3.25.83 \(\int \frac {160+e^x (96-64 x)+32 x-64 x^2+(192 x-128 x^2) \log (x)}{-3 x^3+2 x^4+e^{3 x} (-3+2 x)+e^{2 x} (-9 x+6 x^2)+e^x (-9 x^2+6 x^3)+(-3 x^6+2 x^7) \log ^3(x)+(-9 x^2+6 x^3+e^{2 x} (-9+6 x)+e^x (-18 x+12 x^2)) \log (\frac {1}{-3+2 x})+(-9 x+6 x^2+e^x (-9+6 x)) \log ^2(\frac {1}{-3+2 x})+(-3+2 x) \log ^3(\frac {1}{-3+2 x})+\log ^2(x) (-9 x^5+6 x^6+e^x (-9 x^4+6 x^5)+(-9 x^4+6 x^5) \log (\frac {1}{-3+2 x}))+\log (x) (-9 x^4+6 x^5+e^{2 x} (-9 x^2+6 x^3)+e^x (-18 x^3+12 x^4)+(-18 x^3+12 x^4+e^x (-18 x^2+12 x^3)) \log (\frac {1}{-3+2 x})+(-9 x^2+6 x^3) \log ^2(\frac {1}{-3+2 x}))} \, dx\)

Optimal. Leaf size=23 \[ \frac {16}{\left (e^x+x+x^2 \log (x)+\log \left (\frac {1}{-3+2 x}\right )\right )^2} \]

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Rubi [A]  time = 0.71, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 345, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6688, 12, 6686} \begin {gather*} \frac {16}{\left (x^2 \log (x)+x+e^x+\log \left (\frac {1}{2 x-3}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(160 + E^x*(96 - 64*x) + 32*x - 64*x^2 + (192*x - 128*x^2)*Log[x])/(-3*x^3 + 2*x^4 + E^(3*x)*(-3 + 2*x) +
E^(2*x)*(-9*x + 6*x^2) + E^x*(-9*x^2 + 6*x^3) + (-3*x^6 + 2*x^7)*Log[x]^3 + (-9*x^2 + 6*x^3 + E^(2*x)*(-9 + 6*
x) + E^x*(-18*x + 12*x^2))*Log[(-3 + 2*x)^(-1)] + (-9*x + 6*x^2 + E^x*(-9 + 6*x))*Log[(-3 + 2*x)^(-1)]^2 + (-3
+ 2*x)*Log[(-3 + 2*x)^(-1)]^3 + Log[x]^2*(-9*x^5 + 6*x^6 + E^x*(-9*x^4 + 6*x^5) + (-9*x^4 + 6*x^5)*Log[(-3 +
2*x)^(-1)]) + Log[x]*(-9*x^4 + 6*x^5 + E^(2*x)*(-9*x^2 + 6*x^3) + E^x*(-18*x^3 + 12*x^4) + (-18*x^3 + 12*x^4 +
E^x*(-18*x^2 + 12*x^3))*Log[(-3 + 2*x)^(-1)] + (-9*x^2 + 6*x^3)*Log[(-3 + 2*x)^(-1)]^2)),x]

[Out]

