3.63.37 \(\int \frac {e^{-x+\frac {e^{-x} (2 e^x+5 x)}{\log (-3+2 x)}} (100 e^x+250 x+(375-625 x+250 x^2) \log (-3+2 x))}{(-3+2 x) \log ^2(-3+2 x)} \, dx\)

Optimal. Leaf size=27 \[ 25 \left (4-e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}}\right ) \]

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Rubi [F]  time = 3.32, 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 {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (100 e^x+250 x+\left (375-625 x+250 x^2\right ) \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(100*E^x + 250*x + (375 - 625*x + 250*x^2)*Log[-3 + 2*x]))/((-
3 + 2*x)*Log[-3 + 2*x]^2),x]

[Out]

125*Defer[Int][E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))/Log[-3 + 2*x]^2, x] + 375*Defer[Int][E^(-x + (2*E^x
+ 5*x)/(E^x*Log[-3 + 2*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2), x] + 100*Defer[Int][E^((2*E^x + 5*x)/(E^x*Log[-3 + 2
*x]))/((-3 + 2*x)*Log[-3 + 2*x]^2), x] + (125*Defer[Int][E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))/Log[-3 + 2
*x], x])/2 + (125*Defer[Int][(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(-3 + 2*x))/Log[-3 + 2*x], x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {100 e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)}+\frac {125 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)}\right ) \, dx\\ &=100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) \left (2 x+3 \log (-3+2 x)-5 x \log (-3+2 x)+2 x^2 \log (-3+2 x)\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx\\ &=100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {2 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)}\right ) \, dx\\ &=100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-1+x)}{\log (-3+2 x)} \, dx+250 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) x}{(-3+2 x) \log ^2(-3+2 x)} \, dx\\ &=100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log (-3+2 x)}+\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{2 \log (-3+2 x)}\right ) \, dx+250 \int \left (\frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 \log ^2(-3+2 x)}+\frac {3 \exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{2 (-3+2 x) \log ^2(-3+2 x)}\right ) \, dx\\ &=\frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log (-3+2 x)} \, dx+\frac {125}{2} \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right ) (-3+2 x)}{\log (-3+2 x)} \, dx+100 \int \frac {e^{\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}}}{(-3+2 x) \log ^2(-3+2 x)} \, dx+125 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{\log ^2(-3+2 x)} \, dx+375 \int \frac {\exp \left (-x+\frac {e^{-x} \left (2 e^x+5 x\right )}{\log (-3+2 x)}\right )}{(-3+2 x) \log ^2(-3+2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.25, size = 23, normalized size = 0.85 \begin {gather*} -25 e^{\frac {2+5 e^{-x} x}{\log (-3+2 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + (2*E^x + 5*x)/(E^x*Log[-3 + 2*x]))*(100*E^x + 250*x + (375 - 625*x + 250*x^2)*Log[-3 + 2*x]
))/((-3 + 2*x)*Log[-3 + 2*x]^2),x]

[Out]

-25*E^((2 + (5*x)/E^x)/Log[-3 + 2*x])

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fricas [A]  time = 0.73, size = 37, normalized size = 1.37 \begin {gather*} -25 \, e^{\left (x - \frac {{\left (x e^{x} \log \left (2 \, x - 3\right ) - 5 \, x - 2 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (2 \, x - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x^2-625*x+375)*log(2*x-3)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/exp(x)/log(2*x-3))/(2*x-3)/exp(
x)/log(2*x-3)^2,x, algorithm="fricas")

[Out]

-25*e^(x - (x*e^x*log(2*x - 3) - 5*x - 2*e^x)*e^(-x)/log(2*x - 3))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x^2-625*x+375)*log(2*x-3)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/exp(x)/log(2*x-3))/(2*x-3)/exp(
x)/log(2*x-3)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 145.02Not invertible Error: Bad Argument Value

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maple [A]  time = 0.36, size = 25, normalized size = 0.93

method	result	size
risch	\(-25 \,{\mathrm e}^{\frac {\left (2 \,{\mathrm e}^{x}+5 x \right ) {\mathrm e}^{-x}}{\ln \left (2 x -3\right )}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((250*x^2-625*x+375)*ln(2*x-3)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/exp(x)/ln(2*x-3))/(2*x-3)/exp(x)/ln(2*
x-3)^2,x,method=_RETURNVERBOSE)

[Out]

-25*exp((2*exp(x)+5*x)*exp(-x)/ln(2*x-3))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x^2-625*x+375)*log(2*x-3)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/exp(x)/log(2*x-3))/(2*x-3)/exp(
x)/log(2*x-3)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 4.28, size = 21, normalized size = 0.78 \begin {gather*} -25\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{-x}+2}{\ln \left (2\,x-3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp((exp(-x)*(5*x + 2*exp(x)))/log(2*x - 3))*(250*x + 100*exp(x) + log(2*x - 3)*(250*x^2 - 625*x
+ 375)))/(log(2*x - 3)^2*(2*x - 3)),x)

[Out]

-25*exp((5*x*exp(-x) + 2)/log(2*x - 3))

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sympy [A]  time = 0.77, size = 22, normalized size = 0.81 \begin {gather*} - 25 e^{\frac {\left (5 x + 2 e^{x}\right ) e^{- x}}{\log {\left (2 x - 3 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x**2-625*x+375)*ln(2*x-3)+100*exp(x)+250*x)*exp((2*exp(x)+5*x)/exp(x)/ln(2*x-3))/(2*x-3)/exp(x
)/ln(2*x-3)**2,x)

[Out]

-25*exp((5*x + 2*exp(x))*exp(-x)/log(2*x - 3))

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