3.104.22 \(\int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} (-150+120 x-24 x^2+e^x (-40+5 x+2 x^2) \log (5))}{75-60 x+12 x^2} \, dx\)

Optimal. Leaf size=29 \[ e^{-2 x+\frac {e^x (5+x) \log (5)}{3 (-5+2 x)}}-x \]

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Rubi [B]  time = 2.68, antiderivative size = 99, normalized size of antiderivative = 3.41, number of steps used = 5, number of rules used = 4, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {27, 12, 6742, 2288} \begin {gather*} \frac {5^{-\frac {e^x (x+5)}{3 (5-2 x)}} e^{-2 x} \left (-2 e^x x^2-5 e^x x+40 e^x\right )}{(5-2 x)^2 \left (\frac {e^x (x+5)}{5-2 x}+\frac {2 e^x (x+5)}{(5-2 x)^2}+\frac {e^x}{5-2 x}\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-75 + 60*x - 12*x^2 + E^((30*x - 12*x^2 + E^x*(5 + x)*Log[5])/(-15 + 6*x))*(-150 + 120*x - 24*x^2 + E^x*(
-40 + 5*x + 2*x^2)*Log[5]))/(75 - 60*x + 12*x^2),x]

[Out]

-x + (40*E^x - 5*E^x*x - 2*E^x*x^2)/(5^((E^x*(5 + x))/(3*(5 - 2*x)))*E^(2*x)*(5 - 2*x)^2*(E^x/(5 - 2*x) + (2*E
^x*(5 + x))/(5 - 2*x)^2 + (E^x*(5 + x))/(5 - 2*x)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; 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 \frac {-75+60 x-12 x^2+\exp \left (\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}\right ) \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{3 (-5+2 x)^2} \, dx\\ &=\frac {1}{3} \int \frac {-75+60 x-12 x^2+\exp \left (\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}\right ) \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{(-5+2 x)^2} \, dx\\ &=\frac {1}{3} \int \left (-3+\frac {5^{\frac {e^x (5+x)}{-15+6 x}} e^{-2 x} \left (-150+120 x-24 x^2-40 e^x \log (5)+5 e^x x \log (5)+2 e^x x^2 \log (5)\right )}{(-5+2 x)^2}\right ) \, dx\\ &=-x+\frac {1}{3} \int \frac {5^{\frac {e^x (5+x)}{-15+6 x}} e^{-2 x} \left (-150+120 x-24 x^2-40 e^x \log (5)+5 e^x x \log (5)+2 e^x x^2 \log (5)\right )}{(-5+2 x)^2} \, dx\\ &=-x+\frac {5^{-\frac {e^x (5+x)}{3 (5-2 x)}} e^{-2 x} \left (40 e^x-5 e^x x-2 e^x x^2\right )}{(5-2 x)^2 \left (\frac {e^x}{5-2 x}+\frac {2 e^x (5+x)}{(5-2 x)^2}+\frac {e^x (5+x)}{5-2 x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.75, size = 31, normalized size = 1.07 \begin {gather*} \frac {1}{3} \left (3\ 5^{\frac {e^x (5+x)}{-15+6 x}} e^{-2 x}-3 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-75 + 60*x - 12*x^2 + E^((30*x - 12*x^2 + E^x*(5 + x)*Log[5])/(-15 + 6*x))*(-150 + 120*x - 24*x^2 +
E^x*(-40 + 5*x + 2*x^2)*Log[5]))/(75 - 60*x + 12*x^2),x]

[Out]

((3*5^((E^x*(5 + x))/(-15 + 6*x)))/E^(2*x) - 3*x)/3

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fricas [A]  time = 0.78, size = 31, normalized size = 1.07 \begin {gather*} -x + e^{\left (\frac {{\left (x + 5\right )} e^{x} \log \relax (5) - 12 \, x^{2} + 30 \, x}{3 \, {\left (2 \, x - 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log(5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*
x^2+60*x-75)/(12*x^2-60*x+75),x, algorithm="fricas")

[Out]

-x + e^(1/3*((x + 5)*e^x*log(5) - 12*x^2 + 30*x)/(2*x - 5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {12 \, x^{2} - {\left ({\left (2 \, x^{2} + 5 \, x - 40\right )} e^{x} \log \relax (5) - 24 \, x^{2} + 120 \, x - 150\right )} e^{\left (\frac {{\left (x + 5\right )} e^{x} \log \relax (5) - 12 \, x^{2} + 30 \, x}{3 \, {\left (2 \, x - 5\right )}}\right )} - 60 \, x + 75}{3 \, {\left (4 \, x^{2} - 20 \, x + 25\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log(5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*
x^2+60*x-75)/(12*x^2-60*x+75),x, algorithm="giac")

