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

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

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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} $$\begin{array}{c}\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-\frac{100}{x\log (25x)}+\frac{20}{\log (25x)}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{20(-5+x-5\log (25x))}{x^2{\log }^2(25x)}-\frac{2e^x(5-x+5\log (25x)-5x\log (25x)+x^2\log (25x))}{x^2{\log }^2(25x)})dx\\ & =-(2\int \frac{e^x(5-x+5\log (25x)-5x\log (25x)+x^2\log (25x))}{x^2{\log }^2(25x)}dx)-20\int \frac{-5+x-5\log (25x)}{x^2{\log }^2(25x)}dx\\ & =\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-20\int (\frac{-5+x}{x^2{\log }^2(25x)}-\frac{5}{x^2\log (25x)})dx\\ & =\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-20\int \frac{-5+x}{x^2{\log }^2(25x)}dx+100\int \frac{1}{x^2\log (25x)}dx\\ & =\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-20\int (-\frac{5}{x^2{\log }^2(25x)}+\frac{1}{x{\log }^2(25x)})dx+2500\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (25x))\\ & =2500\text{Ei}(-\log (25x))+\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-20\int \frac{1}{x{\log }^2(25x)}dx+100\int \frac{1}{x^2{\log }^2(25x)}dx\\ & =2500\text{Ei}(-\log (25x))-\frac{100}{x\log (25x)}+\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-20\mathrm{Subst}(\int \frac{1}{x^2}dx,x,\log (25x))-100\int \frac{1}{x^2\log (25x)}dx\\ & =2500\text{Ei}(-\log (25x))+\frac{20}{\log (25x)}-\frac{100}{x\log (25x)}+\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}-2500\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (25x))\\ & =\frac{20}{\log (25x)}-\frac{100}{x\log (25x)}+\frac{2e^x(5x\log (25x)-x^2\log (25x))}{x^2{\log }^2(25x)}\end{array}\end{array}$$

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

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 $$\begin{array}{c}-\frac{2((x-5)e^x-10x+50)}{x\log \left(25x\right)}\end{array}$$

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 $$\begin{array}{c}-\frac{2(x{e}^x-10x-5e^x+50)}{x\log \left(25x\right)}\end{array}$$

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

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 $$\begin{array}{c}-\frac{2(x-5)e^x}{2x\log (5)+x\log (x)}+\frac{20}{\log \left(25x\right)}-2500\mathrm{\Gamma }(-1,\log \left(25x\right))+100\int \frac{1}{2x^2\log (5)+x^2\log (x)}dx\end{array}$$

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 $$\begin{array}{c}-\frac{2({\mathrm{e}}^x-10)(x-5)}{x\ln \left(25x\right)}\end{array}$$

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 $$\begin{array}{c}\frac{(10-2x)e^x}{x\log \left(25x\right)}+\frac{20x-100}{x\log \left(25x\right)}\end{array}$$

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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