3.61.71 \(\int \frac {75 x^2-3 x^3+(75 x^2+e^x (1250-100 x+2 x^2)) \log (x)+e^x (-625+675 x-51 x^2+x^3) \log ^2(x)}{1875 x^2-150 x^3+3 x^4} \, dx\)

Optimal. Leaf size=24 \[ \log (x) \left (-\frac {x}{-25+x}+\frac {e^x \log (x)}{3 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.66, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 7, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1594, 27, 12, 6742, 2314, 31, 2288} \begin {gather*} \frac {e^x \log ^2(x)}{3 x}+\frac {x \log (x)}{25-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*x^2 - 3*x^3 + (75*x^2 + E^x*(1250 - 100*x + 2*x^2))*Log[x] + E^x*(-625 + 675*x - 51*x^2 + x^3)*Log[x]^
2)/(1875*x^2 - 150*x^3 + 3*x^4),x]

[Out]

(x*Log[x])/(25 - x) + (E^x*Log[x]^2)/(3*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

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 x^2-3 x^3+\left (75 x^2+e^x \left (1250-100 x+2 x^2\right )\right ) \log (x)+e^x \left (-625+675 x-51 x^2+x^3\right ) \log ^2(x)}{x^2 \left (1875-150 x+3 x^2\right )} \, dx\\ &=\int \frac {75 x^2-3 x^3+\left (75 x^2+e^x \left (1250-100 x+2 x^2\right )\right ) \log (x)+e^x \left (-625+675 x-51 x^2+x^3\right ) \log ^2(x)}{3 (-25+x)^2 x^2} \, dx\\ &=\frac {1}{3} \int \frac {75 x^2-3 x^3+\left (75 x^2+e^x \left (1250-100 x+2 x^2\right )\right ) \log (x)+e^x \left (-625+675 x-51 x^2+x^3\right ) \log ^2(x)}{(-25+x)^2 x^2} \, dx\\ &=\frac {1}{3} \int \left (-\frac {3 (-25+x-25 \log (x))}{(-25+x)^2}+\frac {e^x \log (x) (2-\log (x)+x \log (x))}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^x \log (x) (2-\log (x)+x \log (x))}{x^2} \, dx-\int \frac {-25+x-25 \log (x)}{(-25+x)^2} \, dx\\ &=\frac {e^x \log ^2(x)}{3 x}-\int \left (\frac {1}{-25+x}-\frac {25 \log (x)}{(-25+x)^2}\right ) \, dx\\ &=-\log (25-x)+\frac {e^x \log ^2(x)}{3 x}+25 \int \frac {\log (x)}{(-25+x)^2} \, dx\\ &=-\log (25-x)+\frac {x \log (x)}{25-x}+\frac {e^x \log ^2(x)}{3 x}+\int \frac {1}{-25+x} \, dx\\ &=\frac {x \log (x)}{25-x}+\frac {e^x \log ^2(x)}{3 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 24, normalized size = 1.00 \begin {gather*} \frac {1}{3} \log (x) \left (-\frac {3 x}{-25+x}+\frac {e^x \log (x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*x^2 - 3*x^3 + (75*x^2 + E^x*(1250 - 100*x + 2*x^2))*Log[x] + E^x*(-625 + 675*x - 51*x^2 + x^3)*L
og[x]^2)/(1875*x^2 - 150*x^3 + 3*x^4),x]

[Out]

(Log[x]*((-3*x)/(-25 + x) + (E^x*Log[x])/x))/3

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 29, normalized size = 1.21 \begin {gather*} \frac {{\left (x - 25\right )} e^{x} \log \relax (x)^{2} - 3 \, x^{2} \log \relax (x)}{3 \, {\left (x^{2} - 25 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-51*x^2+675*x-625)*exp(x)*log(x)^2+((2*x^2-100*x+1250)*exp(x)+75*x^2)*log(x)-3*x^3+75*x^2)/(3*x
^4-150*x^3+1875*x^2),x, algorithm="fricas")

[Out]

1/3*((x - 25)*e^x*log(x)^2 - 3*x^2*log(x))/(x^2 - 25*x)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 35, normalized size = 1.46 \begin {gather*} \frac {x e^{x} \log \relax (x)^{2} - 3 \, x^{2} \log \relax (x) - 25 \, e^{x} \log \relax (x)^{2}}{3 \, {\left (x^{2} - 25 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-51*x^2+675*x-625)*exp(x)*log(x)^2+((2*x^2-100*x+1250)*exp(x)+75*x^2)*log(x)-3*x^3+75*x^2)/(3*x
^4-150*x^3+1875*x^2),x, algorithm="giac")

[Out]

1/3*(x*e^x*log(x)^2 - 3*x^2*log(x) - 25*e^x*log(x)^2)/(x^2 - 25*x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 23, normalized size = 0.96

method	result	size
default	\(\frac {{\mathrm e}^{x} \ln \relax (x )^{2}}{3 x}-\frac {\ln \relax (x ) x}{x -25}\)	\(23\)
risch	\(\frac {{\mathrm e}^{x} \ln \relax (x )^{2}}{3 x}-\frac {25 \ln \relax (x )}{x -25}-\ln \relax (x )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-51*x^2+675*x-625)*exp(x)*ln(x)^2+((2*x^2-100*x+1250)*exp(x)+75*x^2)*ln(x)-3*x^3+75*x^2)/(3*x^4-150*x
^3+1875*x^2),x,method=_RETURNVERBOSE)

[Out]

1/3*exp(x)*ln(x)^2/x-ln(x)*x/(x-25)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 25, normalized size = 1.04 \begin {gather*} \frac {e^{x} \log \relax (x)^{2}}{3 \, x} - \frac {25 \, \log \relax (x)}{x - 25} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-51*x^2+675*x-625)*exp(x)*log(x)^2+((2*x^2-100*x+1250)*exp(x)+75*x^2)*log(x)-3*x^3+75*x^2)/(3*x
^4-150*x^3+1875*x^2),x, algorithm="maxima")

[Out]

1/3*e^x*log(x)^2/x - 25*log(x)/(x - 25) - log(x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (2\,x^2-100\,x+1250\right )+75\,x^2\right )+75\,x^2-3\,x^3+{\mathrm {e}}^x\,{\ln \relax (x)}^2\,\left (x^3-51\,x^2+675\,x-625\right )}{3\,x^4-150\,x^3+1875\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(exp(x)*(2*x^2 - 100*x + 1250) + 75*x^2) + 75*x^2 - 3*x^3 + exp(x)*log(x)^2*(675*x - 51*x^2 + x^3
- 625))/(1875*x^2 - 150*x^3 + 3*x^4),x)

[Out]

int((log(x)*(exp(x)*(2*x^2 - 100*x + 1250) + 75*x^2) + 75*x^2 - 3*x^3 + exp(x)*log(x)^2*(675*x - 51*x^2 + x^3
- 625))/(1875*x^2 - 150*x^3 + 3*x^4), x)

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 22, normalized size = 0.92 \begin {gather*} - \log {\relax (x )} - \frac {25 \log {\relax (x )}}{x - 25} + \frac {e^{x} \log {\relax (x )}^{2}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-51*x**2+675*x-625)*exp(x)*ln(x)**2+((2*x**2-100*x+1250)*exp(x)+75*x**2)*ln(x)-3*x**3+75*x**2)
/(3*x**4-150*x**3+1875*x**2),x)

[Out]

-log(x) - 25*log(x)/(x - 25) + exp(x)*log(x)**2/(3*x)

________________________________________________________________________________________