3.74.96 \(\int \frac {-32+e^x (-72+8 x) \log (x)+e^x (64 x-8 x^2) \log ^2(x)+(-e^{30}+e^{2 x} (144 x-34 x^2+2 x^3)) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx\)

Optimal. Leaf size=30 \[ 3+\frac {\left (e^x (9-x)+\frac {4}{\log (x)}\right )^2}{e^{30}}-\log (3 x) \]

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Rubi [F]  time = 1.02, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32 + E^x*(-72 + 8*x)*Log[x] + E^x*(64*x - 8*x^2)*Log[x]^2 + (-E^30 + E^(2*x)*(144*x - 34*x^2 + 2*x^3))*L
og[x]^3)/(E^30*x*Log[x]^3),x]

[Out]

81*E^(-30 + 2*x) - 18*E^(-30 + 2*x)*x + E^(-30 + 2*x)*x^2 + 16/(E^30*Log[x]^2) - Log[x] + (8*Defer[Int][E^x/Lo
g[x]^2, x])/E^30 - (72*Defer[Int][E^x/(x*Log[x]^2), x])/E^30 + (64*Defer[Int][E^x/Log[x], x])/E^30 - (8*Defer[
Int][(E^x*x)/Log[x], x])/E^30

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{x \log ^3(x)} \, dx}{e^{30}}\\ &=\frac {\int \left (2 e^{2 x} (-9+x) (-8+x)-\frac {8 e^x \left (9-x-8 x \log (x)+x^2 \log (x)\right )}{x \log ^2(x)}+\frac {-32-e^{30} \log ^3(x)}{x \log ^3(x)}\right ) \, dx}{e^{30}}\\ &=\frac {\int \frac {-32-e^{30} \log ^3(x)}{x \log ^3(x)} \, dx}{e^{30}}+\frac {2 \int e^{2 x} (-9+x) (-8+x) \, dx}{e^{30}}-\frac {8 \int \frac {e^x \left (9-x-8 x \log (x)+x^2 \log (x)\right )}{x \log ^2(x)} \, dx}{e^{30}}\\ &=\frac {\operatorname {Subst}\left (\int \frac {-32-e^{30} x^3}{x^3} \, dx,x,\log (x)\right )}{e^{30}}+\frac {2 \int \left (72 e^{2 x}-17 e^{2 x} x+e^{2 x} x^2\right ) \, dx}{e^{30}}-\frac {8 \int \left (\frac {e^x (9-x)}{x \log ^2(x)}+\frac {e^x (-8+x)}{\log (x)}\right ) \, dx}{e^{30}}\\ &=\frac {\operatorname {Subst}\left (\int \left (-e^{30}-\frac {32}{x^3}\right ) \, dx,x,\log (x)\right )}{e^{30}}+\frac {2 \int e^{2 x} x^2 \, dx}{e^{30}}-\frac {8 \int \frac {e^x (9-x)}{x \log ^2(x)} \, dx}{e^{30}}-\frac {8 \int \frac {e^x (-8+x)}{\log (x)} \, dx}{e^{30}}-\frac {34 \int e^{2 x} x \, dx}{e^{30}}+\frac {144 \int e^{2 x} \, dx}{e^{30}}\\ &=72 e^{-30+2 x}-17 e^{-30+2 x} x+e^{-30+2 x} x^2+\frac {16}{e^{30} \log ^2(x)}-\log (x)-\frac {2 \int e^{2 x} x \, dx}{e^{30}}-\frac {8 \int \left (-\frac {e^x}{\log ^2(x)}+\frac {9 e^x}{x \log ^2(x)}\right ) \, dx}{e^{30}}-\frac {8 \int \left (-\frac {8 e^x}{\log (x)}+\frac {e^x x}{\log (x)}\right ) \, dx}{e^{30}}+\frac {17 \int e^{2 x} \, dx}{e^{30}}\\ &=\frac {161}{2} e^{-30+2 x}-18 e^{-30+2 x} x+e^{-30+2 x} x^2+\frac {16}{e^{30} \log ^2(x)}-\log (x)+\frac {\int e^{2 x} \, dx}{e^{30}}+\frac {8 \int \frac {e^x}{\log ^2(x)} \, dx}{e^{30}}-\frac {8 \int \frac {e^x x}{\log (x)} \, dx}{e^{30}}+\frac {64 \int \frac {e^x}{\log (x)} \, dx}{e^{30}}-\frac {72 \int \frac {e^x}{x \log ^2(x)} \, dx}{e^{30}}\\ &=81 e^{-30+2 x}-18 e^{-30+2 x} x+e^{-30+2 x} x^2+\frac {16}{e^{30} \log ^2(x)}-\log (x)+\frac {8 \int \frac {e^x}{\log ^2(x)} \, dx}{e^{30}}-\frac {8 \int \frac {e^x x}{\log (x)} \, dx}{e^{30}}+\frac {64 \int \frac {e^x}{\log (x)} \, dx}{e^{30}}-\frac {72 \int \frac {e^x}{x \log ^2(x)} \, dx}{e^{30}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 42, normalized size = 1.40 \begin {gather*} -\frac {-e^{2 x} (-9+x)^2-\frac {16}{\log ^2(x)}+\frac {8 e^x (-9+x)}{\log (x)}+e^{30} \log (x)}{e^{30}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32 + E^x*(-72 + 8*x)*Log[x] + E^x*(64*x - 8*x^2)*Log[x]^2 + (-E^30 + E^(2*x)*(144*x - 34*x^2 + 2*x
^3))*Log[x]^3)/(E^30*x*Log[x]^3),x]

