3.38.12 \(\int \frac {200 x^4+e^{5 x} (-200 x^4-1250 x^5)+(-750 x^4+750 e^{5 x} x^4) \log (-1+e^{5 x})+(-1250 x^4+1250 e^{5 x} x^4) \log ^2(-1+e^{5 x})}{-1+e^{5 x}+(10-10 e^{5 x}) \log (-1+e^{5 x})+(-25+25 e^{5 x}) \log ^2(-1+e^{5 x})} \, dx\)

Optimal. Leaf size=23 \[ 10 x^4 \left (x+\frac {x}{-\frac {1}{5}+\log \left (-1+e^{5 x}\right )}\right ) \]

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Rubi [F]  time = 5.39, 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 {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(200*x^4 + E^(5*x)*(-200*x^4 - 1250*x^5) + (-750*x^4 + 750*E^(5*x)*x^4)*Log[-1 + E^(5*x)] + (-1250*x^4 + 1
250*E^(5*x)*x^4)*Log[-1 + E^(5*x)]^2)/(-1 + E^(5*x) + (10 - 10*E^(5*x))*Log[-1 + E^(5*x)] + (-25 + 25*E^(5*x))
*Log[-1 + E^(5*x)]^2),x]

[Out]

10*x^5 - 1250*Defer[Int][x^5/(-1 + 5*Log[-1 + E^(5*x)])^2, x] - 250*Defer[Int][x^5/((-1 + E^x)*(-1 + 5*Log[-1
+ E^(5*x)])^2), x] + 1000*Defer[Int][x^5/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*(-1 + 5*Log[-1 + E^(5*x)])^2
), x] + 750*Defer[Int][(E^x*x^5)/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*(-1 + 5*Log[-1 + E^(5*x)])^2), x] +
500*Defer[Int][(E^(2*x)*x^5)/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*(-1 + 5*Log[-1 + E^(5*x)])^2), x] + 250*
Defer[Int][(E^(3*x)*x^5)/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*(-1 + 5*Log[-1 + E^(5*x)])^2), x] + 250*Defe
r[Int][x^4/(-1 + 5*Log[-1 + E^(5*x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 x^4 \left (-4+e^{5 x} (4+25 x)-15 \left (-1+e^{5 x}\right ) \log \left (-1+e^{5 x}\right )-25 \left (-1+e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\\ &=50 \int \frac {x^4 \left (-4+e^{5 x} (4+25 x)-15 \left (-1+e^{5 x}\right ) \log \left (-1+e^{5 x}\right )-25 \left (-1+e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\\ &=50 \int \left (-\frac {5 x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {5 \left (4+3 e^x+2 e^{2 x}+e^{3 x}\right ) x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}-\frac {x^4 \left (4+25 x-15 \log \left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )\right )}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}\right ) \, dx\\ &=-\left (50 \int \frac {x^4 \left (4+25 x-15 \log \left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )\right )}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\right )-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {\left (4+3 e^x+2 e^{2 x}+e^{3 x}\right ) x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\\ &=-\left (50 \int \left (-x^4+\frac {25 x^5}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}-\frac {5 x^4}{-1+5 \log \left (-1+e^{5 x}\right )}\right ) \, dx\right )-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \left (\frac {4 x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {3 e^x x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {2 e^{2 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}+\frac {e^{3 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2}\right ) \, dx\\ &=10 x^5-250 \int \frac {x^5}{\left (-1+e^x\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {e^{3 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+250 \int \frac {x^4}{-1+5 \log \left (-1+e^{5 x}\right )} \, dx+500 \int \frac {e^{2 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+750 \int \frac {e^x x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx+1000 \int \frac {x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx-1250 \int \frac {x^5}{\left (-1+5 \log \left (-1+e^{5 x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.72, size = 29, normalized size = 1.26 \begin {gather*} -50 \left (-\frac {x^5}{5}-\frac {x^5}{-1+5 \log \left (-1+e^{5 x}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(200*x^4 + E^(5*x)*(-200*x^4 - 1250*x^5) + (-750*x^4 + 750*E^(5*x)*x^4)*Log[-1 + E^(5*x)] + (-1250*x
^4 + 1250*E^(5*x)*x^4)*Log[-1 + E^(5*x)]^2)/(-1 + E^(5*x) + (10 - 10*E^(5*x))*Log[-1 + E^(5*x)] + (-25 + 25*E^
(5*x))*Log[-1 + E^(5*x)]^2),x]

