+3 x-e x^3) \log (x)}{256 x^3+32 x^4+x^5} \, dx\)

Optimal. Leaf size=28 \[ \frac {1-\log (5)-e (4-\log (x))-\frac {\log (x)}{x^2}}{16+x} \]

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Rubi [B]  time = 0.39, antiderivative size = 99, normalized size of antiderivative = 3.54, number of steps used = 17, number of rules used = 10, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6, 1594, 27, 6742, 44, 77, 2357, 2304, 2314, 31} \begin {gather*} -\frac {\log (x)}{16 x^2}-\frac {4 e}{x+16}+\frac {\log (x)}{256 x}+\frac {(1-256 e) x \log (x)}{4096 (x+16)}+\frac {1}{16} e \log (x)-\frac {\log (x)}{4096}-\frac {1}{16} e \log (x+16)-\frac {(1-256 e) \log (x+16)}{4096}+\frac {\log (x+16)}{4096}+\frac {1-\log (5)}{x+16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - x - x^3 + E*(16*x^2 + 5*x^3) + x^3*Log[5] + (32 + 3*x - E*x^3)*Log[x])/(256*x^3 + 32*x^4 + x^5),x]

[Out]

(-4*E)/(16 + x) + (1 - Log[5])/(16 + x) - Log[x]/4096 + (E*Log[x])/16 - Log[x]/(16*x^2) + Log[x]/(256*x) + ((1
- 256*E)*x*Log[x])/(4096*(16 + x)) + Log[16 + x]/4096 - ((1 - 256*E)*Log[16 + x])/4096 - (E*Log[16 + x])/16

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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

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

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2304

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

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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 {-16-x+e \left (16 x^2+5 x^3\right )+x^3 (-1+\log (5))+\left (32+3 x-e x^3\right ) \log (x)}{256 x^3+32 x^4+x^5} \, dx\\ &=\int \frac {-16-x+e \left (16 x^2+5 x^3\right )+x^3 (-1+\log (5))+\left (32+3 x-e x^3\right ) \log (x)}{x^3 \left (256+32 x+x^2\right )} \, dx\\ &=\int \frac {-16-x+e \left (16 x^2+5 x^3\right )+x^3 (-1+\log (5))+\left (32+3 x-e x^3\right ) \log (x)}{x^3 (16+x)^2} \, dx\\ &=\int \left (-\frac {16}{x^3 (16+x)^2}-\frac {1}{x^2 (16+x)^2}+\frac {e (16+5 x)}{x (16+x)^2}+\frac {-1+\log (5)}{(16+x)^2}-\frac {\left (-32-3 x+e x^3\right ) \log (x)}{x^3 (16+x)^2}\right ) \, dx\\ &=\frac {1-\log (5)}{16+x}-16 \int \frac {1}{x^3 (16+x)^2} \, dx+e \int \frac {16+5 x}{x (16+x)^2} \, dx-\int \frac {1}{x^2 (16+x)^2} \, dx-\int \frac {\left (-32-3 x+e x^3\right ) \log (x)}{x^3 (16+x)^2} \, dx\\ &=\frac {1-\log (5)}{16+x}-16 \int \left (\frac {1}{256 x^3}-\frac {1}{2048 x^2}+\frac {3}{65536 x}-\frac {1}{4096 (16+x)^2}-\frac {3}{65536 (16+x)}\right ) \, dx+e \int \left (\frac {1}{16 x}+\frac {4}{(16+x)^2}-\frac {1}{16 (16+x)}\right ) \, dx-\int \left (\frac {1}{256 x^2}-\frac {1}{2048 x}+\frac {1}{256 (16+x)^2}+\frac {1}{2048 (16+x)}\right ) \, dx-\int \left (-\frac {\log (x)}{8 x^3}+\frac {\log (x)}{256 x^2}+\frac {(-1+256 e) \log (x)}{256 (16+x)^2}\right ) \, dx\\ &=\frac {1}{32 x^2}-\frac {1}{256 x}-\frac {4 e}{16+x}+\frac {1-\log (5)}{16+x}-\frac {\log (x)}{4096}+\frac {1}{16} e \log (x)+\frac {\log (16+x)}{4096}-\frac {1}{16} e \log (16+x)-\frac {1}{256} \int \frac {\log (x)}{x^2} \, dx+\frac {1}{8} \int \frac {\log (x)}{x^3} \, dx-\frac {1}{256} (-1+256 e) \int \frac {\log (x)}{(16+x)^2} \, dx\\ &=-\frac {4 e}{16+x}+\frac {1-\log (5)}{16+x}-\frac {\log (x)}{4096}+\frac {1}{16} e \log (x)-\frac {\log (x)}{16 x^2}+\frac {\log (x)}{256 x}+\frac {(1-256 e) x \log (x)}{4096 (16+x)}+\frac {\log (16+x)}{4096}-\frac {1}{16} e \log (16+x)-\frac {(1-256 e) \int \frac {1}{16+x} \, dx}{4096}\\ &=-\frac {4 e}{16+x}+\frac {1-\log (5)}{16+x}-\frac {\log (x)}{4096}+\frac {1}{16} e \log (x)-\frac {\log (x)}{16 x^2}+\frac {\log (x)}{256 x}+\frac {(1-256 e) x \log (x)}{4096 (16+x)}+\frac {\log (16+x)}{4096}-\frac {(1-256 e) \log (16+x)}{4096}-\frac {1}{16} e \log (16+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 32, normalized size = 1.14 \begin {gather*} \frac {-x^2 (-1+4 e+\log (5))+\left (-1+e x^2\right ) \log (x)}{x^2 (16+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - x - x^3 + E*(16*x^2 + 5*x^3) + x^3*Log[5] + (32 + 3*x - E*x^3)*Log[x])/(256*x^3 + 32*x^4 + x^
5),x]