[Out]

-((-(E^(2*x)*(-9 + x)^2) - 16/Log[x]^2 + (8*E^x*(-9 + x))/Log[x] + E^30*Log[x])/E^30)

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fricas [A]  time = 0.59, size = 43, normalized size = 1.43 \begin {gather*} \frac {{\left ({\left (x^{2} - 18 \, x + 81\right )} e^{\left (2 \, x\right )} \log \relax (x)^{2} - e^{30} \log \relax (x)^{3} - 8 \, {\left (x - 9\right )} e^{x} \log \relax (x) + 16\right )} e^{\left (-30\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x)*exp(x)*log(x)^2+(8*x-72)*exp(x)*lo
g(x)-32)/x/exp(15)^2/log(x)^3,x, algorithm="fricas")

[Out]

((x^2 - 18*x + 81)*e^(2*x)*log(x)^2 - e^30*log(x)^3 - 8*(x - 9)*e^x*log(x) + 16)*e^(-30)/log(x)^2

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giac [B]  time = 0.12, size = 63, normalized size = 2.10 \begin {gather*} \frac {{\left (x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 18 \, x e^{\left (2 \, x\right )} \log \relax (x)^{2} - e^{30} \log \relax (x)^{3} - 8 \, x e^{x} \log \relax (x) + 81 \, e^{\left (2 \, x\right )} \log \relax (x)^{2} + 72 \, e^{x} \log \relax (x) + 16\right )} e^{\left (-30\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x)*exp(x)*log(x)^2+(8*x-72)*exp(x)*lo
g(x)-32)/x/exp(15)^2/log(x)^3,x, algorithm="giac")

[Out]

(x^2*e^(2*x)*log(x)^2 - 18*x*e^(2*x)*log(x)^2 - e^30*log(x)^3 - 8*x*e^x*log(x) + 81*e^(2*x)*log(x)^2 + 72*e^x*
log(x) + 16)*e^(-30)/log(x)^2

