3.49.37 \(\int \frac {1}{2} (-94-188 x+94 \log (5)+e^x (-47-94 x-47 x^2+(47+47 x) \log (5))+(-94+e^x (-47-47 x)) \log (x)) \, dx\)

Optimal. Leaf size=22 \[ \frac {3}{2}-\frac {47}{2} \left (2+e^x\right ) x (x-\log (5)+\log (x)) \]

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Rubi [B]  time = 0.09, antiderivative size = 75, normalized size of antiderivative = 3.41, number of steps used = 16, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 2196, 2194, 2176, 2554} \begin {gather*} -\frac {47}{2} e^x x^2-47 x^2+47 x-47 x \log (x)-47 x (1-\log (5))+\frac {47}{2} e^x \log (x)-\frac {47}{2} e^x (x+1) \log (x)-\frac {47}{2} e^x \log (5)+\frac {47}{2} e^x (x+1) \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-94 - 188*x + 94*Log[5] + E^x*(-47 - 94*x - 47*x^2 + (47 + 47*x)*Log[5]) + (-94 + E^x*(-47 - 47*x))*Log[x
])/2,x]

[Out]

47*x - 47*x^2 - (47*E^x*x^2)/2 - 47*x*(1 - Log[5]) - (47*E^x*Log[5])/2 + (47*E^x*(1 + x)*Log[5])/2 + (47*E^x*L
og[x])/2 - 47*x*Log[x] - (47*E^x*(1 + x)*Log[x])/2

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (-94-188 x+94 \log (5)+e^x \left (-47-94 x-47 x^2+(47+47 x) \log (5)\right )+\left (-94+e^x (-47-47 x)\right ) \log (x)\right ) \, dx\\ &=-47 x^2-47 x (1-\log (5))+\frac {1}{2} \int e^x \left (-47-94 x-47 x^2+(47+47 x) \log (5)\right ) \, dx+\frac {1}{2} \int \left (-94+e^x (-47-47 x)\right ) \log (x) \, dx\\ &=-47 x^2-47 x (1-\log (5))+\frac {47}{2} e^x \log (x)-47 x \log (x)-\frac {47}{2} e^x (1+x) \log (x)-\frac {1}{2} \int 47 \left (-2-e^x\right ) \, dx+\frac {1}{2} \int \left (-47 e^x-94 e^x x-47 e^x x^2+47 e^x (1+x) \log (5)\right ) \, dx\\ &=-47 x^2-47 x (1-\log (5))+\frac {47}{2} e^x \log (x)-47 x \log (x)-\frac {47}{2} e^x (1+x) \log (x)-\frac {47 \int e^x \, dx}{2}-\frac {47}{2} \int \left (-2-e^x\right ) \, dx-\frac {47}{2} \int e^x x^2 \, dx-47 \int e^x x \, dx+\frac {1}{2} (47 \log (5)) \int e^x (1+x) \, dx\\ &=-\frac {47 e^x}{2}+47 x-47 e^x x-47 x^2-\frac {47 e^x x^2}{2}-47 x (1-\log (5))+\frac {47}{2} e^x (1+x) \log (5)+\frac {47}{2} e^x \log (x)-47 x \log (x)-\frac {47}{2} e^x (1+x) \log (x)+\frac {47 \int e^x \, dx}{2}+47 \int e^x \, dx+47 \int e^x x \, dx-\frac {1}{2} (47 \log (5)) \int e^x \, dx\\ &=47 e^x+47 x-47 x^2-\frac {47 e^x x^2}{2}-47 x (1-\log (5))-\frac {47}{2} e^x \log (5)+\frac {47}{2} e^x (1+x) \log (5)+\frac {47}{2} e^x \log (x)-47 x \log (x)-\frac {47}{2} e^x (1+x) \log (x)-47 \int e^x \, dx\\ &=47 x-47 x^2-\frac {47 e^x x^2}{2}-47 x (1-\log (5))-\frac {47}{2} e^x \log (5)+\frac {47}{2} e^x (1+x) \log (5)+\frac {47}{2} e^x \log (x)-47 x \log (x)-\frac {47}{2} e^x (1+x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 18, normalized size = 0.82 \begin {gather*} -\frac {47}{2} \left (2+e^x\right ) x \left (x+\log \left (\frac {x}{5}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-94 - 188*x + 94*Log[5] + E^x*(-47 - 94*x - 47*x^2 + (47 + 47*x)*Log[5]) + (-94 + E^x*(-47 - 47*x))
*Log[x])/2,x]

[Out]

(-47*(2 + E^x)*x*(x + Log[x/5]))/2

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fricas [B]  time = 0.54, size = 36, normalized size = 1.64 \begin {gather*} -47 \, x^{2} - \frac {47}{2} \, {\left (x^{2} - x \log \relax (5)\right )} e^{x} + 47 \, x \log \relax (5) - \frac {47}{2} \, {\left (x e^{x} + 2 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-47*x-47)*exp(x)-94)*log(x)+1/2*((47*x+47)*log(5)-47*x^2-94*x-47)*exp(x)+47*log(5)-94*x-47,x,
algorithm="fricas")

