3.6.22 \(\int \frac {e^x (-25+235 x+109 x^2+13 x^3+(-20 x-20 x^2-4 x^3) \log (2))+e^x (-25 x-10 x^2-x^3) \log (x)}{125 x+50 x^2+5 x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{5} e^x \left (9+\frac {4 x (1-\log (2))}{5+x}-\log (x)\right ) \]

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Rubi [C]  time = 2.51, antiderivative size = 137, normalized size of antiderivative = 5.07, number of steps used = 28, number of rules used = 11, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {1594, 27, 12, 6741, 6688, 6742, 2178, 2177, 2199, 2194, 2554} \begin {gather*} \frac {6 \text {Ei}(x+5)}{5 e^5}+\frac {(47-\log (16)) \text {Ei}(x+5)}{e^5}+\frac {3 (13-\log (16)) \text {Ei}(x+5)}{e^5}-\frac {4 (109-20 \log (2)) \text {Ei}(x+5)}{5 e^5}-\frac {e^x}{x+5}-\frac {1}{5} e^x \log (x)-\frac {e^x (47-\log (16))}{x+5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{x+5}+\frac {e^x (109-20 \log (2))}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-25 + 235*x + 109*x^2 + 13*x^3 + (-20*x - 20*x^2 - 4*x^3)*Log[2]) + E^x*(-25*x - 10*x^2 - x^3)*Log[x
])/(125*x + 50*x^2 + 5*x^3),x]

[Out]

-(E^x/(5 + x)) + (6*ExpIntegralEi[5 + x])/(5*E^5) + (E^x*(109 - 20*Log[2]))/(5 + x) - (4*ExpIntegralEi[5 + x]*
(109 - 20*Log[2]))/(5*E^5) + (E^x*(13 - Log[16]))/5 - (5*E^x*(13 - Log[16]))/(5 + x) + (3*ExpIntegralEi[5 + x]
*(13 - Log[16]))/E^5 - (E^x*(47 - Log[16]))/(5 + x) + (ExpIntegralEi[5 + x]*(47 - Log[16]))/E^5 - (E^x*Log[x])
/5

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && 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 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 6688

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

Rule 6741

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

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 \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{x \left (125+50 x+5 x^2\right )} \, dx\\ &=\int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{5 x (5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{x (5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (-25+13 x^3 \left (1-\frac {4 \log (2)}{13}\right )+109 x^2 \left (1-\frac {20 \log (2)}{109}\right )+235 x \left (1-\frac {4 \log (2)}{47}\right )-25 x \log (x)-10 x^2 \log (x)-x^3 \log (x)\right )}{x (5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (-25-x^2 (-109+20 \log (2))-5 x (-47+\log (16))-x^3 (-13+\log (16))-x (5+x)^2 \log (x)\right )}{x (5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {25 e^x}{x (5+x)^2}-\frac {e^x x (-109+20 \log (2))}{(5+x)^2}-\frac {5 e^x (-47+\log (16))}{(5+x)^2}-\frac {e^x x^2 (-13+\log (16))}{(5+x)^2}-e^x \log (x)\right ) \, dx\\ &=-\left (\frac {1}{5} \int e^x \log (x) \, dx\right )-5 \int \frac {e^x}{x (5+x)^2} \, dx+\frac {1}{5} (109-20 \log (2)) \int \frac {e^x x}{(5+x)^2} \, dx+\frac {1}{5} (13-\log (16)) \int \frac {e^x x^2}{(5+x)^2} \, dx+(47-\log (16)) \int \frac {e^x}{(5+x)^2} \, dx\\ &=-\frac {e^x (47-\log (16))}{5+x}-\frac {1}{5} e^x \log (x)+\frac {1}{5} \int \frac {e^x}{x} \, dx-5 \int \left (\frac {e^x}{25 x}-\frac {e^x}{5 (5+x)^2}-\frac {e^x}{25 (5+x)}\right ) \, dx+\frac {1}{5} (109-20 \log (2)) \int \left (-\frac {5 e^x}{(5+x)^2}+\frac {e^x}{5+x}\right ) \, dx+\frac {1}{5} (13-\log (16)) \int \left (e^x+\frac {25 e^x}{(5+x)^2}-\frac {10 e^x}{5+x}\right ) \, dx+(47-\log (16)) \int \frac {e^x}{5+x} \, dx\\ &=\frac {\text {Ei}(x)}{5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\text {Ei}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x)-\frac {1}{5} \int \frac {e^x}{x} \, dx+\frac {1}{5} \int \frac {e^x}{5+x} \, dx+\frac {1}{5} (109-20 \log (2)) \int \frac {e^x}{5+x} \, dx+(-109+20 \log (2)) \int \frac {e^x}{(5+x)^2} \, dx+\frac {1}{5} (13-\log (16)) \int e^x \, dx+(5 (13-\log (16))) \int \frac {e^x}{(5+x)^2} \, dx+(2 (-13+\log (16))) \int \frac {e^x}{5+x} \, dx+\int \frac {e^x}{(5+x)^2} \, dx\\ &=-\frac {e^x}{5+x}+\frac {\text {Ei}(5+x)}{5 e^5}+\frac {e^x (109-20 \log (2))}{5+x}+\frac {\text {Ei}(5+x) (109-20 \log (2))}{5 e^5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{5+x}-\frac {2 \text {Ei}(5+x) (13-\log (16))}{e^5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\text {Ei}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x)+(-109+20 \log (2)) \int \frac {e^x}{5+x} \, dx+(5 (13-\log (16))) \int \frac {e^x}{5+x} \, dx+\int \frac {e^x}{5+x} \, dx\\ &=-\frac {e^x}{5+x}+\frac {6 \text {Ei}(5+x)}{5 e^5}+\frac {e^x (109-20 \log (2))}{5+x}-\frac {4 \text {Ei}(5+x) (109-20 \log (2))}{5 e^5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{5+x}+\frac {3 \text {Ei}(5+x) (13-\log (16))}{e^5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\text {Ei}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 26, normalized size = 0.96 \begin {gather*} -\frac {e^x (-45+x (-13+\log (16))+(5+x) \log (x))}{5 (5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-25 + 235*x + 109*x^2 + 13*x^3 + (-20*x - 20*x^2 - 4*x^3)*Log[2]) + E^x*(-25*x - 10*x^2 - x^3)
*Log[x])/(125*x + 50*x^2 + 5*x^3),x]