16/(E^x + x + x^2*Log[x] + Log[(-3 + 2*x)^(-1)])^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 \left (-5-x+2 x^2+e^x (-3+2 x)+2 x (-3+2 x) \log (x)\right )}{(3-2 x) \left (e^x+x+x^2 \log (x)+\log \left (\frac {1}{-3+2 x}\right )\right )^3} \, dx\\ &=32 \int \frac {-5-x+2 x^2+e^x (-3+2 x)+2 x (-3+2 x) \log (x)}{(3-2 x) \left (e^x+x+x^2 \log (x)+\log \left (\frac {1}{-3+2 x}\right )\right )^3} \, dx\\ &=\frac {16}{\left (e^x+x+x^2 \log (x)+\log \left (\frac {1}{-3+2 x}\right )\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 23, normalized size = 1.00 \begin {gather*} \frac {16}{\left (e^x+x+x^2 \log (x)+\log \left (\frac {1}{-3+2 x}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(160 + E^x*(96 - 64*x) + 32*x - 64*x^2 + (192*x - 128*x^2)*Log[x])/(-3*x^3 + 2*x^4 + E^(3*x)*(-3 + 2
*x) + E^(2*x)*(-9*x + 6*x^2) + E^x*(-9*x^2 + 6*x^3) + (-3*x^6 + 2*x^7)*Log[x]^3 + (-9*x^2 + 6*x^3 + E^(2*x)*(-
9 + 6*x) + E^x*(-18*x + 12*x^2))*Log[(-3 + 2*x)^(-1)] + (-9*x + 6*x^2 + E^x*(-9 + 6*x))*Log[(-3 + 2*x)^(-1)]^2
+ (-3 + 2*x)*Log[(-3 + 2*x)^(-1)]^3 + Log[x]^2*(-9*x^5 + 6*x^6 + E^x*(-9*x^4 + 6*x^5) + (-9*x^4 + 6*x^5)*Log[
(-3 + 2*x)^(-1)]) + Log[x]*(-9*x^4 + 6*x^5 + E^(2*x)*(-9*x^2 + 6*x^3) + E^x*(-18*x^3 + 12*x^4) + (-18*x^3 + 12
*x^4 + E^x*(-18*x^2 + 12*x^3))*Log[(-3 + 2*x)^(-1)] + (-9*x^2 + 6*x^3)*Log[(-3 + 2*x)^(-1)]^2)),x]

[Out]

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

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fricas [B]  time = 0.94, size = 75, normalized size = 3.26 \begin {gather*} \frac {16}{x^{4} \log \relax (x)^{2} + x^{2} + 2 \, x e^{x} + 2 \, {\left (x^{3} + x^{2} e^{x} + x^{2} \log \left (\frac {1}{2 \, x - 3}\right )\right )} \log \relax (x) + 2 \, {\left (x + e^{x}\right )} \log \left (\frac {1}{2 \, x - 3}\right ) + \log \left (\frac {1}{2 \, x - 3}\right )^{2} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x^2+192*x)*log(x)+(-64*x+96)*exp(x)-64*x^2+32*x+160)/((2*x^7-3*x^6)*log(x)^3+((6*x^5-9*x^4)*l
og(1/(2*x-3))+(6*x^5-9*x^4)*exp(x)+6*x^6-9*x^5)*log(x)^2+((6*x^3-9*x^2)*log(1/(2*x-3))^2+((12*x^3-18*x^2)*exp(
x)+12*x^4-18*x^3)*log(1/(2*x-3))+(6*x^3-9*x^2)*exp(x)^2+(12*x^4-18*x^3)*exp(x)+6*x^5-9*x^4)*log(x)+(2*x-3)*log
(1/(2*x-3))^3+((6*x-9)*exp(x)+6*x^2-9*x)*log(1/(2*x-3))^2+((6*x-9)*exp(x)^2+(12*x^2-18*x)*exp(x)+6*x^3-9*x^2)*
log(1/(2*x-3))+(2*x-3)*exp(x)^3+(6*x^2-9*x)*exp(x)^2+(6*x^3-9*x^2)*exp(x)+2*x^4-3*x^3),x, algorithm="fricas")

[Out]

16/(x^4*log(x)^2 + x^2 + 2*x*e^x + 2*(x^3 + x^2*e^x + x^2*log(1/(2*x - 3)))*log(x) + 2*(x + e^x)*log(1/(2*x -
3)) + log(1/(2*x - 3))^2 + e^(2*x))