[Out]

-47*x^2 - 47/2*(x^2 - x*log(5))*e^x + 47*x*log(5) - 47/2*(x*e^x + 2*x)*log(x)

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giac [B]  time = 0.12, size = 41, normalized size = 1.86 \begin {gather*} -47 \, x^{2} - \frac {47}{2} \, {\left (x^{2} - x \log \relax (5) + 1\right )} e^{x} + 47 \, x \log \relax (5) - \frac {47}{2} \, {\left (x e^{x} + 2 \, x\right )} \log \relax (x) + \frac {47}{2} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-47*x-47)*exp(x)-94)*log(x)+1/2*((47*x+47)*log(5)-47*x^2-94*x-47)*exp(x)+47*log(5)-94*x-47,x,
algorithm="giac")

[Out]

-47*x^2 - 47/2*(x^2 - x*log(5) + 1)*e^x + 47*x*log(5) - 47/2*(x*e^x + 2*x)*log(x) + 47/2*e^x

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maple [A]  time = 0.06, size = 38, normalized size = 1.73

method	result	size
default	\(-47 x^{2}-\frac {47 x \,{\mathrm e}^{x} \ln \relax (x )}{2}-47 x \ln \relax (x )-\frac {47 \,{\mathrm e}^{x} x^{2}}{2}+\frac {47 x \,{\mathrm e}^{x} \ln \relax (5)}{2}+47 x \ln \relax (5)\)	\(38\)
norman	\(-47 x^{2}-\frac {47 x \,{\mathrm e}^{x} \ln \relax (x )}{2}-47 x \ln \relax (x )-\frac {47 \,{\mathrm e}^{x} x^{2}}{2}+\frac {47 x \,{\mathrm e}^{x} \ln \relax (5)}{2}+47 x \ln \relax (5)\)	\(38\)
risch	\(-47 x^{2}-\frac {47 x \,{\mathrm e}^{x} \ln \relax (x )}{2}-47 x \ln \relax (x )-\frac {47 \,{\mathrm e}^{x} x^{2}}{2}+\frac {47 x \,{\mathrm e}^{x} \ln \relax (5)}{2}+47 x \ln \relax (5)\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-47*x-47)*exp(x)-94)*ln(x)+1/2*((47*x+47)*ln(5)-47*x^2-94*x-47)*exp(x)+47*ln(5)-94*x-47,x,method=_RE
TURNVERBOSE)

[Out]

-47*x^2-47/2*x*exp(x)*ln(x)-47*x*ln(x)-47/2*exp(x)*x^2+47/2*x*exp(x)*ln(5)+47*x*ln(5)

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maxima [B]  time = 0.45, size = 41, normalized size = 1.86 \begin {gather*} -47 \, x^{2} - \frac {47}{2} \, {\left (x^{2} - x \log \relax (5) + 1\right )} e^{x} + 47 \, x \log \relax (5) - \frac {47}{2} \, {\left (x e^{x} + 2 \, x\right )} \log \relax (x) + \frac {47}{2} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-47*x-47)*exp(x)-94)*log(x)+1/2*((47*x+47)*log(5)-47*x^2-94*x-47)*exp(x)+47*log(5)-94*x-47,x,
algorithm="maxima")

[Out]

-47*x^2 - 47/2*(x^2 - x*log(5) + 1)*e^x + 47*x*log(5) - 47/2*(x*e^x + 2*x)*log(x) + 47/2*e^x

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mupad [B]  time = 3.33, size = 13, normalized size = 0.59 \begin {gather*} -\frac {47\,x\,\left (x+\ln \left (\frac {x}{5}\right )\right )\,\left ({\mathrm {e}}^x+2\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(47*log(5) - 94*x - (log(x)*(exp(x)*(47*x + 47) + 94))/2 - (exp(x)*(94*x - log(5)*(47*x + 47) + 47*x^2 + 47
))/2 - 47,x)

[Out]

-(47*x*(x + log(x/5))*(exp(x) + 2))/2

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sympy [A]  time = 0.38, size = 41, normalized size = 1.86 \begin {gather*} - 47 x^{2} - 47 x \log {\relax (x )} + 47 x \log {\relax (5 )} + \frac {\left (- 47 x^{2} - 47 x \log {\relax (x )} + 47 x \log {\relax (5 )}\right ) e^{x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-47*x-47)*exp(x)-94)*ln(x)+1/2*((47*x+47)*ln(5)-47*x**2-94*x-47)*exp(x)+47*ln(5)-94*x-47,x)

[Out]

-47*x**2 - 47*x*log(x) + 47*x*log(5) + (-47*x**2 - 47*x*log(x) + 47*x*log(5))*exp(x)/2

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