3.75.64 \(\int \frac {e (192 x+96 x^2+12 x^3)+(e^5 (-96-24 x)+e (192 x+48 x^2)) \log (e^4-2 x)}{(e^4 x^3-2 x^4) \log ^3(e^4-2 x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \]

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Rubi [F]  time = 1.16, 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 {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48*x^2))*Log[E^4 - 2*x])/((E^4*x^3 - 2*x^4)*
Log[E^4 - 2*x]^3),x]

[Out]

(3*(8 + E^4)^2)/(E^7*Log[E^4 - 2*x]^2) + (192*Defer[Int][1/(x^2*Log[E^4 - 2*x]^3), x])/E^3 + (96*(4 + E^4)*Def
er[Int][1/(x*Log[E^4 - 2*x]^3), x])/E^7 - 96*E*Defer[Int][1/(x^3*Log[E^4 - 2*x]^2), x] - 24*E*Defer[Int][1/(x^
2*Log[E^4 - 2*x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx\\ &=\int \frac {12 e (4+x) \left (x (4+x)-2 \left (e^4-2 x\right ) \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx\\ &=(12 e) \int \frac {(4+x) \left (x (4+x)-2 \left (e^4-2 x\right ) \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx\\ &=(12 e) \int \left (\frac {(4+x)^2}{\left (e^4-2 x\right ) x^2 \log ^3\left (e^4-2 x\right )}-\frac {2 (4+x)}{x^3 \log ^2\left (e^4-2 x\right )}\right ) \, dx\\ &=(12 e) \int \frac {(4+x)^2}{\left (e^4-2 x\right ) x^2 \log ^3\left (e^4-2 x\right )} \, dx-(24 e) \int \frac {4+x}{x^3 \log ^2\left (e^4-2 x\right )} \, dx\\ &=(12 e) \int \left (\frac {\left (8+e^4\right )^2}{e^8 \left (e^4-2 x\right ) \log ^3\left (e^4-2 x\right )}+\frac {16}{e^4 x^2 \log ^3\left (e^4-2 x\right )}+\frac {8 \left (4+e^4\right )}{e^8 x \log ^3\left (e^4-2 x\right )}\right ) \, dx-(24 e) \int \left (\frac {4}{x^3 \log ^2\left (e^4-2 x\right )}+\frac {1}{x^2 \log ^2\left (e^4-2 x\right )}\right ) \, dx\\ &=\frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}+\frac {\left (12 \left (8+e^4\right )^2\right ) \int \frac {1}{\left (e^4-2 x\right ) \log ^3\left (e^4-2 x\right )} \, dx}{e^7}\\ &=\frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}-\frac {\left (6 \left (8+e^4\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,e^4-2 x\right )}{e^7}\\ &=\frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}-\frac {\left (6 \left (8+e^4\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (e^4-2 x\right )\right )}{e^7}\\ &=\frac {3 \left (8+e^4\right )^2}{e^7 \log ^2\left (e^4-2 x\right )}+\frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 21, normalized size = 1.00 \begin {gather*} \frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48*x^2))*Log[E^4 - 2*x])/((E^4*x^3 - 2
*x^4)*Log[E^4 - 2*x]^3),x]

[Out]

(3*E*(4 + x)^2)/(x^2*Log[E^4 - 2*x]^2)

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fricas [A]  time = 0.82, size = 24, normalized size = 1.14 \begin {gather*} \frac {3 \, {\left (x^{2} + 8 \, x + 16\right )} e}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="fricas")

[Out]

3*(x^2 + 8*x + 16)*e/(x^2*log(-2*x + e^4)^2)

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giac [A]  time = 0.16, size = 30, normalized size = 1.43 \begin {gather*} \frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="giac")

[Out]

3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)

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maple [A]  time = 0.38, size = 25, normalized size = 1.19




method result size



risch \(\frac {3 \,{\mathrm e} \left (x^{2}+8 x +16\right )}{x^{2} \ln \left ({\mathrm e}^{4}-2 x \right )^{2}}\) \(25\)
norman \(\frac {24 x \,{\mathrm e}+3 x^{2} {\mathrm e}+48 \,{\mathrm e}}{\ln \left ({\mathrm e}^{4}-2 x \right )^{2} x^{2}}\) \(31\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*ln(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)
-2*x^4)/ln(exp(4)-2*x)^3,x,method=_RETURNVERBOSE)

[Out]

3*exp(1)*(x^2+8*x+16)/x^2/ln(exp(4)-2*x)^2

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maxima [A]  time = 0.43, size = 30, normalized size = 1.43 \begin {gather*} \frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="maxima")

[Out]

3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)

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mupad [B]  time = 4.65, size = 21, normalized size = 1.00 \begin {gather*} \frac {3\,\mathrm {e}\,{\left (x+4\right )}^2}{x^2\,{\ln \left ({\mathrm {e}}^4-2\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*(192*x + 96*x^2 + 12*x^3) + log(exp(4) - 2*x)*(exp(1)*(192*x + 48*x^2) - exp(5)*(24*x + 96)))/(log
(exp(4) - 2*x)^3*(x^3*exp(4) - 2*x^4)),x)

[Out]

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

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sympy [A]  time = 0.16, size = 32, normalized size = 1.52 \begin {gather*} \frac {3 e x^{2} + 24 e x + 48 e}{x^{2} \log {\left (- 2 x + e^{4} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x**2+192*x)*exp(1))*ln(exp(4)-2*x)+(12*x**3+96*x**2+192*x)*exp(1))/(x
**3*exp(4)-2*x**4)/ln(exp(4)-2*x)**3,x)

[Out]

(3*E*x**2 + 24*E*x + 48*E)/(x**2*log(-2*x + exp(4))**2)

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