[Out]

(-(x^2*(-1 + 4*E + Log[5])) + (-1 + E*x^2)*Log[x])/(x^2*(16 + x))

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fricas [A]  time = 0.63, size = 44, normalized size = 1.57 \begin {gather*} -\frac {4 \, x^{2} e + x^{2} \log \relax (5) - x^{2} - {\left (x^{2} e - 1\right )} \log \relax (x)}{x^{3} + 16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16)/(x^5+32*x^4+256*x^3),x, algo
rithm="fricas")

[Out]

-(4*x^2*e + x^2*log(5) - x^2 - (x^2*e - 1)*log(x))/(x^3 + 16*x^2)

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giac [A]  time = 0.19, size = 42, normalized size = 1.50 \begin {gather*} \frac {x^{2} e \log \relax (x) - 4 \, x^{2} e - x^{2} \log \relax (5) + x^{2} - \log \relax (x)}{x^{3} + 16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16)/(x^5+32*x^4+256*x^3),x, algo
rithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 37, normalized size = 1.32

method	result	size
norman	\(\frac {\left (1-4 \,{\mathrm e}-\ln \relax (5)\right ) x^{2}+x^{2} {\mathrm e} \ln \relax (x )-\ln \relax (x )}{x^{2} \left (x +16\right )}\)	\(37\)
risch	\(\frac {\left (x^{2} {\mathrm e}-1\right ) \ln \relax (x )}{x^{2} \left (x +16\right )}+\frac {1}{x +16}-\frac {4 \,{\mathrm e}}{x +16}-\frac {\ln \relax (5)}{x +16}\)	\(44\)
default	\(-\frac {\ln \relax (5)}{x +16}-\frac {4 \,{\mathrm e}}{x +16}+\frac {1}{x +16}-\frac {\ln \relax (x )}{4096}+\frac {{\mathrm e} \ln \relax (x )}{16}-\frac {\ln \relax (x )}{16 x^{2}}+\frac {\ln \relax (x )}{256 x}-\frac {{\mathrm e} \ln \relax (x ) x}{16 \left (x +16\right )}+\frac {x \ln \relax (x )}{4096 x +65536}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3*exp(1)+3*x+32)*ln(x)+x^3*ln(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16)/(x^5+32*x^4+256*x^3),x,method=_RETUR
NVERBOSE)