[Out]

-1/5*(E^x*(-45 + x*(-13 + Log[16]) + (5 + x)*Log[x]))/(5 + x)

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fricas [A]  time = 0.61, size = 29, normalized size = 1.07 \begin {gather*} -\frac {{\left (x + 5\right )} e^{x} \log \relax (x) + {\left (4 \, x \log \relax (2) - 13 \, x - 45\right )} e^{x}}{5 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-10*x^2-25*x)*exp(x)*log(x)+((-4*x^3-20*x^2-20*x)*log(2)+13*x^3+109*x^2+235*x-25)*exp(x))/(5*x
^3+50*x^2+125*x),x, algorithm="fricas")

[Out]

-1/5*((x + 5)*e^x*log(x) + (4*x*log(2) - 13*x - 45)*e^x)/(x + 5)

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giac [A]  time = 0.43, size = 36, normalized size = 1.33 \begin {gather*} -\frac {4 \, x e^{x} \log \relax (2) + x e^{x} \log \relax (x) - 13 \, x e^{x} + 5 \, e^{x} \log \relax (x) - 45 \, e^{x}}{5 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-10*x^2-25*x)*exp(x)*log(x)+((-4*x^3-20*x^2-20*x)*log(2)+13*x^3+109*x^2+235*x-25)*exp(x))/(5*x
^3+50*x^2+125*x),x, algorithm="giac")

[Out]

-1/5*(4*x*e^x*log(2) + x*e^x*log(x) - 13*x*e^x + 5*e^x*log(x) - 45*e^x)/(x + 5)

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maple [A]  time = 0.20, size = 27, normalized size = 1.00

method	result	size
risch	\(-\frac {{\mathrm e}^{x} \ln \relax (x )}{5}-\frac {\left (4 x \ln \relax (2)-13 x -45\right ) {\mathrm e}^{x}}{5 \left (5+x \right )}\)	\(27\)
norman	\(\frac {\left (-\frac {4 \ln \relax (2)}{5}+\frac {13}{5}\right ) x \,{\mathrm e}^{x}-{\mathrm e}^{x} \ln \relax (x )-\frac {x \,{\mathrm e}^{x} \ln \relax (x )}{5}+9 \,{\mathrm e}^{x}}{5+x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-10*x^2-25*x)*exp(x)*ln(x)+((-4*x^3-20*x^2-20*x)*ln(2)+13*x^3+109*x^2+235*x-25)*exp(x))/(5*x^3+50*x^
2+125*x),x,method=_RETURNVERBOSE)

[Out]

-1/5*exp(x)*ln(x)-1/5*(4*x*ln(2)-13*x-45)/(5+x)*exp(x)

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maxima [A]  time = 0.51, size = 25, normalized size = 0.93 \begin {gather*} -\frac {{\left (x {\left (4 \, \log \relax (2) - 13\right )} + {\left (x + 5\right )} \log \relax (x) - 45\right )} e^{x}}{5 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-10*x^2-25*x)*exp(x)*log(x)+((-4*x^3-20*x^2-20*x)*log(2)+13*x^3+109*x^2+235*x-25)*exp(x))/(5*x
^3+50*x^2+125*x),x, algorithm="maxima")

[Out]

-1/5*(x*(4*log(2) - 13) + (x + 5)*log(x) - 45)*e^x/(x + 5)

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mupad [B]  time = 1.04, size = 29, normalized size = 1.07 \begin {gather*} \frac {45\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\left (\ln \left (16\right )-13\right )}{5\,x+25}-\frac {{\mathrm {e}}^x\,\ln \relax (x)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(235*x - log(2)*(20*x + 20*x^2 + 4*x^3) + 109*x^2 + 13*x^3 - 25) - exp(x)*log(x)*(25*x + 10*x^2 +
x^3))/(125*x + 50*x^2 + 5*x^3),x)

[Out]

(45*exp(x) - x*exp(x)*(log(16) - 13))/(5*x + 25) - (exp(x)*log(x))/5

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sympy [A]  time = 0.42, size = 29, normalized size = 1.07 \begin {gather*} \frac {\left (- x \log {\relax (x )} - 4 x \log {\relax (2 )} + 13 x - 5 \log {\relax (x )} + 45\right ) e^{x}}{5 x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-10*x**2-25*x)*exp(x)*ln(x)+((-4*x**3-20*x**2-20*x)*ln(2)+13*x**3+109*x**2+235*x-25)*exp(x))/
(5*x**3+50*x**2+125*x),x)

[Out]

(-x*log(x) - 4*x*log(2) + 13*x - 5*log(x) + 45)*exp(x)/(5*x + 25)

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