3.90.30 \(\int \frac {120-115 x+e^{-1+3 x} (-80+260 x-180 x^2)+e^{-2+6 x} (-120 x+120 x^2)+(-80+80 x+e^{-1+3 x} (-120 x+120 x^2)) \log (x^2)}{-x+x^2} \, dx\)

Optimal. Leaf size=27 \[ 5 \left (\log (2-2 x)+\left (3-2 \left (e^{-1+3 x}+\log \left (x^2\right )\right )\right )^2\right ) \]

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Rubi [A]  time = 1.15, antiderivative size = 53, normalized size of antiderivative = 1.96, number of steps used = 10, number of rules used = 6, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {1593, 6742, 2194, 2288, 72, 2301} \begin {gather*} 20 \log ^2\left (x^2\right )-\frac {20 e^{3 x-1} \left (3 x-2 x \log \left (x^2\right )\right )}{x}+20 e^{6 x-2}+5 \log (1-x)-120 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120 - 115*x + E^(-1 + 3*x)*(-80 + 260*x - 180*x^2) + E^(-2 + 6*x)*(-120*x + 120*x^2) + (-80 + 80*x + E^(-
1 + 3*x)*(-120*x + 120*x^2))*Log[x^2])/(-x + x^2),x]

[Out]

20*E^(-2 + 6*x) + 5*Log[1 - x] - 120*Log[x] + 20*Log[x^2]^2 - (20*E^(-1 + 3*x)*(3*x - 2*x*Log[x^2]))/x

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{(-1+x) x} \, dx\\ &=\int \left (120 e^{-2+6 x}+\frac {20 e^{-1+3 x} \left (4-9 x+6 x \log \left (x^2\right )\right )}{x}+\frac {5 \left (24-23 x-16 \log \left (x^2\right )+16 x \log \left (x^2\right )\right )}{(-1+x) x}\right ) \, dx\\ &=5 \int \frac {24-23 x-16 \log \left (x^2\right )+16 x \log \left (x^2\right )}{(-1+x) x} \, dx+20 \int \frac {e^{-1+3 x} \left (4-9 x+6 x \log \left (x^2\right )\right )}{x} \, dx+120 \int e^{-2+6 x} \, dx\\ &=20 e^{-2+6 x}-\frac {20 e^{-1+3 x} \left (3 x-2 x \log \left (x^2\right )\right )}{x}+5 \int \left (\frac {24-23 x}{(-1+x) x}+\frac {16 \log \left (x^2\right )}{x}\right ) \, dx\\ &=20 e^{-2+6 x}-\frac {20 e^{-1+3 x} \left (3 x-2 x \log \left (x^2\right )\right )}{x}+5 \int \frac {24-23 x}{(-1+x) x} \, dx+80 \int \frac {\log \left (x^2\right )}{x} \, dx\\ &=20 e^{-2+6 x}+20 \log ^2\left (x^2\right )-\frac {20 e^{-1+3 x} \left (3 x-2 x \log \left (x^2\right )\right )}{x}+5 \int \left (\frac {1}{-1+x}-\frac {24}{x}\right ) \, dx\\ &=20 e^{-2+6 x}+5 \log (1-x)-120 \log (x)+20 \log ^2\left (x^2\right )-\frac {20 e^{-1+3 x} \left (3 x-2 x \log \left (x^2\right )\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 36, normalized size = 1.33 \begin {gather*} \frac {5 \left (e^2 \log (1-x)+\left (-3 e+2 e^{3 x}+2 e \log \left (x^2\right )\right )^2\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120 - 115*x + E^(-1 + 3*x)*(-80 + 260*x - 180*x^2) + E^(-2 + 6*x)*(-120*x + 120*x^2) + (-80 + 80*x
+ E^(-1 + 3*x)*(-120*x + 120*x^2))*Log[x^2])/(-x + x^2),x]

[Out]

(5*(E^2*Log[1 - x] + (-3*E + 2*E^(3*x) + 2*E*Log[x^2])^2))/E^2

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fricas [A]  time = 0.51, size = 47, normalized size = 1.74 \begin {gather*} 40 \, e^{\left (3 \, x - 1\right )} \log \left (x^{2}\right ) + 20 \, \log \left (x^{2}\right )^{2} + 20 \, e^{\left (6 \, x - 2\right )} - 60 \, e^{\left (3 \, x - 1\right )} + 5 \, \log \left (x - 1\right ) - 120 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((120*x^2-120*x)*exp(3*x-1)+80*x-80)*log(x^2)+(120*x^2-120*x)*exp(3*x-1)^2+(-180*x^2+260*x-80)*exp(
3*x-1)-115*x+120)/(x^2-x),x, algorithm="fricas")

[Out]

40*e^(3*x - 1)*log(x^2) + 20*log(x^2)^2 + 20*e^(6*x - 2) - 60*e^(3*x - 1) + 5*log(x - 1) - 120*log(x)