[Out]

-50*(-1/5*x^5 - x^5/(-1 + 5*Log[-1 + E^(5*x)]))

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fricas [A]  time = 0.85, size = 33, normalized size = 1.43 \begin {gather*} \frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x)-750*x^4)*log(exp(5*x)-1)+(-1250*x^
5-200*x^4)*exp(5*x)+200*x^4)/((25*exp(5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1)
,x, algorithm="fricas")

[Out]

10*(5*x^5*log(e^(5*x) - 1) + 4*x^5)/(5*log(e^(5*x) - 1) - 1)

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giac [A]  time = 0.41, size = 33, normalized size = 1.43 \begin {gather*} \frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x)-750*x^4)*log(exp(5*x)-1)+(-1250*x^
5-200*x^4)*exp(5*x)+200*x^4)/((25*exp(5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1)
,x, algorithm="giac")

[Out]

10*(5*x^5*log(e^(5*x) - 1) + 4*x^5)/(5*log(e^(5*x) - 1) - 1)

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maple [A]  time = 0.03, size = 25, normalized size = 1.09

method	result	size
risch	\(10 x^{5}+\frac {50 x^{5}}{-1+5 \ln \left ({\mathrm e}^{5 x}-1\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1250*x^4*exp(5*x)-1250*x^4)*ln(exp(5*x)-1)^2+(750*x^4*exp(5*x)-750*x^4)*ln(exp(5*x)-1)+(-1250*x^5-200*x^
4)*exp(5*x)+200*x^4)/((25*exp(5*x)-25)*ln(exp(5*x)-1)^2+(-10*exp(5*x)+10)*ln(exp(5*x)-1)+exp(5*x)-1),x,method=
_RETURNVERBOSE)

[Out]

10*x^5+50*x^5/(-1+5*ln(exp(5*x)-1))

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maxima [B]  time = 0.55, size = 70, normalized size = 3.04 \begin {gather*} \frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, x^{5} \log \left (e^{x} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, \log \left (e^{x} - 1\right ) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x)-750*x^4)*log(exp(5*x)-1)+(-1250*x^
5-200*x^4)*exp(5*x)+200*x^4)/((25*exp(5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1)
,x, algorithm="maxima")

[Out]

10*(5*x^5*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) + 5*x^5*log(e^x - 1) + 4*x^5)/(5*log(e^(4*x) + e^(3*x) +
e^(2*x) + e^x + 1) + 5*log(e^x - 1) - 1)

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mupad [B]  time = 2.31, size = 79, normalized size = 3.43 \begin {gather*} \frac {10\,x^4\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}+5\,x\,{\mathrm {e}}^{5\,x}-1\right )-50\,x^4\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}-1\right )}{5\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )-1}-10\,x^4\,{\mathrm {e}}^{-5\,x}+10\,x^4+10\,x^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(5*x) - 1)*(750*x^4*exp(5*x) - 750*x^4) - exp(5*x)*(200*x^4 + 1250*x^5) + log(exp(5*x) - 1)^2*(125
0*x^4*exp(5*x) - 1250*x^4) + 200*x^4)/(exp(5*x) - log(exp(5*x) - 1)*(10*exp(5*x) - 10) + log(exp(5*x) - 1)^2*(
25*exp(5*x) - 25) - 1),x)

[Out]

(10*x^4*exp(-5*x)*(exp(5*x) + 5*x*exp(5*x) - 1) - 50*x^4*log(exp(5*x) - 1)*exp(-5*x)*(exp(5*x) - 1))/(5*log(ex
p(5*x) - 1) - 1) - 10*x^4*exp(-5*x) + 10*x^4 + 10*x^5

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sympy [A]  time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} 10 x^{5} + \frac {50 x^{5}}{5 \log {\left (e^{5 x} - 1 \right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1250*x**4*exp(5*x)-1250*x**4)*ln(exp(5*x)-1)**2+(750*x**4*exp(5*x)-750*x**4)*ln(exp(5*x)-1)+(-1250
*x**5-200*x**4)*exp(5*x)+200*x**4)/((25*exp(5*x)-25)*ln(exp(5*x)-1)**2+(-10*exp(5*x)+10)*ln(exp(5*x)-1)+exp(5*
x)-1),x)

[Out]

10*x**5 + 50*x**5/(5*log(exp(5*x) - 1) - 1)

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