3.2.100 \(\int \frac {e^{1+x} (10+2 x) \log (5)+e^{1+x} (8 x+2 x^2) \log (5) \log (x)}{e^{20} (25 x+10 x^2+x^3)} \, dx\)

Optimal. Leaf size=16 \[ \frac {2 e^{-19+x} \log (5) \log (x)}{5+x} \]

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Rubi [A]  time = 0.87, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.157, Rules used = {12, 1594, 27, 6688, 6742, 2178, 2197, 2554} \begin {gather*} \frac {2 e^{x-19} \log (5) \log (x)}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + x)*(10 + 2*x)*Log[5] + E^(1 + x)*(8*x + 2*x^2)*Log[5]*Log[x])/(E^20*(25*x + 10*x^2 + x^3)),x]

[Out]

(2*E^(-19 + x)*Log[5]*Log[x])/(5 + x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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^{1+x} (10+2 x) \log (5)+e^{1+x} \left (8 x+2 x^2\right ) \log (5) \log (x)}{25 x+10 x^2+x^3} \, dx}{e^{20}}\\ &=\frac {\int \frac {e^{1+x} (10+2 x) \log (5)+e^{1+x} \left (8 x+2 x^2\right ) \log (5) \log (x)}{x \left (25+10 x+x^2\right )} \, dx}{e^{20}}\\ &=\frac {\int \frac {e^{1+x} (10+2 x) \log (5)+e^{1+x} \left (8 x+2 x^2\right ) \log (5) \log (x)}{x (5+x)^2} \, dx}{e^{20}}\\ &=\frac {\int \frac {2 e^{1+x} \log (5) (5+x+x (4+x) \log (x))}{x (5+x)^2} \, dx}{e^{20}}\\ &=\frac {(2 \log (5)) \int \frac {e^{1+x} (5+x+x (4+x) \log (x))}{x (5+x)^2} \, dx}{e^{20}}\\ &=\frac {(2 \log (5)) \int \left (\frac {e^{1+x}}{x (5+x)}+\frac {e^{1+x} (4+x) \log (x)}{(5+x)^2}\right ) \, dx}{e^{20}}\\ &=\frac {(2 \log (5)) \int \frac {e^{1+x}}{x (5+x)} \, dx}{e^{20}}+\frac {(2 \log (5)) \int \frac {e^{1+x} (4+x) \log (x)}{(5+x)^2} \, dx}{e^{20}}\\ &=\frac {2 e^{-19+x} \log (5) \log (x)}{5+x}-\frac {(2 \log (5)) \int \frac {e^{1+x}}{x (5+x)} \, dx}{e^{20}}+\frac {(2 \log (5)) \int \left (\frac {e^{1+x}}{5 x}-\frac {e^{1+x}}{5 (5+x)}\right ) \, dx}{e^{20}}\\ &=\frac {2 e^{-19+x} \log (5) \log (x)}{5+x}+\frac {(2 \log (5)) \int \frac {e^{1+x}}{x} \, dx}{5 e^{20}}-\frac {(2 \log (5)) \int \frac {e^{1+x}}{5+x} \, dx}{5 e^{20}}-\frac {(2 \log (5)) \int \left (\frac {e^{1+x}}{5 x}-\frac {e^{1+x}}{5 (5+x)}\right ) \, dx}{e^{20}}\\ &=\frac {2 \text {Ei}(x) \log (5)}{5 e^{19}}-\frac {2 \text {Ei}(5+x) \log (5)}{5 e^{24}}+\frac {2 e^{-19+x} \log (5) \log (x)}{5+x}-\frac {(2 \log (5)) \int \frac {e^{1+x}}{x} \, dx}{5 e^{20}}+\frac {(2 \log (5)) \int \frac {e^{1+x}}{5+x} \, dx}{5 e^{20}}\\ &=\frac {2 e^{-19+x} \log (5) \log (x)}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 16, normalized size = 1.00 \begin {gather*} \frac {2 e^{-19+x} \log (5) \log (x)}{5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + x)*(10 + 2*x)*Log[5] + E^(1 + x)*(8*x + 2*x^2)*Log[5]*Log[x])/(E^20*(25*x + 10*x^2 + x^3)),x
]

[Out]

(2*E^(-19 + x)*Log[5]*Log[x])/(5 + x)

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fricas [A]  time = 0.72, size = 15, normalized size = 0.94 \begin {gather*} \frac {2 \, e^{\left (x - 19\right )} \log \relax (5) \log \relax (x)}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*log(5)*exp(x+1)*log(x)+(2*x+10)*log(5)*exp(x+1))/(x^3+10*x^2+25*x)/exp(10)^2,x, algorit
hm="fricas")

[Out]

2*e^(x - 19)*log(5)*log(x)/(x + 5)

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giac [A]  time = 0.36, size = 15, normalized size = 0.94 \begin {gather*} \frac {2 \, e^{\left (x - 19\right )} \log \relax (5) \log \relax (x)}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*log(5)*exp(x+1)*log(x)+(2*x+10)*log(5)*exp(x+1))/(x^3+10*x^2+25*x)/exp(10)^2,x, algorit
hm="giac")

[Out]

2*e^(x - 19)*log(5)*log(x)/(x + 5)

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maple [A]  time = 0.10, size = 16, normalized size = 1.00

method	result	size
risch	\(\frac {2 \ln \relax (5) \ln \relax (x ) {\mathrm e}^{x -19}}{5+x}\)	\(16\)
norman	\(\frac {2 \ln \relax (x ) {\mathrm e}^{x +1} {\mathrm e}^{-20} \ln \relax (5)}{5+x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+8*x)*ln(5)*exp(x+1)*ln(x)+(2*x+10)*ln(5)*exp(x+1))/(x^3+10*x^2+25*x)/exp(10)^2,x,method=_RETURNVER
BOSE)

[Out]

2*ln(5)/(5+x)*ln(x)*exp(x-19)

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maxima [A]  time = 0.54, size = 15, normalized size = 0.94 \begin {gather*} \frac {2 \, e^{\left (x - 19\right )} \log \relax (5) \log \relax (x)}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*log(5)*exp(x+1)*log(x)+(2*x+10)*log(5)*exp(x+1))/(x^3+10*x^2+25*x)/exp(10)^2,x, algorit
hm="maxima")

[Out]

2*e^(x - 19)*log(5)*log(x)/(x + 5)

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mupad [B]  time = 0.47, size = 29, normalized size = 1.81 \begin {gather*} \frac {2\,x^2\,\mathrm {e}\,{\mathrm {e}}^x\,\ln \relax (5)\,\ln \relax (x)}{{\mathrm {e}}^{20}\,x^3+5\,{\mathrm {e}}^{20}\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-20)*(exp(x + 1)*log(5)*(2*x + 10) + exp(x + 1)*log(5)*log(x)*(8*x + 2*x^2)))/(25*x + 10*x^2 + x^3),x
)

[Out]

(2*x^2*exp(1)*exp(x)*log(5)*log(x))/(5*x^2*exp(20) + x^3*exp(20))

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sympy [A]  time = 0.32, size = 22, normalized size = 1.38 \begin {gather*} \frac {2 e^{x + 1} \log {\relax (5 )} \log {\relax (x )}}{x e^{20} + 5 e^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*exp(x + 1)*log(5)*log(x)/(x*exp(20) + 5*exp(20))

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