3.52.16 \(\int \frac {e^x (-4+(-6 x-4 x^2) \log (3)) (2+(3 x+2 x^2) \log (3)+(2 x+(3 x+7 x^2+2 x^3) \log (3)) \log (x))}{\log (3) (2 x+(3 x^2+2 x^3) \log (3))} \, dx\)

Optimal. Leaf size=22 \[ e^x \left (2 x-4 x (2+x)-\frac {4}{\log (3)}\right ) \log (x) \]

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Rubi [B]  time = 0.41, antiderivative size = 74, normalized size of antiderivative = 3.36, number of steps used = 18, number of rules used = 9, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {12, 1586, 6742, 2199, 2178, 2176, 2194, 2196, 2554} \begin {gather*} -\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-14 e^x x \log (x)+\frac {4 e^x x \log (9) \log (x)}{\log (3)}+14 e^x \log (x)-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}-\frac {4 e^x \log (9) \log (x)}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-4 + (-6*x - 4*x^2)*Log[3])*(2 + (3*x + 2*x^2)*Log[3] + (2*x + (3*x + 7*x^2 + 2*x^3)*Log[3])*Log[x])
)/(Log[3]*(2*x + (3*x^2 + 2*x^3)*Log[3])),x]

[Out]

14*E^x*Log[x] - 14*E^x*x*Log[x] - (4*E^x*Log[9]*Log[x])/Log[3] + (4*E^x*x*Log[9]*Log[x])/Log[3] - (2*E^x*x^2*L
og[9]*Log[x])/Log[3] - (2*E^x*(2 + Log[27])*Log[x])/Log[3]

Rule 12

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^x \left (-4+\left (-6 x-4 x^2\right ) \log (3)\right ) \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{2 x+\left (3 x^2+2 x^3\right ) \log (3)} \, dx}{\log (3)}\\ &=\frac {\int -\frac {2 e^x \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{x} \, dx}{\log (3)}\\ &=-\frac {2 \int \frac {e^x \left (2+\left (3 x+2 x^2\right ) \log (3)+\left (2 x+\left (3 x+7 x^2+2 x^3\right ) \log (3)\right ) \log (x)\right )}{x} \, dx}{\log (3)}\\ &=-\frac {2 \int \left (\frac {e^x \left (2+x^2 \log (9)+x \log (27)\right )}{x}+e^x \left (2+7 x \log (3)+x^2 \log (9)+\log (27)\right ) \log (x)\right ) \, dx}{\log (3)}\\ &=-\frac {2 \int \frac {e^x \left (2+x^2 \log (9)+x \log (27)\right )}{x} \, dx}{\log (3)}-\frac {2 \int e^x \left (2+7 x \log (3)+x^2 \log (9)+\log (27)\right ) \log (x) \, dx}{\log (3)}\\ &=14 e^x \log (x)-14 e^x x \log (x)-\frac {4 e^x \log (9) \log (x)}{\log (3)}+\frac {4 e^x x \log (9) \log (x)}{\log (3)}-\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}-\frac {2 \int \left (\frac {2 e^x}{x}+e^x x \log (9)+e^x \log (27)\right ) \, dx}{\log (3)}+\frac {2 \int \frac {e^x \left (2+x^2 \log (9)+x \log (27)\right )}{x} \, dx}{\log (3)}\\ &=14 e^x \log (x)-14 e^x x \log (x)-\frac {4 e^x \log (9) \log (x)}{\log (3)}+\frac {4 e^x x \log (9) \log (x)}{\log (3)}-\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}+\frac {2 \int \left (\frac {2 e^x}{x}+e^x x \log (9)+e^x \log (27)\right ) \, dx}{\log (3)}-\frac {4 \int \frac {e^x}{x} \, dx}{\log (3)}-\frac {(2 \log (9)) \int e^x x \, dx}{\log (3)}-\frac {(2 \log (27)) \int e^x \, dx}{\log (3)}\\ &=-\frac {4 \text {Ei}(x)}{\log (3)}-\frac {2 e^x x \log (9)}{\log (3)}-\frac {2 e^x \log (27)}{\log (3)}+14 e^x \log (x)-14 e^x x \log (x)-\frac {4 e^x \log (9) \log (x)}{\log (3)}+\frac {4 e^x x \log (9) \log (x)}{\log (3)}-\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}+\frac {4 \int \frac {e^x}{x} \, dx}{\log (3)}+\frac {(2 \log (9)) \int e^x \, dx}{\log (3)}+\frac {(2 \log (9)) \int e^x x \, dx}{\log (3)}+\frac {(2 \log (27)) \int e^x \, dx}{\log (3)}\\ &=\frac {2 e^x \log (9)}{\log (3)}+14 e^x \log (x)-14 e^x x \log (x)-\frac {4 e^x \log (9) \log (x)}{\log (3)}+\frac {4 e^x x \log (9) \log (x)}{\log (3)}-\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}-\frac {(2 \log (9)) \int e^x \, dx}{\log (3)}\\ &=14 e^x \log (x)-14 e^x x \log (x)-\frac {4 e^x \log (9) \log (x)}{\log (3)}+\frac {4 e^x x \log (9) \log (x)}{\log (3)}-\frac {2 e^x x^2 \log (9) \log (x)}{\log (3)}-\frac {2 e^x (2+\log (27)) \log (x)}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.05 \begin {gather*} -\frac {2 e^x \left (2+x^2 \log (9)+x \log (27)\right ) \log (x)}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-4 + (-6*x - 4*x^2)*Log[3])*(2 + (3*x + 2*x^2)*Log[3] + (2*x + (3*x + 7*x^2 + 2*x^3)*Log[3])*L
og[x]))/(Log[3]*(2*x + (3*x^2 + 2*x^3)*Log[3])),x]

