3.2.39 \(\int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x (-1-3 x^3-8 x^4-5 x^5)+(-2 x-e^x x) \log (\frac {16 x}{25})}{2 x+e^x x} \, dx\)

Optimal. Leaf size=28 \[ \log \left (\frac {2+e^x}{x}\right )-x \left (\left (x+x^2\right )^2+\log \left (\frac {16 x}{25}\right )\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 35, normalized size of antiderivative = 1.25, number of steps used = 11, number of rules used = 7, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6742, 2282, 36, 29, 31, 14, 2295} \begin {gather*} -x^5-2 x^4-x^3-x \log \left (\frac {16 x}{25}\right )+\log \left (e^x+2\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 2*x - 6*x^3 - 16*x^4 - 10*x^5 + E^x*(-1 - 3*x^3 - 8*x^4 - 5*x^5) + (-2*x - E^x*x)*Log[(16*x)/25])/(2
*x + E^x*x),x]

[Out]

-x^3 - 2*x^4 - x^5 + Log[2 + E^x] - x*Log[(16*x)/25] - Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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 {2}{2+e^x}+\frac {-1-3 x^3-8 x^4-5 x^5-x \log \left (\frac {16 x}{25}\right )}{x}\right ) \, dx\\ &=-\left (2 \int \frac {1}{2+e^x} \, dx\right )+\int \frac {-1-3 x^3-8 x^4-5 x^5-x \log \left (\frac {16 x}{25}\right )}{x} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^x\right )\right )+\int \left (\frac {-1-3 x^3-8 x^4-5 x^5}{x}-\log \left (\frac {16 x}{25}\right )\right ) \, dx\\ &=\int \frac {-1-3 x^3-8 x^4-5 x^5}{x} \, dx-\int \log \left (\frac {16 x}{25}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^x\right )\\ &=\log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )+\int \left (-\frac {1}{x}-3 x^2-8 x^3-5 x^4\right ) \, dx\\ &=-x^3-2 x^4-x^5+\log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 35, normalized size = 1.25 \begin {gather*} -x^3-2 x^4-x^5+\log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 2*x - 6*x^3 - 16*x^4 - 10*x^5 + E^x*(-1 - 3*x^3 - 8*x^4 - 5*x^5) + (-2*x - E^x*x)*Log[(16*x)/2
5])/(2*x + E^x*x),x]

[Out]

-x^3 - 2*x^4 - x^5 + Log[2 + E^x] - x*Log[(16*x)/25] - Log[x]

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fricas [A]  time = 0.73, size = 32, normalized size = 1.14 \begin {gather*} -x^{5} - 2 \, x^{4} - x^{3} - x \log \left (\frac {16}{25} \, x\right ) - \log \relax (x) + \log \left (e^{x} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-2*x)*log(16/25*x)+(-5*x^5-8*x^4-3*x^3-1)*exp(x)-10*x^5-16*x^4-6*x^3-2*x-2)/(exp(x)*x+2*x
),x, algorithm="fricas")

[Out]

-x^5 - 2*x^4 - x^3 - x*log(16/25*x) - log(x) + log(e^x + 2)

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giac [A]  time = 0.47, size = 32, normalized size = 1.14 \begin {gather*} -x^{5} - 2 \, x^{4} - x^{3} - x \log \left (\frac {16}{25} \, x\right ) - \log \relax (x) + \log \left (e^{x} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-2*x)*log(16/25*x)+(-5*x^5-8*x^4-3*x^3-1)*exp(x)-10*x^5-16*x^4-6*x^3-2*x-2)/(exp(x)*x+2*x
),x, algorithm="giac")

[Out]

-x^5 - 2*x^4 - x^3 - x*log(16/25*x) - log(x) + log(e^x + 2)

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maple [A]  time = 0.03, size = 33, normalized size = 1.18

method	result	size
risch	\(-\ln \left (\frac {16 x}{25}\right ) x -x^{5}-2 x^{4}-x^{3}-\ln \relax (x )+\ln \left ({\mathrm e}^{x}+2\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x-2*x)*ln(16/25*x)+(-5*x^5-8*x^4-3*x^3-1)*exp(x)-10*x^5-16*x^4-6*x^3-2*x-2)/(exp(x)*x+2*x),x,met
hod=_RETURNVERBOSE)

[Out]

-ln(16/25*x)*x-x^5-2*x^4-x^3-ln(x)+ln(exp(x)+2)

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maxima [A]  time = 0.52, size = 38, normalized size = 1.36 \begin {gather*} -x^{5} - 2 \, x^{4} - x^{3} + 2 \, x {\left (\log \relax (5) - 2 \, \log \relax (2)\right )} - {\left (x + 1\right )} \log \relax (x) + \log \left (e^{x} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-2*x)*log(16/25*x)+(-5*x^5-8*x^4-3*x^3-1)*exp(x)-10*x^5-16*x^4-6*x^3-2*x-2)/(exp(x)*x+2*x
),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.36, size = 32, normalized size = 1.14 \begin {gather*} \ln \left ({\mathrm {e}}^x+2\right )-\ln \relax (x)-x\,\ln \left (\frac {16\,x}{25}\right )-x^3-2\,x^4-x^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + exp(x)*(3*x^3 + 8*x^4 + 5*x^5 + 1) + log((16*x)/25)*(2*x + x*exp(x)) + 6*x^3 + 16*x^4 + 10*x^5 + 2
)/(2*x + x*exp(x)),x)

[Out]

log(exp(x) + 2) - log(x) - x*log((16*x)/25) - x^3 - 2*x^4 - x^5

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sympy [A]  time = 0.31, size = 29, normalized size = 1.04 \begin {gather*} - x^{5} - 2 x^{4} - x^{3} - x \log {\left (\frac {16 x}{25} \right )} - \log {\relax (x )} + \log {\left (e^{x} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-2*x)*ln(16/25*x)+(-5*x**5-8*x**4-3*x**3-1)*exp(x)-10*x**5-16*x**4-6*x**3-2*x-2)/(exp(x)*
x+2*x),x)

[Out]

-x**5 - 2*x**4 - x**3 - x*log(16*x/25) - log(x) + log(exp(x) + 2)

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