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giac [B]  time = 0.52, size = 81, normalized size = 3.52 \begin {gather*} \frac {16}{x^{4} \log \relax (x)^{2} + 2 \, x^{3} \log \relax (x) + 2 \, x^{2} e^{x} \log \relax (x) - 2 \, x^{2} \log \left (2 \, x - 3\right ) \log \relax (x) + x^{2} + 2 \, x e^{x} - 2 \, x \log \left (2 \, x - 3\right ) - 2 \, e^{x} \log \left (2 \, x - 3\right ) + \log \left (2 \, x - 3\right )^{2} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x^2+192*x)*log(x)+(-64*x+96)*exp(x)-64*x^2+32*x+160)/((2*x^7-3*x^6)*log(x)^3+((6*x^5-9*x^4)*l
og(1/(2*x-3))+(6*x^5-9*x^4)*exp(x)+6*x^6-9*x^5)*log(x)^2+((6*x^3-9*x^2)*log(1/(2*x-3))^2+((12*x^3-18*x^2)*exp(
x)+12*x^4-18*x^3)*log(1/(2*x-3))+(6*x^3-9*x^2)*exp(x)^2+(12*x^4-18*x^3)*exp(x)+6*x^5-9*x^4)*log(x)+(2*x-3)*log
(1/(2*x-3))^3+((6*x-9)*exp(x)+6*x^2-9*x)*log(1/(2*x-3))^2+((6*x-9)*exp(x)^2+(12*x^2-18*x)*exp(x)+6*x^3-9*x^2)*
log(1/(2*x-3))+(2*x-3)*exp(x)^3+(6*x^2-9*x)*exp(x)^2+(6*x^3-9*x^2)*exp(x)+2*x^4-3*x^3),x, algorithm="giac")

[Out]

16/(x^4*log(x)^2 + 2*x^3*log(x) + 2*x^2*e^x*log(x) - 2*x^2*log(2*x - 3)*log(x) + x^2 + 2*x*e^x - 2*x*log(2*x -
3) - 2*e^x*log(2*x - 3) + log(2*x - 3)^2 + e^(2*x))

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maple [C]  time = 0.77, size = 35, normalized size = 1.52

method	result	size
risch	\(-\frac {64}{\left (2 i x^{2} \ln \relax (x )-2 i \ln \relax (2)+2 i x +2 i {\mathrm e}^{x}-2 i \ln \left (x -\frac {3}{2}\right )\right )^{2}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-128*x^2+192*x)*ln(x)+(-64*x+96)*exp(x)-64*x^2+32*x+160)/((2*x^7-3*x^6)*ln(x)^3+((6*x^5-9*x^4)*ln(1/(2*x
-3))+(6*x^5-9*x^4)*exp(x)+6*x^6-9*x^5)*ln(x)^2+((6*x^3-9*x^2)*ln(1/(2*x-3))^2+((12*x^3-18*x^2)*exp(x)+12*x^4-1
8*x^3)*ln(1/(2*x-3))+(6*x^3-9*x^2)*exp(x)^2+(12*x^4-18*x^3)*exp(x)+6*x^5-9*x^4)*ln(x)+(2*x-3)*ln(1/(2*x-3))^3+
((6*x-9)*exp(x)+6*x^2-9*x)*ln(1/(2*x-3))^2+((6*x-9)*exp(x)^2+(12*x^2-18*x)*exp(x)+6*x^3-9*x^2)*ln(1/(2*x-3))+(
2*x-3)*exp(x)^3+(6*x^2-9*x)*exp(x)^2+(6*x^3-9*x^2)*exp(x)+2*x^4-3*x^3),x,method=_RETURNVERBOSE)

[Out]