[Out]

((1-4*exp(1)-ln(5))*x^2+x^2*exp(1)*ln(x)-ln(x))/x^2/(x+16)

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maxima [B]  time = 0.41, size = 147, normalized size = 5.25 \begin {gather*} \frac {1}{16} \, {\left (\frac {16}{x + 16} - \log \left (x + 16\right ) + \log \relax (x)\right )} e + \frac {1}{4096} \, {\left (256 \, e - 1\right )} \log \left (x + 16\right ) + \frac {16 \, x^{2} - {\left (x^{3} {\left (256 \, e - 1\right )} - 16 \, x^{2} + 4096\right )} \log \relax (x) + 128 \, x - 2048}{4096 \, {\left (x^{3} + 16 \, x^{2}\right )}} - \frac {3 \, x^{2} + 24 \, x - 128}{256 \, {\left (x^{3} + 16 \, x^{2}\right )}} + \frac {x + 8}{128 \, {\left (x^{2} + 16 \, x\right )}} - \frac {5 \, e}{x + 16} - \frac {\log \relax (5)}{x + 16} + \frac {1}{x + 16} + \frac {1}{4096} \, \log \left (x + 16\right ) - \frac {1}{4096} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3*exp(1)+3*x+32)*log(x)+x^3*log(5)+(5*x^3+16*x^2)*exp(1)-x^3-x-16)/(x^5+32*x^4+256*x^3),x, algo
rithm="maxima")

[Out]

1/16*(16/(x + 16) - log(x + 16) + log(x))*e + 1/4096*(256*e - 1)*log(x + 16) + 1/4096*(16*x^2 - (x^3*(256*e -
1) - 16*x^2 + 4096)*log(x) + 128*x - 2048)/(x^3 + 16*x^2) - 1/256*(3*x^2 + 24*x - 128)/(x^3 + 16*x^2) + 1/128*
(x + 8)/(x^2 + 16*x) - 5*e/(x + 16) - log(5)/(x + 16) + 1/(x + 16) + 1/4096*log(x + 16) - 1/4096*log(x)

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mupad [B]  time = 4.35, size = 40, normalized size = 1.43 \begin {gather*} \frac {x^4\,\left (\frac {\mathrm {e}}{4}+\frac {\ln \relax (5)}{16}-\frac {1}{16}\right )-x\,\ln \relax (x)+x^3\,\mathrm {e}\,\ln \relax (x)}{x^4+16\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - log(x)*(3*x - x^3*exp(1) + 32) - exp(1)*(16*x^2 + 5*x^3) - x^3*log(5) + x^3 + 16)/(256*x^3 + 32*x^4
+ x^5),x)

[Out]

(x^4*(exp(1)/4 + log(5)/16 - 1/16) - x*log(x) + x^3*exp(1)*log(x))/(16*x^3 + x^4)

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sympy [A]  time = 0.25, size = 32, normalized size = 1.14 \begin {gather*} \frac {\left (e x^{2} - 1\right ) \log {\relax (x )}}{x^{3} + 16 x^{2}} - \frac {-1 + \log {\relax (5 )} + 4 e}{x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3*exp(1)+3*x+32)*ln(x)+x**3*ln(5)+(5*x**3+16*x**2)*exp(1)-x**3-x-16)/(x**5+32*x**4+256*x**3),x
)

[Out]

(E*x**2 - 1)*log(x)/(x**3 + 16*x**2) - (-1 + log(5) + 4*E)/(x + 16)

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