3.43.80 \(\int \frac {18 x-24 x^2+8 x^3+e^{\frac {x}{-3+2 x+\log (25)}} (-3+\log (25))+(-12 x+8 x^2) \log (25)+2 x \log ^2(25)}{9-12 x+4 x^2+(-6+4 x) \log (25)+\log ^2(25)} \, dx\)

Optimal. Leaf size=18 \[ 4+e^{\frac {x}{-3+2 x+\log (25)}}+x^2 \]

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Rubi [A]  time = 0.26, antiderivative size = 31, normalized size of antiderivative = 1.72, number of steps used = 5, number of rules used = 4, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6, 6688, 2230, 2209} \begin {gather*} x^2+e^{\frac {1}{2}-\frac {3-\log (25)}{2 (-2 x+3-\log (25))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*x - 24*x^2 + 8*x^3 + E^(x/(-3 + 2*x + Log[25]))*(-3 + Log[25]) + (-12*x + 8*x^2)*Log[25] + 2*x*Log[25]
^2)/(9 - 12*x + 4*x^2 + (-6 + 4*x)*Log[25] + Log[25]^2),x]

[Out]

E^(1/2 - (3 - Log[25])/(2*(3 - 2*x - Log[25]))) + x^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

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 {-24 x^2+8 x^3+e^{\frac {x}{-3+2 x+\log (25)}} (-3+\log (25))+\left (-12 x+8 x^2\right ) \log (25)+x \left (18+2 \log ^2(25)\right )}{9-12 x+4 x^2+(-6+4 x) \log (25)+\log ^2(25)} \, dx\\ &=\int \left (2 x+\frac {e^{\frac {x}{-3+2 x+\log (25)}} (-3+\log (25))}{(-3+2 x+\log (25))^2}\right ) \, dx\\ &=x^2+(-3+\log (25)) \int \frac {e^{\frac {x}{-3+2 x+\log (25)}}}{(-3+2 x+\log (25))^2} \, dx\\ &=x^2+(-3+\log (25)) \int \frac {e^{\frac {1}{2}-\frac {-3+\log (25)}{2 (-3+2 x+\log (25))}}}{(-3+2 x+\log (25))^2} \, dx\\ &=e^{\frac {1}{2}-\frac {3-\log (25)}{2 (3-2 x-\log (25))}}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 37, normalized size = 2.06 \begin {gather*} 5^{\frac {1}{3-2 x-\log (25)}} e^{\frac {1}{2}+\frac {3}{2 (-3+2 x+\log (25))}}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*x - 24*x^2 + 8*x^3 + E^(x/(-3 + 2*x + Log[25]))*(-3 + Log[25]) + (-12*x + 8*x^2)*Log[25] + 2*x*L
og[25]^2)/(9 - 12*x + 4*x^2 + (-6 + 4*x)*Log[25] + Log[25]^2),x]

[Out]

5^(3 - 2*x - Log[25])^(-1)*E^(1/2 + 3/(2*(-3 + 2*x + Log[25]))) + x^2

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fricas [A]  time = 0.89, size = 18, normalized size = 1.00 \begin {gather*} x^{2} + e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)-3)*exp(x/(2*log(5)+2*x-3))+8*x*log(5)^2+2*(8*x^2-12*x)*log(5)+8*x^3-24*x^2+18*x)/(4*log(5
)^2+2*(4*x-6)*log(5)+4*x^2-12*x+9),x, algorithm="fricas")

[Out]

x^2 + e^(x/(2*x + 2*log(5) - 3))