[Out]

integrate(-1/3*(12*x^2 - ((2*x^2 + 5*x - 40)*e^x*log(5) - 24*x^2 + 120*x - 150)*e^(1/3*((x + 5)*e^x*log(5) - 1
2*x^2 + 30*x)/(2*x - 5)) - 60*x + 75)/(4*x^2 - 20*x + 25), x)

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maple [A]  time = 0.22, size = 37, normalized size = 1.28

method	result	size
risch	\(-x +5^{\frac {x \,{\mathrm e}^{x}}{6 x -15}} 5^{\frac {5 \,{\mathrm e}^{x}}{3 \left (2 x -5\right )}} {\mathrm e}^{-2 x}\)	\(37\)
norman	\(\frac {-2 x^{2}+2 x \,{\mathrm e}^{\frac {\left (5+x \right ) \ln \relax (5) {\mathrm e}^{x}-12 x^{2}+30 x}{6 x -15}}-5 \,{\mathrm e}^{\frac {\left (5+x \right ) \ln \relax (5) {\mathrm e}^{x}-12 x^{2}+30 x}{6 x -15}}+\frac {25}{2}}{2 x -5}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+5*x-40)*ln(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*ln(5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*x^2+60*x
-75)/(12*x^2-60*x+75),x,method=_RETURNVERBOSE)

[Out]

-x+5^(1/3*exp(x)*x/(2*x-5))*5^(5/3*exp(x)/(2*x-5))*exp(-2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x + \frac {1}{3} \, \int -\frac {{\left (24 \, x^{2} - {\left (2 \, x^{2} \log \relax (5) + 5 \, x \log \relax (5) - 40 \, \log \relax (5)\right )} e^{x} - 120 \, x + 150\right )} e^{\left (\frac {1}{6} \, e^{x} \log \relax (5) - 2 \, x + \frac {5 \, e^{x} \log \relax (5)}{2 \, {\left (2 \, x - 5\right )}}\right )}}{4 \, x^{2} - 20 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log(5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*
x^2+60*x-75)/(12*x^2-60*x+75),x, algorithm="maxima")

[Out]

-x + 1/3*integrate(-(24*x^2 - (2*x^2*log(5) + 5*x*log(5) - 40*log(5))*e^x - 120*x + 150)*e^(1/6*e^x*log(5) - 2
*x + 5/2*e^x*log(5)/(2*x - 5))/(4*x^2 - 20*x + 25), x)

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mupad [B]  time = 6.57, size = 55, normalized size = 1.90 \begin {gather*} 5^{\frac {5\,{\mathrm {e}}^x}{6\,x-15}}\,5^{\frac {x\,{\mathrm {e}}^x}{6\,x-15}}\,{\mathrm {e}}^{-\frac {12\,x^2}{6\,x-15}}\,{\mathrm {e}}^{\frac {30\,x}{6\,x-15}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x + exp((30*x - 12*x^2 + exp(x)*log(5)*(x + 5))/(6*x - 15))*(120*x - 24*x^2 + exp(x)*log(5)*(5*x + 2*x
^2 - 40) - 150) - 12*x^2 - 75)/(12*x^2 - 60*x + 75),x)

[Out]

5^((5*exp(x))/(6*x - 15))*5^((x*exp(x))/(6*x - 15))*exp(-(12*x^2)/(6*x - 15))*exp((30*x)/(6*x - 15)) - x

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sympy [A]  time = 0.40, size = 26, normalized size = 0.90 \begin {gather*} - x + e^{\frac {- 12 x^{2} + 30 x + \left (x + 5\right ) e^{x} \log {\relax (5 )}}{6 x - 15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+5*x-40)*ln(5)*exp(x)-24*x**2+120*x-150)*exp(((5+x)*ln(5)*exp(x)-12*x**2+30*x)/(6*x-15))-12
*x**2+60*x-75)/(12*x**2-60*x+75),x)

[Out]

-x + exp((-12*x**2 + 30*x + (x + 5)*exp(x)*log(5))/(6*x - 15))

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