3.98.75 \(\int \frac {(e^{10}-e^5 x) \log ^2(e^{10}-2 e^5 x+x^2)+e^{\frac {4 x^3}{e^5 \log (e^{10}-2 e^5 x+x^2)}} (8 x^3+(12 e^5 x^2-12 x^3) \log (e^{10}-2 e^5 x+x^2))}{(e^{10}-e^5 x) \log ^2(e^{10}-2 e^5 x+x^2)} \, dx\)

Optimal. Leaf size=25 \[ 1+e^{\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}}+x \]

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Rubi [A]  time = 2.19, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6742, 6688, 6706} \begin {gather*} e^{\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((E^10 - E^5*x)*Log[E^10 - 2*E^5*x + x^2]^2 + E^((4*x^3)/(E^5*Log[E^10 - 2*E^5*x + x^2]))*(8*x^3 + (12*E^5
*x^2 - 12*x^3)*Log[E^10 - 2*E^5*x + x^2]))/((E^10 - E^5*x)*Log[E^10 - 2*E^5*x + x^2]^2),x]

[Out]

E^((4*x^3)/(E^5*Log[(E^5 - x)^2])) + x

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {4 e^{-5+\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}} x^2 \left (2 x+3 e^5 \log \left (\left (e^5-x\right )^2\right )-3 x \log \left (\left (e^5-x\right )^2\right )\right )}{\left (e^5-x\right ) \log ^2\left (\left (e^5-x\right )^2\right )}\right ) \, dx\\ &=x+4 \int \frac {e^{-5+\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}} x^2 \left (2 x+3 e^5 \log \left (\left (e^5-x\right )^2\right )-3 x \log \left (\left (e^5-x\right )^2\right )\right )}{\left (e^5-x\right ) \log ^2\left (\left (e^5-x\right )^2\right )} \, dx\\ &=x+4 \int \frac {e^{-5+\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}} x^2 \left (2 x+3 \left (e^5-x\right ) \log \left (\left (e^5-x\right )^2\right )\right )}{\left (e^5-x\right ) \log ^2\left (\left (e^5-x\right )^2\right )} \, dx\\ &=e^{\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.27, size = 29, normalized size = 1.16 \begin {gather*} -e^5+e^{\frac {4 x^3}{e^5 \log \left (\left (e^5-x\right )^2\right )}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^10 - E^5*x)*Log[E^10 - 2*E^5*x + x^2]^2 + E^((4*x^3)/(E^5*Log[E^10 - 2*E^5*x + x^2]))*(8*x^3 + (
12*E^5*x^2 - 12*x^3)*Log[E^10 - 2*E^5*x + x^2]))/((E^10 - E^5*x)*Log[E^10 - 2*E^5*x + x^2]^2),x]

[Out]

-E^5 + E^((4*x^3)/(E^5*Log[(E^5 - x)^2])) + x

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fricas [A]  time = 0.76, size = 24, normalized size = 0.96 \begin {gather*} x + e^{\left (\frac {4 \, x^{3} e^{\left (-5\right )}}{\log \left (x^{2} - 2 \, x e^{5} + e^{10}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2*exp(5)-12*x^3)*log(exp(5)^2-2*x*exp(5)+x^2)+8*x^3)*exp(4*x^3/exp(5)/log(exp(5)^2-2*x*exp(5
)+x^2))+(exp(5)^2-x*exp(5))*log(exp(5)^2-2*x*exp(5)+x^2)^2)/(exp(5)^2-x*exp(5))/log(exp(5)^2-2*x*exp(5)+x^2)^2
,x, algorithm="fricas")

[Out]

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

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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((((12*x^2*exp(5)-12*x^3)*log(exp(5)^2-2*x*exp(5)+x^2)+8*x^3)*exp(4*x^3/exp(5)/log(exp(5)^2-2*x*exp(5
)+x^2))+(exp(5)^2-x*exp(5))*log(exp(5)^2-2*x*exp(5)+x^2)^2)/(exp(5)^2-x*exp(5))/log(exp(5)^2-2*x*exp(5)+x^2)^2
,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 1.81Unable to divide, perhaps due to rounding error%%%{-196608,[1,22,6]%%%}+%%%{786432,[1,
21,7]%%%}+%

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maple [A]  time = 0.41, size = 25, normalized size = 1.00

method	result	size
risch	\(x +{\mathrm e}^{\frac {4 x^{3} {\mathrm e}^{-5}}{\ln \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x^2*exp(5)-12*x^3)*ln(exp(5)^2-2*x*exp(5)+x^2)+8*x^3)*exp(4*x^3/exp(5)/ln(exp(5)^2-2*x*exp(5)+x^2))+
(exp(5)^2-x*exp(5))*ln(exp(5)^2-2*x*exp(5)+x^2)^2)/(exp(5)^2-x*exp(5))/ln(exp(5)^2-2*x*exp(5)+x^2)^2,x,method=
_RETURNVERBOSE)

[Out]

x+exp(4*x^3*exp(-5)/ln(exp(10)-2*x*exp(5)+x^2))

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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((((12*x^2*exp(5)-12*x^3)*log(exp(5)^2-2*x*exp(5)+x^2)+8*x^3)*exp(4*x^3/exp(5)/log(exp(5)^2-2*x*exp(5
)+x^2))+(exp(5)^2-x*exp(5))*log(exp(5)^2-2*x*exp(5)+x^2)^2)/(exp(5)^2-x*exp(5))/log(exp(5)^2-2*x*exp(5)+x^2)^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 = 6.05, size = 24, normalized size = 0.96 \begin {gather*} x+{\mathrm {e}}^{\frac {4\,x^3\,{\mathrm {e}}^{-5}}{\ln \left (x^2-2\,{\mathrm {e}}^5\,x+{\mathrm {e}}^{10}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(10) - 2*x*exp(5) + x^2)^2*(exp(10) - x*exp(5)) + exp((4*x^3*exp(-5))/log(exp(10) - 2*x*exp(5) + x
^2))*(8*x^3 + log(exp(10) - 2*x*exp(5) + x^2)*(12*x^2*exp(5) - 12*x^3)))/(log(exp(10) - 2*x*exp(5) + x^2)^2*(e
xp(10) - x*exp(5))),x)

[Out]

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

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sympy [A]  time = 0.58, size = 26, normalized size = 1.04 \begin {gather*} x + e^{\frac {4 x^{3}}{e^{5} \log {\left (x^{2} - 2 x e^{5} + e^{10} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x**2*exp(5)-12*x**3)*ln(exp(5)**2-2*x*exp(5)+x**2)+8*x**3)*exp(4*x**3/exp(5)/ln(exp(5)**2-2*x*
exp(5)+x**2))+(exp(5)**2-x*exp(5))*ln(exp(5)**2-2*x*exp(5)+x**2)**2)/(exp(5)**2-x*exp(5))/ln(exp(5)**2-2*x*exp
(5)+x**2)**2,x)

[Out]

x + exp(4*x**3*exp(-5)/log(x**2 - 2*x*exp(5) + exp(10)))

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