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giac [B]  time = 0.36, size = 282, normalized size = 15.67 \begin {gather*} -\frac {\frac {32 \, x \log \relax (5)^{3}}{2 \, x + 2 \, \log \relax (5) - 3} - 8 \, \log \relax (5)^{3} - \frac {32 \, x e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )} \log \relax (5)}{2 \, x + 2 \, \log \relax (5) - 3} + \frac {32 \, x^{2} e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )} \log \relax (5)}{{\left (2 \, x + 2 \, \log \relax (5) - 3\right )}^{2}} + 8 \, e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )} \log \relax (5) - \frac {144 \, x \log \relax (5)^{2}}{2 \, x + 2 \, \log \relax (5) - 3} + 36 \, \log \relax (5)^{2} + \frac {48 \, x e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )}}{2 \, x + 2 \, \log \relax (5) - 3} - \frac {48 \, x^{2} e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )}}{{\left (2 \, x + 2 \, \log \relax (5) - 3\right )}^{2}} + \frac {216 \, x \log \relax (5)}{2 \, x + 2 \, \log \relax (5) - 3} - \frac {108 \, x}{2 \, x + 2 \, \log \relax (5) - 3} - 12 \, e^{\left (\frac {x}{2 \, x + 2 \, \log \relax (5) - 3}\right )} - 54 \, \log \relax (5) + 27}{4 \, {\left (\frac {4 \, x}{2 \, x + 2 \, \log \relax (5) - 3} - \frac {4 \, x^{2}}{{\left (2 \, x + 2 \, \log \relax (5) - 3\right )}^{2}} - 1\right )} {\left (2 \, \log \relax (5) - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)-3)*exp(x/(2*log(5)+2*x-3))+8*x*log(5)^2+2*(8*x^2-12*x)*log(5)+8*x^3-24*x^2+18*x)/(4*log(5
)^2+2*(4*x-6)*log(5)+4*x^2-12*x+9),x, algorithm="giac")

[Out]

-1/4*(32*x*log(5)^3/(2*x + 2*log(5) - 3) - 8*log(5)^3 - 32*x*e^(x/(2*x + 2*log(5) - 3))*log(5)/(2*x + 2*log(5)
- 3) + 32*x^2*e^(x/(2*x + 2*log(5) - 3))*log(5)/(2*x + 2*log(5) - 3)^2 + 8*e^(x/(2*x + 2*log(5) - 3))*log(5)
- 144*x*log(5)^2/(2*x + 2*log(5) - 3) + 36*log(5)^2 + 48*x*e^(x/(2*x + 2*log(5) - 3))/(2*x + 2*log(5) - 3) - 4
8*x^2*e^(x/(2*x + 2*log(5) - 3))/(2*x + 2*log(5) - 3)^2 + 216*x*log(5)/(2*x + 2*log(5) - 3) - 108*x/(2*x + 2*l
og(5) - 3) - 12*e^(x/(2*x + 2*log(5) - 3)) - 54*log(5) + 27)/((4*x/(2*x + 2*log(5) - 3) - 4*x^2/(2*x + 2*log(5
) - 3)^2 - 1)*(2*log(5) - 3))

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maple [A]  time = 0.14, size = 19, normalized size = 1.06

