3.57.94 \(\int \frac {e^x (-20+4 x) \log (-5+x)+(10 x-2 x^2) \log (x)+(2 e^x x+e^x (-10 x+2 x^2) \log (-5+x)) \log (x^2)}{-5 x+x^2} \, dx\)

Optimal. Leaf size=24 \[ 2 x \left (1-\log (x)+\frac {e^x \log (-5+x) \log \left (x^2\right )}{x}\right ) \]

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Rubi [A]  time = 0.79, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 14, number of rules used = 10, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1593, 6688, 2295, 2178, 2554, 12, 2288, 6742, 2194, 2557} \begin {gather*} 2 e^x \log (x-5) \log \left (x^2\right )+2 x-2 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-20 + 4*x)*Log[-5 + x] + (10*x - 2*x^2)*Log[x] + (2*E^x*x + E^x*(-10*x + 2*x^2)*Log[-5 + x])*Log[x^2
])/(-5*x + x^2),x]

[Out]

2*x - 2*x*Log[x] + 2*E^x*Log[-5 + x]*Log[x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 6688

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

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 (-20+4 x) \log (-5+x)+\left (10 x-2 x^2\right ) \log (x)+\left (2 e^x x+e^x \left (-10 x+2 x^2\right ) \log (-5+x)\right ) \log \left (x^2\right )}{(-5+x) x} \, dx\\ &=\int \left (-2 \log (x)+\frac {2 e^x \log \left (x^2\right )}{-5+x}+\frac {2 e^x \log (-5+x) \left (2+x \log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=-(2 \int \log (x) \, dx)+2 \int \frac {e^x \log \left (x^2\right )}{-5+x} \, dx+2 \int \frac {e^x \log (-5+x) \left (2+x \log \left (x^2\right )\right )}{x} \, dx\\ &=2 x-2 x \log (x)+2 e^5 \text {Ei}(-5+x) \log \left (x^2\right )-2 \int \frac {2 e^5 \text {Ei}(-5+x)}{x} \, dx+2 \int \left (\frac {2 e^x \log (-5+x)}{x}+e^x \log (-5+x) \log \left (x^2\right )\right ) \, dx\\ &=2 x-2 x \log (x)+2 e^5 \text {Ei}(-5+x) \log \left (x^2\right )+2 \int e^x \log (-5+x) \log \left (x^2\right ) \, dx+4 \int \frac {e^x \log (-5+x)}{x} \, dx-\left (4 e^5\right ) \int \frac {\text {Ei}(-5+x)}{x} \, dx\\ &=2 x+4 \text {Ei}(x) \log (-5+x)-2 x \log (x)+2 e^5 \text {Ei}(-5+x) \log \left (x^2\right )+2 e^x \log (-5+x) \log \left (x^2\right )-2 \int \frac {2 e^x \log (-5+x)}{x} \, dx-2 \int \frac {e^x \log \left (x^2\right )}{-5+x} \, dx-4 \int \frac {\text {Ei}(x)}{-5+x} \, dx-\left (4 e^5\right ) \int \frac {\text {Ei}(-5+x)}{x} \, dx\\ &=2 x+4 \text {Ei}(x) \log (-5+x)-2 x \log (x)+2 e^x \log (-5+x) \log \left (x^2\right )+2 \int \frac {2 e^5 \text {Ei}(-5+x)}{x} \, dx-4 \int \frac {\text {Ei}(x)}{-5+x} \, dx-4 \int \frac {e^x \log (-5+x)}{x} \, dx-\left (4 e^5\right ) \int \frac {\text {Ei}(-5+x)}{x} \, dx\\ &=2 x-2 x \log (x)+2 e^x \log (-5+x) \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 21, normalized size = 0.88 \begin {gather*} 2 \left (x-x \log (x)+e^x \log (-5+x) \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-20 + 4*x)*Log[-5 + x] + (10*x - 2*x^2)*Log[x] + (2*E^x*x + E^x*(-10*x + 2*x^2)*Log[-5 + x])*L
og[x^2])/(-5*x + x^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 20, normalized size = 0.83 \begin {gather*} 2 \, {\left (2 \, e^{x} \log \left (x - 5\right ) - x\right )} \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x)*exp(x)*log(x-5)+2*exp(x)*x)*log(x^2)+(-2*x^2+10*x)*log(x)+(4*x-20)*exp(x)*log(x-5))/(
x^2-5*x),x, algorithm="fricas")

[Out]

2*(2*e^x*log(x - 5) - x)*log(x) + 2*x

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giac [A]  time = 0.29, size = 21, normalized size = 0.88 \begin {gather*} 2 \, e^{x} \log \left (x^{2}\right ) \log \left (x - 5\right ) - 2 \, x \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x)*exp(x)*log(x-5)+2*exp(x)*x)*log(x^2)+(-2*x^2+10*x)*log(x)+(4*x-20)*exp(x)*log(x-5))/(
x^2-5*x),x, algorithm="giac")

[Out]

2*e^x*log(x^2)*log(x - 5) - 2*x*log(x) + 2*x

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maple [C]  time = 0.32, size = 77, normalized size = 3.21

method	result	size
risch	\(\left (-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}+4 \,{\mathrm e}^{x} \ln \relax (x )\right ) \ln \left (x -5\right )-2 x \ln \relax (x )+2 x\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-10*x)*exp(x)*ln(x-5)+2*exp(x)*x)*ln(x^2)+(-2*x^2+10*x)*ln(x)+(4*x-20)*exp(x)*ln(x-5))/(x^2-5*x),x
,method=_RETURNVERBOSE)

[Out]

(-I*Pi*csgn(I*x)^2*csgn(I*x^2)*exp(x)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2*exp(x)-I*Pi*csgn(I*x^2)^3*exp(x)+4*exp(x)
*ln(x))*ln(x-5)-2*x*ln(x)+2*x

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maxima [A]  time = 0.42, size = 19, normalized size = 0.79 \begin {gather*} 4 \, e^{x} \log \left (x - 5\right ) \log \relax (x) - 2 \, x \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x)*exp(x)*log(x-5)+2*exp(x)*x)*log(x^2)+(-2*x^2+10*x)*log(x)+(4*x-20)*exp(x)*log(x-5))/(
x^2-5*x),x, algorithm="maxima")

[Out]

4*e^x*log(x - 5)*log(x) - 2*x*log(x) + 2*x

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mupad [B]  time = 4.05, size = 21, normalized size = 0.88 \begin {gather*} 2\,x-2\,x\,\ln \relax (x)+2\,\ln \left (x-5\right )\,\ln \left (x^2\right )\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(2*x*exp(x) - log(x - 5)*exp(x)*(10*x - 2*x^2)) + log(x)*(10*x - 2*x^2) + log(x - 5)*exp(x)*(4*
x - 20))/(5*x - x^2),x)

[Out]

2*x - 2*x*log(x) + 2*log(x - 5)*log(x^2)*exp(x)

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sympy [A]  time = 5.89, size = 22, normalized size = 0.92 \begin {gather*} - 2 x \log {\relax (x )} + 2 x + 4 e^{x} \log {\relax (x )} \log {\left (x - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-10*x)*exp(x)*ln(x-5)+2*exp(x)*x)*ln(x**2)+(-2*x**2+10*x)*ln(x)+(4*x-20)*exp(x)*ln(x-5))/(x
**2-5*x),x)

[Out]

-2*x*log(x) + 2*x + 4*exp(x)*log(x)*log(x - 5)

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