-64/(2*I*x^2*ln(x)-2*I*ln(2)+2*I*x+2*I*exp(x)-2*I*ln(x-3/2))^2

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maxima [B]  time = 0.99, size = 65, normalized size = 2.83 \begin {gather*} \frac {16}{x^{4} \log \relax (x)^{2} + 2 \, x^{3} \log \relax (x) + x^{2} + 2 \, {\left (x^{2} \log \relax (x) + x\right )} e^{x} - 2 \, {\left (x^{2} \log \relax (x) + x + e^{x}\right )} \log \left (2 \, x - 3\right ) + \log \left (2 \, x - 3\right )^{2} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x^2+192*x)*log(x)+(-64*x+96)*exp(x)-64*x^2+32*x+160)/((2*x^7-3*x^6)*log(x)^3+((6*x^5-9*x^4)*l
og(1/(2*x-3))+(6*x^5-9*x^4)*exp(x)+6*x^6-9*x^5)*log(x)^2+((6*x^3-9*x^2)*log(1/(2*x-3))^2+((12*x^3-18*x^2)*exp(
x)+12*x^4-18*x^3)*log(1/(2*x-3))+(6*x^3-9*x^2)*exp(x)^2+(12*x^4-18*x^3)*exp(x)+6*x^5-9*x^4)*log(x)+(2*x-3)*log
(1/(2*x-3))^3+((6*x-9)*exp(x)+6*x^2-9*x)*log(1/(2*x-3))^2+((6*x-9)*exp(x)^2+(12*x^2-18*x)*exp(x)+6*x^3-9*x^2)*
log(1/(2*x-3))+(2*x-3)*exp(x)^3+(6*x^2-9*x)*exp(x)^2+(6*x^3-9*x^2)*exp(x)+2*x^4-3*x^3),x, algorithm="maxima")

[Out]

16/(x^4*log(x)^2 + 2*x^3*log(x) + x^2 + 2*(x^2*log(x) + x)*e^x - 2*(x^2*log(x) + x + e^x)*log(2*x - 3) + log(2
*x - 3)^2 + e^(2*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {32\,x-{\mathrm {e}}^x\,\left (64\,x-96\right )+\ln \relax (x)\,\left (192\,x-128\,x^2\right )-64\,x^2+160}{{\mathrm {e}}^{2\,x}\,\left (9\,x-6\,x^2\right )+{\mathrm {e}}^x\,\left (9\,x^2-6\,x^3\right )+\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (18\,x^3-12\,x^4\right )+\ln \left (\frac {1}{2\,x-3}\right )\,\left ({\mathrm {e}}^x\,\left (18\,x^2-12\,x^3\right )+18\,x^3-12\,x^4\right )+{\mathrm {e}}^{2\,x}\,\left (9\,x^2-6\,x^3\right )+{\ln \left (\frac {1}{2\,x-3}\right )}^2\,\left (9\,x^2-6\,x^3\right )+9\,x^4-6\,x^5\right )+{\ln \relax (x)}^3\,\left (3\,x^6-2\,x^7\right )-{\ln \left (\frac {1}{2\,x-3}\right )}^2\,\left ({\mathrm {e}}^x\,\left (6\,x-9\right )-9\,x+6\,x^2\right )+{\ln \relax (x)}^2\,\left ({\mathrm {e}}^x\,\left (9\,x^4-6\,x^5\right )+\ln \left (\frac {1}{2\,x-3}\right )\,\left (9\,x^4-6\,x^5\right )+9\,x^5-6\,x^6\right )+\ln \left (\frac {1}{2\,x-3}\right )\,\left ({\mathrm {e}}^x\,\left (18\,x-12\,x^2\right )-{\mathrm {e}}^{2\,x}\,\left (6\,x-9\right )+9\,x^2-6\,x^3\right )-{\mathrm {e}}^{3\,x}\,\left (2\,x-3\right )+3\,x^3-2\,x^4-{\ln \left (\frac {1}{2\,x-3}\right )}^3\,\left (2\,x-3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x - exp(x)*(64*x - 96) + log(x)*(192*x - 128*x^2) - 64*x^2 + 160)/(exp(2*x)*(9*x - 6*x^2) + exp(x)*(9
*x^2 - 6*x^3) + log(x)*(exp(x)*(18*x^3 - 12*x^4) + log(1/(2*x - 3))*(exp(x)*(18*x^2 - 12*x^3) + 18*x^3 - 12*x^
4) + exp(2*x)*(9*x^2 - 6*x^3) + log(1/(2*x - 3))^2*(9*x^2 - 6*x^3) + 9*x^4 - 6*x^5) + log(x)^3*(3*x^6 - 2*x^7)
- log(1/(2*x - 3))^2*(exp(x)*(6*x - 9) - 9*x + 6*x^2) + log(x)^2*(exp(x)*(9*x^4 - 6*x^5) + log(1/(2*x - 3))*(
9*x^4 - 6*x^5) + 9*x^5 - 6*x^6) + log(1/(2*x - 3))*(exp(x)*(18*x - 12*x^2) - exp(2*x)*(6*x - 9) + 9*x^2 - 6*x^
3) - exp(3*x)*(2*x - 3) + 3*x^3 - 2*x^4 - log(1/(2*x - 3))^3*(2*x - 3)),x)