method	result	size
risch	\(x^{2}+{\mathrm e}^{\frac {x}{2 \ln \relax (5)+2 x -3}}\)	\(19\)
norman	\(\frac {\left (2 \ln \relax (5)-3\right ) x^{2}+\left (2 \ln \relax (5)-3\right ) {\mathrm e}^{\frac {x}{2 \ln \relax (5)+2 x -3}}+2 x^{3}+2 \,{\mathrm e}^{\frac {x}{2 \ln \relax (5)+2 x -3}} x}{2 \ln \relax (5)+2 x -3}\)	\(67\)
derivativedivides	\(-\frac {\left (-4 \ln \relax (5)+6\right ) \left (-\frac {27 \left (2 \ln \relax (5)+2 x -3\right )^{2}}{4 \left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}-\frac {27 \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}-\frac {3 \,{\mathrm e}^{\frac {1}{2}+\frac {-\ln \relax (5)+\frac {3}{2}}{2 \ln \relax (5)+2 x -3}}}{4 \ln \relax (5)^{2}-12 \ln \relax (5)+9}+\frac {27 \ln \relax (5) \left (2 \ln \relax (5)+2 x -3\right )^{2}}{2 \left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}-\frac {9 \ln \relax (5)^{2} \left (2 \ln \relax (5)+2 x -3\right )^{2}}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}+\frac {2 \ln \relax (5)^{3} \left (2 \ln \relax (5)+2 x -3\right )^{2}}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}+\frac {54 \ln \relax (5) \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}-\frac {36 \ln \relax (5)^{2} \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}+\frac {8 \ln \relax (5)^{3} \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}+\frac {2 \ln \relax (5) {\mathrm e}^{\frac {1}{2}+\frac {-\ln \relax (5)+\frac {3}{2}}{2 \ln \relax (5)+2 x -3}}}{4 \ln \relax (5)^{2}-12 \ln \relax (5)+9}\right )}{2}\)	\(378\)
default	\(-\frac {\left (-4 \ln \relax (5)+6\right ) \left (-\frac {27 \left (2 \ln \relax (5)+2 x -3\right )^{2}}{4 \left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}-\frac {27 \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}-\frac {3 \,{\mathrm e}^{\frac {1}{2}+\frac {-\ln \relax (5)+\frac {3}{2}}{2 \ln \relax (5)+2 x -3}}}{4 \ln \relax (5)^{2}-12 \ln \relax (5)+9}+\frac {27 \ln \relax (5) \left (2 \ln \relax (5)+2 x -3\right )^{2}}{2 \left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}-\frac {9 \ln \relax (5)^{2} \left (2 \ln \relax (5)+2 x -3\right )^{2}}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}+\frac {2 \ln \relax (5)^{3} \left (2 \ln \relax (5)+2 x -3\right )^{2}}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )^{2}}+\frac {54 \ln \relax (5) \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}-\frac {36 \ln \relax (5)^{2} \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}+\frac {8 \ln \relax (5)^{3} \left (2 \ln \relax (5)+2 x -3\right )}{\left (16 \ln \relax (5)^{2}-48 \ln \relax (5)+36\right ) \left (-\ln \relax (5)+\frac {3}{2}\right )}+\frac {2 \ln \relax (5) {\mathrm e}^{\frac {1}{2}+\frac {-\ln \relax (5)+\frac {3}{2}}{2 \ln \relax (5)+2 x -3}}}{4 \ln \relax (5)^{2}-12 \ln \relax (5)+9}\right )}{2}\)	\(378\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(5)-3)*exp(x/(2*ln(5)+2*x-3))+8*x*ln(5)^2+2*(8*x^2-12*x)*ln(5)+8*x^3-24*x^2+18*x)/(4*ln(5)^2+2*(4*x-
6)*ln(5)+4*x^2-12*x+9),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.45, size = 347, normalized size = 19.28 \begin {gather*} 2 \, {\left (\frac {2 \, \log \relax (5) - 3}{2 \, x + 2 \, \log \relax (5) - 3} + \log \left (2 \, x + 2 \, \log \relax (5) - 3\right )\right )} \log \relax (5)^{2} + x^{2} - 2 \, x {\left (2 \, \log \relax (5) - 3\right )} - 2 \, {\left (2 \, {\left (2 \, \log \relax (5) - 3\right )} \log \left (2 \, x + 2 \, \log \relax (5) - 3\right ) - 2 \, x + \frac {4 \, \log \relax (5)^{2} - 12 \, \log \relax (5) + 9}{2 \, x + 2 \, \log \relax (5) - 3}\right )} \log \relax (5) - 6 \, {\left (\frac {2 \, \log \relax (5) - 3}{2 \, x + 2 \, \log \relax (5) - 3} + \log \left (2 \, x + 2 \, \log \relax (5) - 3\right )\right )} \log \relax (5) + \frac {3}{2} \, {\left (4 \, \log \relax (5)^{2} - 12 \, \log \relax (5) + 9\right )} \log \left (2 \, x + 2 \, \log \relax (5) - 3\right ) + 6 \, {\left (2 \, \log \relax (5) - 3\right )} \log \left (2 \, x + 2 \, \log \relax (5) - 3\right ) - 6 \, x + \frac {2 \, e^{\left (-\frac {\log \relax (5)}{2 \, x + 2 \, \log \relax (5) - 3} + \frac {3}{2 \, {\left (2 \, x + 2 \, \log \relax (5) - 3\right )}} + \frac {1}{2}\right )} \log \relax (5)}{2 \, \log \relax (5) - 3} + \frac {8 \, \log \relax (5)^{3} - 36 \, \log \relax (5)^{2} + 54 \, \log \relax (5) - 27}{2 \, {\left (2 \, x + 2 \, \log \relax (5) - 3\right )}} + \frac {3 \, {\left (4 \, \log \relax (5)^{2} - 12 \, \log \relax (5) + 9\right )}}{2 \, x + 2 \, \log \relax (5) - 3} + \frac {9 \, {\left (2 \, \log \relax (5) - 3\right )}}{2 \, {\left (2 \, x + 2 \, \log \relax (5) - 3\right )}} - \frac {3 \, e^{\left (-\frac {\log \relax (5)}{2 \, x + 2 \, \log \relax (5) - 3} + \frac {3}{2 \, {\left (2 \, x + 2 \, \log \relax (5) - 3\right )}} + \frac {1}{2}\right )}}{2 \, \log \relax (5) - 3} + \frac {9}{2} \, \log \left (2 \, x + 2 \, \log \relax (5) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(5)-3)*exp(x/(2*log(5)+2*x-3))+8*x*log(5)^2+2*(8*x^2-12*x)*log(5)+8*x^3-24*x^2+18*x)/(4*log(5
)^2+2*(4*x-6)*log(5)+4*x^2-12*x+9),x, algorithm="maxima")

