3.85.65 \(\int \frac {100-20 x+e^x (-10+2 x)+(100+e^x (-10+10 x-2 x^2)) \log (25 x)}{x^2 \log ^2(25 x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {2 \left (-10+e^x\right ) (5-x)}{x \log (25 x)} \]

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Rubi [B]  time = 0.81, antiderivative size = 51, normalized size of antiderivative = 2.43, number of steps used = 14, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {6742, 2353, 2306, 2309, 2178, 2302, 30, 2288} \begin {gather*} \frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-\frac {100}{x \log (25 x)}+\frac {20}{\log (25 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 - 20*x + E^x*(-10 + 2*x) + (100 + E^x*(-10 + 10*x - 2*x^2))*Log[25*x])/(x^2*Log[25*x]^2),x]

[Out]

20/Log[25*x] - 100/(x*Log[25*x]) + (2*E^x*(5*x*Log[25*x] - x^2*Log[25*x]))/(x^2*Log[25*x]^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

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

Rule 2306

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

Rule 2309

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

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 \left (-\frac {20 (-5+x-5 \log (25 x))}{x^2 \log ^2(25 x)}-\frac {2 e^x \left (5-x+5 \log (25 x)-5 x \log (25 x)+x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^x \left (5-x+5 \log (25 x)-5 x \log (25 x)+x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)} \, dx\right )-20 \int \frac {-5+x-5 \log (25 x)}{x^2 \log ^2(25 x)} \, dx\\ &=\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \left (\frac {-5+x}{x^2 \log ^2(25 x)}-\frac {5}{x^2 \log (25 x)}\right ) \, dx\\ &=\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \frac {-5+x}{x^2 \log ^2(25 x)} \, dx+100 \int \frac {1}{x^2 \log (25 x)} \, dx\\ &=\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \left (-\frac {5}{x^2 \log ^2(25 x)}+\frac {1}{x \log ^2(25 x)}\right ) \, dx+2500 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (25 x)\right )\\ &=2500 \text {Ei}(-\log (25 x))+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \int \frac {1}{x \log ^2(25 x)} \, dx+100 \int \frac {1}{x^2 \log ^2(25 x)} \, dx\\ &=2500 \text {Ei}(-\log (25 x))-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-20 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (25 x)\right )-100 \int \frac {1}{x^2 \log (25 x)} \, dx\\ &=2500 \text {Ei}(-\log (25 x))+\frac {20}{\log (25 x)}-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}-2500 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (25 x)\right )\\ &=\frac {20}{\log (25 x)}-\frac {100}{x \log (25 x)}+\frac {2 e^x \left (5 x \log (25 x)-x^2 \log (25 x)\right )}{x^2 \log ^2(25 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 19, normalized size = 0.90 \begin {gather*} -\frac {2 \left (-10+e^x\right ) (-5+x)}{x \log (25 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 20*x + E^x*(-10 + 2*x) + (100 + E^x*(-10 + 10*x - 2*x^2))*Log[25*x])/(x^2*Log[25*x]^2),x]

[Out]

(-2*(-10 + E^x)*(-5 + x))/(x*Log[25*x])

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fricas [A]  time = 0.83, size = 22, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left ({\left (x - 5\right )} e^{x} - 10 \, x + 50\right )}}{x \log \left (25 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+10*x-10)*exp(x)+100)*log(25*x)+(2*x-10)*exp(x)-20*x+100)/x^2/log(25*x)^2,x, algorithm="fri
cas")

[Out]

-2*((x - 5)*e^x - 10*x + 50)/(x*log(25*x))

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giac [A]  time = 0.24, size = 24, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (x e^{x} - 10 \, x - 5 \, e^{x} + 50\right )}}{x \log \left (25 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+10*x-10)*exp(x)+100)*log(25*x)+(2*x-10)*exp(x)-20*x+100)/x^2/log(25*x)^2,x, algorithm="gia
c")

[Out]

-2*(x*e^x - 10*x - 5*e^x + 50)/(x*log(25*x))

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maple [A]  time = 0.06, size = 25, normalized size = 1.19

method	result	size
norman	\(\frac {-100+20 x -2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}{x \ln \left (25 x \right )}\)	\(25\)
risch	\(-\frac {2 \left ({\mathrm e}^{x} x -10 x -5 \,{\mathrm e}^{x}+50\right )}{x \ln \left (25 x \right )}\)	\(25\)
default	\(\frac {-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}{\ln \left (25 x \right ) x}+\frac {20}{\ln \left (25 x \right )}-\frac {100}{x \ln \left (25 x \right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+10*x-10)*exp(x)+100)*ln(25*x)+(2*x-10)*exp(x)-20*x+100)/x^2/ln(25*x)^2,x,method=_RETURNVERBOSE)

[Out]

(-100+20*x-2*exp(x)*x+10*exp(x))/x/ln(25*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2 \, {\left (x - 5\right )} e^{x}}{2 \, x \log \relax (5) + x \log \relax (x)} + \frac {20}{\log \left (25 \, x\right )} - 2500 \, \Gamma \left (-1, \log \left (25 \, x\right )\right ) + 100 \, \int \frac {1}{2 \, x^{2} \log \relax (5) + x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+10*x-10)*exp(x)+100)*log(25*x)+(2*x-10)*exp(x)-20*x+100)/x^2/log(25*x)^2,x, algorithm="max
ima")

[Out]

-2*(x - 5)*e^x/(2*x*log(5) + x*log(x)) + 20/log(25*x) - 2500*gamma(-1, log(25*x)) + 100*integrate(1/(2*x^2*log
(5) + x^2*log(x)), x)

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mupad [B]  time = 5.26, size = 18, normalized size = 0.86 \begin {gather*} -\frac {2\,\left ({\mathrm {e}}^x-10\right )\,\left (x-5\right )}{x\,\ln \left (25\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - exp(x)*(2*x - 10) + log(25*x)*(exp(x)*(2*x^2 - 10*x + 10) - 100) - 100)/(x^2*log(25*x)^2),x)

[Out]

-(2*(exp(x) - 10)*(x - 5))/(x*log(25*x))

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sympy [A]  time = 0.31, size = 26, normalized size = 1.24 \begin {gather*} \frac {\left (10 - 2 x\right ) e^{x}}{x \log {\left (25 x \right )}} + \frac {20 x - 100}{x \log {\left (25 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+10*x-10)*exp(x)+100)*ln(25*x)+(2*x-10)*exp(x)-20*x+100)/x**2/ln(25*x)**2,x)

[Out]

(10 - 2*x)*exp(x)/(x*log(25*x)) + (20*x - 100)/(x*log(25*x))

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