3.22.75 \(\int \frac {-5 x-e^x x \log (2)-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x \log (2)} \, dx\)

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

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2194, 2302, 30} \begin {gather*} -2 x^2-e^x+2 \log ^4(x)-\frac {5 x}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x - E^x*x*Log[2] - 4*x^2*Log[2] + 8*Log[2]*Log[x]^3)/(x*Log[2]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 30

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

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-5 x-e^x x \log (2)-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x} \, dx}{\log (2)}\\ &=\frac {\int \left (-e^x \log (2)+\frac {-5 x-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x}\right ) \, dx}{\log (2)}\\ &=\frac {\int \frac {-5 x-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x} \, dx}{\log (2)}-\int e^x \, dx\\ &=-e^x+\frac {\int \left (-5-4 x \log (2)+\frac {8 \log (2) \log ^3(x)}{x}\right ) \, dx}{\log (2)}\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+8 \int \frac {\log ^3(x)}{x} \, dx\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+8 \operatorname {Subst}\left (\int x^3 \, dx,x,\log (x)\right )\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+2 \log ^4(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 32, normalized size = 1.28 \begin {gather*} -e^x-\frac {5 x}{\log (2)}-\frac {x^2 \log (16)}{2 \log (2)}+2 \log ^4(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)*log(x)^3-x*log(2)*exp(x)-4*x^2*log(2)-5*x)/x/log(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)*log(x)^3-x*log(2)*exp(x)-4*x^2*log(2)-5*x)/x/log(2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 24, normalized size = 0.96

method	result	size
risch	\(-\frac {5 x}{\ln \relax (2)}-2 x^{2}-{\mathrm e}^{x}+2 \ln \relax (x )^{4}\)	\(24\)
default	\(\frac {-5 x -2 x^{2} \ln \relax (2)-{\mathrm e}^{x} \ln \relax (2)+2 \ln \relax (2) \ln \relax (x )^{4}}{\ln \relax (2)}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*ln(2)*ln(x)^3-x*ln(2)*exp(x)-4*x^2*ln(2)-5*x)/x/ln(2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)*log(x)^3-x*log(2)*exp(x)-4*x^2*log(2)-5*x)/x/log(2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.28, size = 23, normalized size = 0.92 \begin {gather*} 2\,{\ln \relax (x)}^4-{\mathrm {e}}^x-\frac {5\,x}{\ln \relax (2)}-2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - 8*log(2)*log(x)^3 + 4*x^2*log(2) + x*exp(x)*log(2))/(x*log(2)),x)

[Out]

2*log(x)^4 - exp(x) - (5*x)/log(2) - 2*x^2

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sympy [A]  time = 0.28, size = 20, normalized size = 0.80 \begin {gather*} - 2 x^{2} - \frac {5 x}{\log {\relax (2 )}} - e^{x} + 2 \log {\relax (x )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*ln(2)*ln(x)**3-x*ln(2)*exp(x)-4*x**2*ln(2)-5*x)/x/ln(2),x)

[Out]

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

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