[Out]

2*((2*log(5) - 3)/(2*x + 2*log(5) - 3) + log(2*x + 2*log(5) - 3))*log(5)^2 + x^2 - 2*x*(2*log(5) - 3) - 2*(2*(
2*log(5) - 3)*log(2*x + 2*log(5) - 3) - 2*x + (4*log(5)^2 - 12*log(5) + 9)/(2*x + 2*log(5) - 3))*log(5) - 6*((
2*log(5) - 3)/(2*x + 2*log(5) - 3) + log(2*x + 2*log(5) - 3))*log(5) + 3/2*(4*log(5)^2 - 12*log(5) + 9)*log(2*
x + 2*log(5) - 3) + 6*(2*log(5) - 3)*log(2*x + 2*log(5) - 3) - 6*x + 2*e^(-log(5)/(2*x + 2*log(5) - 3) + 3/2/(
2*x + 2*log(5) - 3) + 1/2)*log(5)/(2*log(5) - 3) + 1/2*(8*log(5)^3 - 36*log(5)^2 + 54*log(5) - 27)/(2*x + 2*lo
g(5) - 3) + 3*(4*log(5)^2 - 12*log(5) + 9)/(2*x + 2*log(5) - 3) + 9/2*(2*log(5) - 3)/(2*x + 2*log(5) - 3) - 3*
e^(-log(5)/(2*x + 2*log(5) - 3) + 3/2/(2*x + 2*log(5) - 3) + 1/2)/(2*log(5) - 3) + 9/2*log(2*x + 2*log(5) - 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {18\,x+{\mathrm {e}}^{\frac {x}{2\,x+2\,\ln \relax (5)-3}}\,\left (2\,\ln \relax (5)-3\right )-2\,\ln \relax (5)\,\left (12\,x-8\,x^2\right )+8\,x\,{\ln \relax (5)}^2-24\,x^2+8\,x^3}{2\,\ln \relax (5)\,\left (4\,x-6\right )-12\,x+4\,{\ln \relax (5)}^2+4\,x^2+9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x + exp(x/(2*x + 2*log(5) - 3))*(2*log(5) - 3) - 2*log(5)*(12*x - 8*x^2) + 8*x*log(5)^2 - 24*x^2 + 8*x
^3)/(2*log(5)*(4*x - 6) - 12*x + 4*log(5)^2 + 4*x^2 + 9),x)

[Out]

int((18*x + exp(x/(2*x + 2*log(5) - 3))*(2*log(5) - 3) - 2*log(5)*(12*x - 8*x^2) + 8*x*log(5)^2 - 24*x^2 + 8*x
^3)/(2*log(5)*(4*x - 6) - 12*x + 4*log(5)^2 + 4*x^2 + 9), x)

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sympy [A]  time = 0.29, size = 15, normalized size = 0.83 \begin {gather*} x^{2} + e^{\frac {x}{2 x - 3 + 2 \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(5)-3)*exp(x/(2*ln(5)+2*x-3))+8*x*ln(5)**2+2*(8*x**2-12*x)*ln(5)+8*x**3-24*x**2+18*x)/(4*ln(5)
**2+2*(4*x-6)*ln(5)+4*x**2-12*x+9),x)

[Out]

x**2 + exp(x/(2*x - 3 + 2*log(5)))

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