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giac [A]  time = 0.16, size = 56, normalized size = 2.07 \begin {gather*} 5 \, {\left (4 \, e^{3} \log \left (x^{2}\right )^{2} + 8 \, e^{\left (3 \, x + 2\right )} \log \left (x^{2}\right ) + e^{3} \log \left (x - 1\right ) - 24 \, e^{3} \log \relax (x) + 4 \, e^{\left (6 \, x + 1\right )} - 12 \, e^{\left (3 \, x + 2\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((120*x^2-120*x)*exp(3*x-1)+80*x-80)*log(x^2)+(120*x^2-120*x)*exp(3*x-1)^2+(-180*x^2+260*x-80)*exp(
3*x-1)-115*x+120)/(x^2-x),x, algorithm="giac")

[Out]

5*(4*e^3*log(x^2)^2 + 8*e^(3*x + 2)*log(x^2) + e^3*log(x - 1) - 24*e^3*log(x) + 4*e^(6*x + 1) - 12*e^(3*x + 2)
)*e^(-3)

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maple [A]  time = 0.50, size = 52, normalized size = 1.93

method	result	size
norman	\(-60 \ln \left (x^{2}\right )+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}+40 \,{\mathrm e}^{3 x -1} \ln \left (x^{2}\right )-60 \,{\mathrm e}^{3 x -1}+5 \ln \left (x -1\right )\)	\(52\)
default	\(40 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) {\mathrm e}^{3 x -1}-60 \,{\mathrm e}^{3 x -1}+80 \ln \relax (x ) {\mathrm e}^{3 x -1}+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}-120 \ln \relax (x )+5 \ln \left (x -1\right )\)	\(65\)
risch	\(80 \ln \relax (x )^{2}+80 \ln \relax (x ) {\mathrm e}^{3 x -1}-40 i \pi \ln \left (\left (-8 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-8 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+80 i \pi \ln \left (\left (-8 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-8 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-40 i \pi \ln \left (\left (-8 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-8 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \mathrm {csgn}\left (i x^{2}\right )^{3}-120 \ln \left (\left (-8 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-8 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right )+5 \ln \left (1-x \right )+20 \,{\mathrm e}^{6 x -2}-20 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{3 x -1}+40 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{3 x -1}-20 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{3 x -1}-60 \,{\mathrm e}^{3 x -1}\)	\(368\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((120*x^2-120*x)*exp(3*x-1)+80*x-80)*ln(x^2)+(120*x^2-120*x)*exp(3*x-1)^2+(-180*x^2+260*x-80)*exp(3*x-1)-
115*x+120)/(x^2-x),x,method=_RETURNVERBOSE)

[Out]

-60*ln(x^2)+20*exp(3*x-1)^2+20*ln(x^2)^2+40*exp(3*x-1)*ln(x^2)-60*exp(3*x-1)+5*ln(x-1)

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maxima [A]  time = 0.45, size = 44, normalized size = 1.63 \begin {gather*} 20 \, {\left (4 \, e^{2} \log \relax (x)^{2} + {\left (4 \, e \log \relax (x) - 3 \, e\right )} e^{\left (3 \, x\right )} + e^{\left (6 \, x\right )}\right )} e^{\left (-2\right )} + 5 \, \log \left (x - 1\right ) - 120 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((120*x^2-120*x)*exp(3*x-1)+80*x-80)*log(x^2)+(120*x^2-120*x)*exp(3*x-1)^2+(-180*x^2+260*x-80)*exp(
3*x-1)-115*x+120)/(x^2-x),x, algorithm="maxima")

[Out]

20*(4*e^2*log(x)^2 + (4*e*log(x) - 3*e)*e^(3*x) + e^(6*x))*e^(-2) + 5*log(x - 1) - 120*log(x)

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mupad [B]  time = 7.72, size = 49, normalized size = 1.81 \begin {gather*} 5\,\ln \left (x-1\right )-60\,\ln \left (x^2\right )-60\,{\mathrm {e}}^{3\,x-1}+20\,{\mathrm {e}}^{6\,x-2}+40\,\ln \left (x^2\right )\,{\mathrm {e}}^{3\,x-1}+20\,{\ln \left (x^2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((115*x + log(x^2)*(exp(3*x - 1)*(120*x - 120*x^2) - 80*x + 80) + exp(6*x - 2)*(120*x - 120*x^2) + exp(3*x
- 1)*(180*x^2 - 260*x + 80) - 120)/(x - x^2),x)

[Out]

5*log(x - 1) - 60*log(x^2) - 60*exp(3*x - 1) + 20*exp(6*x - 2) + 40*log(x^2)*exp(3*x - 1) + 20*log(x^2)^2

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sympy [A]  time = 0.45, size = 42, normalized size = 1.56 \begin {gather*} \left (40 \log {\left (x^{2} \right )} - 60\right ) e^{3 x - 1} + 20 e^{6 x - 2} - 120 \log {\relax (x )} + 20 \log {\left (x^{2} \right )}^{2} + 5 \log {\left (x - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((120*x**2-120*x)*exp(3*x-1)+80*x-80)*ln(x**2)+(120*x**2-120*x)*exp(3*x-1)**2+(-180*x**2+260*x-80)*
exp(3*x-1)-115*x+120)/(x**2-x),x)

[Out]

(40*log(x**2) - 60)*exp(3*x - 1) + 20*exp(6*x - 2) - 120*log(x) + 20*log(x**2)**2 + 5*log(x - 1)

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