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maple [A]  time = 0.04, size = 55, normalized size = 1.83

method	result	size
risch	\(x^{2} {\mathrm e}^{-30+2 x}-\ln \relax (x )-18 x \,{\mathrm e}^{-30+2 x}+81 \,{\mathrm e}^{-30+2 x}-\frac {8 \,{\mathrm e}^{-30} \left (x \,{\mathrm e}^{x} \ln \relax (x )-9 \,{\mathrm e}^{x} \ln \relax (x )-2\right )}{\ln \relax (x )^{2}}\)	\(55\)
default	\({\mathrm e}^{-30} \left (\frac {-8 \,{\mathrm e}^{x} x +72 \,{\mathrm e}^{x}}{\ln \relax (x )}+{\mathrm e}^{2 x} x^{2}-18 x \,{\mathrm e}^{2 x}+81 \,{\mathrm e}^{2 x}+\frac {16}{\ln \relax (x )^{2}}-{\mathrm e}^{30} \ln \relax (x )\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*ln(x)^3+(-8*x^2+64*x)*exp(x)*ln(x)^2+(8*x-72)*exp(x)*ln(x)-32)/
x/exp(15)^2/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(-30+2*x)-ln(x)-18*x*exp(-30+2*x)+81*exp(-30+2*x)-8*exp(-30)*(x*exp(x)*ln(x)-9*exp(x)*ln(x)-2)/ln(x)^2

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maxima [B]  time = 0.40, size = 60, normalized size = 2.00 \begin {gather*} \frac {1}{2} \, {\left ({\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 17 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 2 \, e^{30} \log \relax (x) - \frac {16 \, {\left (x - 9\right )} e^{x}}{\log \relax (x)} + \frac {32}{\log \relax (x)^{2}} + 144 \, e^{\left (2 \, x\right )}\right )} e^{\left (-30\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x)*exp(x)*log(x)^2+(8*x-72)*exp(x)*lo
g(x)-32)/x/exp(15)^2/log(x)^3,x, algorithm="maxima")

[Out]

1/2*((2*x^2 - 2*x + 1)*e^(2*x) - 17*(2*x - 1)*e^(2*x) - 2*e^30*log(x) - 16*(x - 9)*e^x/log(x) + 32/log(x)^2 +
144*e^(2*x))*e^(-30)

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mupad [B]  time = 5.23, size = 61, normalized size = 2.03 \begin {gather*} 81\,{\mathrm {e}}^{2\,x-30}-\ln \relax (x)+\frac {16\,{\mathrm {e}}^{-30}}{{\ln \relax (x)}^2}-18\,x\,{\mathrm {e}}^{2\,x-30}+x^2\,{\mathrm {e}}^{2\,x-30}+\frac {72\,{\mathrm {e}}^{x-30}}{\ln \relax (x)}-\frac {8\,x\,{\mathrm {e}}^{x-30}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-30)*(log(x)^3*(exp(30) - exp(2*x)*(144*x - 34*x^2 + 2*x^3)) - exp(x)*log(x)*(8*x - 72) - exp(x)*log
(x)^2*(64*x - 8*x^2) + 32))/(x*log(x)^3),x)

[Out]

81*exp(2*x - 30) - log(x) + (16*exp(-30))/log(x)^2 - 18*x*exp(2*x - 30) + x^2*exp(2*x - 30) + (72*exp(x - 30))
/log(x) - (8*x*exp(x - 30))/log(x)

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sympy [B]  time = 0.56, size = 68, normalized size = 2.27 \begin {gather*} \frac {\left (- 8 x e^{30} + 72 e^{30}\right ) e^{x} + \left (x^{2} e^{30} \log {\relax (x )} - 18 x e^{30} \log {\relax (x )} + 81 e^{30} \log {\relax (x )}\right ) e^{2 x}}{e^{60} \log {\relax (x )}} - \log {\relax (x )} + \frac {16}{e^{30} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-34*x**2+144*x)*exp(x)**2-exp(15)**2)*ln(x)**3+(-8*x**2+64*x)*exp(x)*ln(x)**2+(8*x-72)*exp(
x)*ln(x)-32)/x/exp(15)**2/ln(x)**3,x)

[Out]

((-8*x*exp(30) + 72*exp(30))*exp(x) + (x**2*exp(30)*log(x) - 18*x*exp(30)*log(x) + 81*exp(30)*log(x))*exp(2*x)
)*exp(-60)/log(x) - log(x) + 16*exp(-30)/log(x)**2

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