3.103.98 \(\int \frac {e^{2 x} (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5))}{81-36 x^2+4 x^4+(18-18 x-4 x^2+4 x^3) \log (3)+(1-2 x+x^2) \log ^2(3)+(18-4 x^2+(2-2 x) \log (3)) \log (5)+\log ^2(5)} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^{2 x}}{9+\log (3)-x (2 x+\log (3))+\log (5)} \]

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Rubi [A]  time = 0.75, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 2289} \begin {gather*} \frac {e^{2 x}}{-2 x^2-x \log (3)+9+\log (15)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(18 + 4*x - 4*x^2 + (3 - 2*x)*Log[3] + 2*Log[5]))/(81 - 36*x^2 + 4*x^4 + (18 - 18*x - 4*x^2 + 4*x
^3)*Log[3] + (1 - 2*x + x^2)*Log[3]^2 + (18 - 4*x^2 + (2 - 2*x)*Log[3])*Log[5] + Log[5]^2),x]

[Out]

E^(2*x)/(9 - 2*x^2 - x*Log[3] + Log[15])

Rule 2289

Int[(F_)^(u_)*(v_)^(n_.)*(w_), x_Symbol] :> With[{z = Log[F]*v*D[u, x] + (n + 1)*D[v, x]}, Simp[(Coefficient[w
, x, Exponent[w, x]]*F^u*v^(n + 1))/Coefficient[z, x, Exponent[z, x]], x] /; EqQ[Exponent[w, x], Exponent[z, x
]] && EqQ[w*Coefficient[z, x, Exponent[z, x]], z*Coefficient[w, x, Exponent[w, x]]]] /; FreeQ[{F, n}, x] && Po
lynomialQ[u, x] && PolynomialQ[v, x] && PolynomialQ[w, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (18-4 x^2+x (4-\log (9))+\log (675)\right )}{\left (9-2 x^2-x \log (3)+\log (15)\right )^2} \, dx\\ &=\frac {e^{2 x}}{9-2 x^2-x \log (3)+\log (15)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{2 x} \left (18+4 x-4 x^2+(3-2 x) \log (3)+2 \log (5)\right )}{81-36 x^2+4 x^4+\left (18-18 x-4 x^2+4 x^3\right ) \log (3)+\left (1-2 x+x^2\right ) \log ^2(3)+\left (18-4 x^2+(2-2 x) \log (3)\right ) \log (5)+\log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(2*x)*(18 + 4*x - 4*x^2 + (3 - 2*x)*Log[3] + 2*Log[5]))/(81 - 36*x^2 + 4*x^4 + (18 - 18*x - 4*x^2
+ 4*x^3)*Log[3] + (1 - 2*x + x^2)*Log[3]^2 + (18 - 4*x^2 + (2 - 2*x)*Log[3])*Log[5] + Log[5]^2),x]

[Out]

Integrate[(E^(2*x)*(18 + 4*x - 4*x^2 + (3 - 2*x)*Log[3] + 2*Log[5]))/(81 - 36*x^2 + 4*x^4 + (18 - 18*x - 4*x^2
+ 4*x^3)*Log[3] + (1 - 2*x + x^2)*Log[3]^2 + (18 - 4*x^2 + (2 - 2*x)*Log[3])*Log[5] + Log[5]^2), x]

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fricas [A]  time = 0.74, size = 25, normalized size = 1.09 \begin {gather*} -\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + {\left (x - 1\right )} \log \relax (3) - \log \relax (5) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5)+(3-2*x)*log(3)-4*x^2+4*x+18)*exp(x)^2/(log(5)^2+((-2*x+2)*log(3)-4*x^2+18)*log(5)+(x^2-2*x
+1)*log(3)^2+(4*x^3-4*x^2-18*x+18)*log(3)+4*x^4-36*x^2+81),x, algorithm="fricas")

[Out]

-e^(2*x)/(2*x^2 + (x - 1)*log(3) - log(5) - 9)

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giac [A]  time = 0.17, size = 27, normalized size = 1.17 \begin {gather*} -\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + x \log \relax (3) - \log \relax (5) - \log \relax (3) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5)+(3-2*x)*log(3)-4*x^2+4*x+18)*exp(x)^2/(log(5)^2+((-2*x+2)*log(3)-4*x^2+18)*log(5)+(x^2-2*x
+1)*log(3)^2+(4*x^3-4*x^2-18*x+18)*log(3)+4*x^4-36*x^2+81),x, algorithm="giac")

[Out]

-e^(2*x)/(2*x^2 + x*log(3) - log(5) - log(3) - 9)

