3.67.94 \(\int \frac {75 e^x-75 x+(e^x (30-75 x)+45 x) \log (x)+(-62 x+e^x (-13+75 x)) \log ^2(x)+(-23 x+23 e^x x) \log ^3(x)+(-x+e^x x) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {\left (-e^x+x\right ) \left (-x+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}\right )}{x} \]

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Rubi [B]  time = 1.72, antiderivative size = 65, normalized size of antiderivative = 2.10, number of steps used = 39, number of rules used = 8, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6741, 6742, 2297, 2299, 2178, 2360, 2298, 2288} \begin {gather*} -x-\frac {e^x \left (-x \log ^4(x)-23 x \log ^3(x)-75 x \log ^2(x)+75 x \log (x)\right )}{x \log ^2(x) (\log (x)+5)^2}+\frac {3 x}{\log (x)}-\frac {16 x}{\log (x)+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*E^x - 75*x + (E^x*(30 - 75*x) + 45*x)*Log[x] + (-62*x + E^x*(-13 + 75*x))*Log[x]^2 + (-23*x + 23*E^x*x
)*Log[x]^3 + (-x + E^x*x)*Log[x]^4)/(25*x*Log[x]^2 + 10*x*Log[x]^3 + x*Log[x]^4),x]

[Out]

-x + (3*x)/Log[x] - (16*x)/(5 + Log[x]) - (E^x*(75*x*Log[x] - 75*x*Log[x]^2 - 23*x*Log[x]^3 - x*Log[x]^4))/(x*
Log[x]^2*(5 + Log[x])^2)

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

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 {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{x \log ^2(x) (5+\log (x))^2} \, dx\\ &=\int \left (-\frac {62}{(5+\log (x))^2}-\frac {75}{\log ^2(x) (5+\log (x))^2}+\frac {45}{\log (x) (5+\log (x))^2}-\frac {23 \log (x)}{(5+\log (x))^2}-\frac {\log ^2(x)}{(5+\log (x))^2}+\frac {e^x \left (75+30 \log (x)-75 x \log (x)-13 \log ^2(x)+75 x \log ^2(x)+23 x \log ^3(x)+x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}\right ) \, dx\\ &=-\left (23 \int \frac {\log (x)}{(5+\log (x))^2} \, dx\right )+45 \int \frac {1}{\log (x) (5+\log (x))^2} \, dx-62 \int \frac {1}{(5+\log (x))^2} \, dx-75 \int \frac {1}{\log ^2(x) (5+\log (x))^2} \, dx-\int \frac {\log ^2(x)}{(5+\log (x))^2} \, dx+\int \frac {e^x \left (75+30 \log (x)-75 x \log (x)-13 \log ^2(x)+75 x \log ^2(x)+23 x \log ^3(x)+x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2} \, dx\\ &=\frac {62 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}-23 \int \left (-\frac {5}{(5+\log (x))^2}+\frac {1}{5+\log (x)}\right ) \, dx+45 \int \left (\frac {1}{25 \log (x)}-\frac {1}{5 (5+\log (x))^2}-\frac {1}{25 (5+\log (x))}\right ) \, dx-62 \int \frac {1}{5+\log (x)} \, dx-75 \int \left (\frac {1}{25 \log ^2(x)}-\frac {2}{125 \log (x)}+\frac {1}{25 (5+\log (x))^2}+\frac {2}{125 (5+\log (x))}\right ) \, dx-\int \left (1+\frac {25}{(5+\log (x))^2}-\frac {10}{5+\log (x)}\right ) \, dx\\ &=-x+\frac {62 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}+\frac {6}{5} \int \frac {1}{\log (x)} \, dx-\frac {6}{5} \int \frac {1}{5+\log (x)} \, dx+\frac {9}{5} \int \frac {1}{\log (x)} \, dx-\frac {9}{5} \int \frac {1}{5+\log (x)} \, dx-3 \int \frac {1}{\log ^2(x)} \, dx-3 \int \frac {1}{(5+\log (x))^2} \, dx-9 \int \frac {1}{(5+\log (x))^2} \, dx+10 \int \frac {1}{5+\log (x)} \, dx-23 \int \frac {1}{5+\log (x)} \, dx-25 \int \frac {1}{(5+\log (x))^2} \, dx-62 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )+115 \int \frac {1}{(5+\log (x))^2} \, dx\\ &=-x-\frac {62 \text {Ei}(5+\log (x))}{e^5}+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}+3 \text {li}(x)-\frac {6}{5} \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-\frac {9}{5} \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-3 \int \frac {1}{\log (x)} \, dx-3 \int \frac {1}{5+\log (x)} \, dx-9 \int \frac {1}{5+\log (x)} \, dx+10 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-23 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-25 \int \frac {1}{5+\log (x)} \, dx+115 \int \frac {1}{5+\log (x)} \, dx\\ &=-x-\frac {78 \text {Ei}(5+\log (x))}{e^5}+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}-3 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-9 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )-25 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )+115 \operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (x)\right )\\ &=-x+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}-\frac {e^x \left (75 x \log (x)-75 x \log ^2(x)-23 x \log ^3(x)-x \log ^4(x)\right )}{x \log ^2(x) (5+\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 28, normalized size = 0.90 \begin {gather*} \frac {\left (e^x-x\right ) \left (-15+18 \log (x)+\log ^2(x)\right )}{\log (x) (5+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*E^x - 75*x + (E^x*(30 - 75*x) + 45*x)*Log[x] + (-62*x + E^x*(-13 + 75*x))*Log[x]^2 + (-23*x + 23
*E^x*x)*Log[x]^3 + (-x + E^x*x)*Log[x]^4)/(25*x*Log[x]^2 + 10*x*Log[x]^3 + x*Log[x]^4),x]