[Out]

(-2*E^x*(2 + x^2*Log[9] + x*Log[27])*Log[x])/Log[3]

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fricas [A]  time = 0.53, size = 26, normalized size = 1.18 \begin {gather*} e^{\left (x + \log \left (-\frac {2 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \relax (3) + 2\right )} \log \relax (x)}{\log \relax (3)}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp(log(((-4*x^2-6*x)*log(3)-4)*log(x)/
log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/log(x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 33, normalized size = 1.50 \begin {gather*} -\frac {2 \, {\left (2 \, x^{2} e^{x} \log \relax (3) \log \relax (x) + 3 \, x e^{x} \log \relax (3) \log \relax (x) + 2 \, e^{x} \log \relax (x)\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp(log(((-4*x^2-6*x)*log(3)-4)*log(x)/
log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/log(x),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.11, size = 186, normalized size = 8.45

method	result	size
risch	\(-\frac {4 \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right ) \ln \relax (x ) {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )}{2}+\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )^{2} \mathrm {csgn}\left (i \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i \ln \relax (x ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )}{2}-i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (1+\left (x^{2}+\frac {3}{2} x \right ) \ln \relax (3)\right )\right )^{2}+x}}{\ln \relax (3)}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3+7*x^2+3*x)*ln(3)+2*x)*ln(x)+(2*x^2+3*x)*ln(3)+2)*exp(ln(((-4*x^2-6*x)*ln(3)-4)*ln(x)/ln(3))+x)/((
2*x^3+3*x^2)*ln(3)+2*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-4/ln(3)*(1+(x^2+3/2*x)*ln(3))*ln(x)*exp(1/2*I*Pi*csgn(I*ln(x)*(1+(x^2+3/2*x)*ln(3)))^3+1/2*I*Pi*csgn(I*ln(x)*
(1+(x^2+3/2*x)*ln(3)))^2*csgn(I*ln(x))+1/2*I*Pi*csgn(I*ln(x)*(1+(x^2+3/2*x)*ln(3)))^2*csgn(I*(1+(x^2+3/2*x)*ln
(3)))-1/2*I*Pi*csgn(I*ln(x)*(1+(x^2+3/2*x)*ln(3)))*csgn(I*ln(x))*csgn(I*(1+(x^2+3/2*x)*ln(3)))-I*Pi*csgn(I*ln(
x)*(1+(x^2+3/2*x)*ln(3)))^2+x)

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maxima [A]  time = 0.48, size = 24, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (2 \, x^{2} \log \relax (3) + 3 \, x \log \relax (3) + 2\right )} e^{x} \log \relax (x)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+7*x^2+3*x)*log(3)+2*x)*log(x)+(2*x^2+3*x)*log(3)+2)*exp(log(((-4*x^2-6*x)*log(3)-4)*log(x)/
log(3))+x)/((2*x^3+3*x^2)*log(3)+2*x)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.18, size = 24, normalized size = 1.09 \begin {gather*} -\frac {2\,{\mathrm {e}}^x\,\ln \relax (x)\,\left (2\,\ln \relax (3)\,x^2+3\,\ln \relax (3)\,x+2\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(-(log(x)*(log(3)*(6*x + 4*x^2) + 4))/log(3)))*(log(3)*(3*x + 2*x^2) + log(x)*(2*x + log(3)*(3
*x + 7*x^2 + 2*x^3)) + 2))/(log(x)*(2*x + log(3)*(3*x^2 + 2*x^3))),x)

[Out]

-(2*exp(x)*log(x)*(3*x*log(3) + 2*x^2*log(3) + 2))/log(3)

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sympy [A]  time = 0.43, size = 34, normalized size = 1.55 \begin {gather*} \frac {\left (- 4 x^{2} \log {\relax (3 )} \log {\relax (x )} - 6 x \log {\relax (3 )} \log {\relax (x )} - 4 \log {\relax (x )}\right ) e^{x}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3+7*x**2+3*x)*ln(3)+2*x)*ln(x)+(2*x**2+3*x)*ln(3)+2)*exp(ln(((-4*x**2-6*x)*ln(3)-4)*ln(x)/ln
(3))+x)/((2*x**3+3*x**2)*ln(3)+2*x)/ln(x),x)

[Out]

(-4*x**2*log(3)*log(x) - 6*x*log(3)*log(x) - 4*log(x))*exp(x)/log(3)

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