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maple [A]  time = 0.24, size = 28, normalized size = 1.22

method	result	size
gosper	\(-\frac {{\mathrm e}^{2 x}}{x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9}\)	\(28\)
norman	\(-\frac {{\mathrm e}^{2 x}}{x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9}\)	\(28\)
risch	\(-\frac {{\mathrm e}^{2 x}}{x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9}\)	\(28\)
default	\(\frac {36 \,{\mathrm e}^{2 x} \ln \relax (3)}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}+\frac {{\mathrm e}^{2 x} \ln \relax (3)^{2}}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}+\frac {2 \,{\mathrm e}^{2 x} \left (x \ln \relax (3)^{2}-\ln \relax (3)^{2}-\ln \relax (3) \ln \relax (5)+4 x \ln \relax (3)+4 x \ln \relax (5)-9 \ln \relax (3)+36 x \right )}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}+\frac {4 \,{\mathrm e}^{2 x} \left (x \ln \relax (3)-2 \ln \relax (3)-2 \ln \relax (5)-18\right )}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}-\frac {18 \,{\mathrm e}^{2 x} \left (\ln \relax (3)+4 x \right )}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}-\frac {12 \ln \relax (3) {\mathrm e}^{2 x} x}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}+\frac {2 \ln \relax (5) {\mathrm e}^{2 x} \ln \relax (3)}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}-\frac {8 \ln \relax (5) {\mathrm e}^{2 x} x}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}-\frac {2 \ln \relax (3)^{2} {\mathrm e}^{2 x} x}{\left (\ln \relax (3)^{2}+8 \ln \relax (3)+8 \ln \relax (5)+72\right ) \left (x \ln \relax (3)+2 x^{2}-\ln \relax (3)-\ln \relax (5)-9\right )}\)	\(465\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(5)+(3-2*x)*ln(3)-4*x^2+4*x+18)*exp(x)^2/(ln(5)^2+((-2*x+2)*ln(3)-4*x^2+18)*ln(5)+(x^2-2*x+1)*ln(3)^2
+(4*x^3-4*x^2-18*x+18)*ln(3)+4*x^4-36*x^2+81),x,method=_RETURNVERBOSE)

[Out]

-exp(x)^2/(x*ln(3)+2*x^2-ln(3)-ln(5)-9)

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maxima [A]  time = 0.48, size = 27, normalized size = 1.17 \begin {gather*} -\frac {e^{\left (2 \, x\right )}}{2 \, x^{2} + x \log \relax (3) - \log \relax (5) - \log \relax (3) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5)+(3-2*x)*log(3)-4*x^2+4*x+18)*exp(x)^2/(log(5)^2+((-2*x+2)*log(3)-4*x^2+18)*log(5)+(x^2-2*x
+1)*log(3)^2+(4*x^3-4*x^2-18*x+18)*log(3)+4*x^4-36*x^2+81),x, algorithm="maxima")

[Out]

-e^(2*x)/(2*x^2 + x*log(3) - log(5) - log(3) - 9)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,x}\,\left (4\,x+2\,\ln \relax (5)-\ln \relax (3)\,\left (2\,x-3\right )-4\,x^2+18\right )}{{\ln \relax (3)}^2\,\left (x^2-2\,x+1\right )-\ln \relax (5)\,\left (\ln \relax (3)\,\left (2\,x-2\right )+4\,x^2-18\right )-\ln \relax (3)\,\left (-4\,x^3+4\,x^2+18\,x-18\right )+{\ln \relax (5)}^2-36\,x^2+4\,x^4+81} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(4*x + 2*log(5) - log(3)*(2*x - 3) - 4*x^2 + 18))/(log(3)^2*(x^2 - 2*x + 1) - log(5)*(log(3)*(2*
x - 2) + 4*x^2 - 18) - log(3)*(18*x + 4*x^2 - 4*x^3 - 18) + log(5)^2 - 36*x^2 + 4*x^4 + 81),x)

[Out]

int((exp(2*x)*(4*x + 2*log(5) - log(3)*(2*x - 3) - 4*x^2 + 18))/(log(3)^2*(x^2 - 2*x + 1) - log(5)*(log(3)*(2*
x - 2) + 4*x^2 - 18) - log(3)*(18*x + 4*x^2 - 4*x^3 - 18) + log(5)^2 - 36*x^2 + 4*x^4 + 81), x)

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sympy [A]  time = 0.24, size = 24, normalized size = 1.04 \begin {gather*} - \frac {e^{2 x}}{2 x^{2} + x \log {\relax (3 )} - 9 - \log {\relax (5 )} - \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(5)+(3-2*x)*ln(3)-4*x**2+4*x+18)*exp(x)**2/(ln(5)**2+((-2*x+2)*ln(3)-4*x**2+18)*ln(5)+(x**2-2*x
+1)*ln(3)**2+(4*x**3-4*x**2-18*x+18)*ln(3)+4*x**4-36*x**2+81),x)

[Out]

-exp(2*x)/(2*x**2 + x*log(3) - 9 - log(5) - log(3))

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