[Out]

int(-(32*x - exp(x)*(64*x - 96) + log(x)*(192*x - 128*x^2) - 64*x^2 + 160)/(exp(2*x)*(9*x - 6*x^2) + exp(x)*(9
*x^2 - 6*x^3) + log(x)*(exp(x)*(18*x^3 - 12*x^4) + log(1/(2*x - 3))*(exp(x)*(18*x^2 - 12*x^3) + 18*x^3 - 12*x^
4) + exp(2*x)*(9*x^2 - 6*x^3) + log(1/(2*x - 3))^2*(9*x^2 - 6*x^3) + 9*x^4 - 6*x^5) + log(x)^3*(3*x^6 - 2*x^7)
- log(1/(2*x - 3))^2*(exp(x)*(6*x - 9) - 9*x + 6*x^2) + log(x)^2*(exp(x)*(9*x^4 - 6*x^5) + log(1/(2*x - 3))*(
9*x^4 - 6*x^5) + 9*x^5 - 6*x^6) + log(1/(2*x - 3))*(exp(x)*(18*x - 12*x^2) - exp(2*x)*(6*x - 9) + 9*x^2 - 6*x^
3) - exp(3*x)*(2*x - 3) + 3*x^3 - 2*x^4 - log(1/(2*x - 3))^3*(2*x - 3)), x)

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sympy [B]  time = 0.63, size = 90, normalized size = 3.91 \begin {gather*} \frac {16}{x^{4} \log {\relax (x )}^{2} + 2 x^{3} \log {\relax (x )} + 2 x^{2} \log {\relax (x )} \log {\left (\frac {1}{2 x - 3} \right )} + x^{2} + 2 x \log {\left (\frac {1}{2 x - 3} \right )} + \left (2 x^{2} \log {\relax (x )} + 2 x + 2 \log {\left (\frac {1}{2 x - 3} \right )}\right ) e^{x} + e^{2 x} + \log {\left (\frac {1}{2 x - 3} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x**2+192*x)*ln(x)+(-64*x+96)*exp(x)-64*x**2+32*x+160)/((2*x**7-3*x**6)*ln(x)**3+((6*x**5-9*x*
*4)*ln(1/(2*x-3))+(6*x**5-9*x**4)*exp(x)+6*x**6-9*x**5)*ln(x)**2+((6*x**3-9*x**2)*ln(1/(2*x-3))**2+((12*x**3-1
8*x**2)*exp(x)+12*x**4-18*x**3)*ln(1/(2*x-3))+(6*x**3-9*x**2)*exp(x)**2+(12*x**4-18*x**3)*exp(x)+6*x**5-9*x**4
)*ln(x)+(2*x-3)*ln(1/(2*x-3))**3+((6*x-9)*exp(x)+6*x**2-9*x)*ln(1/(2*x-3))**2+((6*x-9)*exp(x)**2+(12*x**2-18*x
)*exp(x)+6*x**3-9*x**2)*ln(1/(2*x-3))+(2*x-3)*exp(x)**3+(6*x**2-9*x)*exp(x)**2+(6*x**3-9*x**2)*exp(x)+2*x**4-3
*x**3),x)

[Out]

16/(x**4*log(x)**2 + 2*x**3*log(x) + 2*x**2*log(x)*log(1/(2*x - 3)) + x**2 + 2*x*log(1/(2*x - 3)) + (2*x**2*lo
g(x) + 2*x + 2*log(1/(2*x - 3)))*exp(x) + exp(2*x) + log(1/(2*x - 3))**2)

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