[Out]

((E^x - x)*(-15 + 18*Log[x] + Log[x]^2))/(Log[x]*(5 + Log[x]))

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fricas [A]  time = 0.45, size = 42, normalized size = 1.35 \begin {gather*} -\frac {{\left (x - e^{x}\right )} \log \relax (x)^{2} + 18 \, {\left (x - e^{x}\right )} \log \relax (x) - 15 \, x + 15 \, e^{x}}{\log \relax (x)^{2} + 5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*exp(x)-62*x)*log(x)^2+((-75*x+30)*exp(
x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log(x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="fricas")

[Out]

-((x - e^x)*log(x)^2 + 18*(x - e^x)*log(x) - 15*x + 15*e^x)/(log(x)^2 + 5*log(x))

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giac [A]  time = 0.25, size = 46, normalized size = 1.48 \begin {gather*} -\frac {x \log \relax (x)^{2} - e^{x} \log \relax (x)^{2} + 18 \, x \log \relax (x) - 18 \, e^{x} \log \relax (x) - 15 \, x + 15 \, e^{x}}{\log \relax (x)^{2} + 5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*exp(x)-62*x)*log(x)^2+((-75*x+30)*exp(
x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log(x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="giac")

[Out]

-(x*log(x)^2 - e^x*log(x)^2 + 18*x*log(x) - 18*e^x*log(x) - 15*x + 15*e^x)/(log(x)^2 + 5*log(x))

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maple [A]  time = 0.16, size = 31, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{x}-x -\frac {\left (x -{\mathrm e}^{x}\right ) \left (13 \ln \relax (x )-15\right )}{\ln \relax (x ) \left (5+\ln \relax (x )\right )}\)	\(31\)
norman	\(\frac {{\mathrm e}^{x} \ln \relax (x )^{2}+15 x -18 x \ln \relax (x )-x \ln \relax (x )^{2}+18 \,{\mathrm e}^{x} \ln \relax (x )-15 \,{\mathrm e}^{x}}{\ln \relax (x ) \left (5+\ln \relax (x )\right )}\)	\(45\)
default	\(-\frac {x \left (\ln \relax (x )^{2}+18 \ln \relax (x )-15\right )}{\left (5+\ln \relax (x )\right ) \ln \relax (x )}+\frac {{\mathrm e}^{x} \ln \relax (x )^{2}+18 \,{\mathrm e}^{x} \ln \relax (x )-15 \,{\mathrm e}^{x}}{\ln \relax (x ) \left (5+\ln \relax (x )\right )}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x-x)*ln(x)^4+(23*exp(x)*x-23*x)*ln(x)^3+((75*x-13)*exp(x)-62*x)*ln(x)^2+((-75*x+30)*exp(x)+45*x)*
ln(x)+75*exp(x)-75*x)/(x*ln(x)^4+10*x*ln(x)^3+25*x*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

exp(x)-x-(x-exp(x))*(13*ln(x)-15)/ln(x)/(5+ln(x))

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maxima [A]  time = 0.41, size = 42, normalized size = 1.35 \begin {gather*} -\frac {x \log \relax (x)^{2} - {\left (\log \relax (x)^{2} + 18 \, \log \relax (x) - 15\right )} e^{x} + 18 \, x \log \relax (x) - 15 \, x}{\log \relax (x)^{2} + 5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*exp(x)-62*x)*log(x)^2+((-75*x+30)*exp(
x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log(x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="maxima")

[Out]

-(x*log(x)^2 - (log(x)^2 + 18*log(x) - 15)*e^x + 18*x*log(x) - 15*x)/(log(x)^2 + 5*log(x))

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mupad [B]  time = 4.21, size = 28, normalized size = 0.90 \begin {gather*} -\frac {\left (x-{\mathrm {e}}^x\right )\,\left ({\ln \relax (x)}^2+18\,\ln \relax (x)-15\right )}{\ln \relax (x)\,\left (\ln \relax (x)+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(75*x - 75*exp(x) + log(x)^4*(x - x*exp(x)) + log(x)^2*(62*x - exp(x)*(75*x - 13)) + log(x)^3*(23*x - 23*
x*exp(x)) - log(x)*(45*x - exp(x)*(75*x - 30)))/(25*x*log(x)^2 + 10*x*log(x)^3 + x*log(x)^4),x)

[Out]

-((x - exp(x))*(18*log(x) + log(x)^2 - 15))/(log(x)*(log(x) + 5))

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sympy [B]  time = 0.36, size = 46, normalized size = 1.48 \begin {gather*} - x + \frac {- 13 x \log {\relax (x )} + 15 x}{\log {\relax (x )}^{2} + 5 \log {\relax (x )}} + \frac {\left (\log {\relax (x )}^{2} + 18 \log {\relax (x )} - 15\right ) e^{x}}{\log {\relax (x )}^{2} + 5 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x)*ln(x)**4+(23*exp(x)*x-23*x)*ln(x)**3+((75*x-13)*exp(x)-62*x)*ln(x)**2+((-75*x+30)*exp(
x)+45*x)*ln(x)+75*exp(x)-75*x)/(x*ln(x)**4+10*x*ln(x)**3+25*x*ln(x)**2),x)

[Out]

-x + (-13*x*log(x) + 15*x)/(log(x)**2 + 5*log(x)) + (log(x)**2 + 18*log(x) - 15)*exp(x)/(log(x)**2 + 5*log(x))

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