3.41.51 \(\int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} (-4 x^2-x^2 \log (169)+(-400 x+8 x^2+(-100 x+2 x^2) \log (169)) \log (50-x))}{(-1000+20 x) \log ^2(50-x)} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {x^2 (4+\log (169))}{20 \log (50-x)}} \]

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Rubi [A]  time = 1.07, antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 4, number of rules used = 4, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6, 6741, 12, 6706} \begin {gather*} 13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4*x^2 + x^2*Log[169])/(20*Log[50 - x]))*(-4*x^2 - x^2*Log[169] + (-400*x + 8*x^2 + (-100*x + 2*x^2)*L
og[169])*Log[50 - x]))/((-1000 + 20*x)*Log[50 - x]^2),x]

[Out]

13^(x^2/(10*Log[50 - x]))*E^(x^2/(5*Log[50 - x]))

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (x^2 (-4-\log (169))+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx\\ &=\int \frac {e^{\frac {x^2 (4+\log (169))}{20 \log (50-x)}} x (4+\log (169)) (x+100 \log (50-x)-2 x \log (50-x))}{(1000-20 x) \log ^2(50-x)} \, dx\\ &=(4+\log (169)) \int \frac {e^{\frac {x^2 (4+\log (169))}{20 \log (50-x)}} x (x+100 \log (50-x)-2 x \log (50-x))}{(1000-20 x) \log ^2(50-x)} \, dx\\ &=13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.82, size = 48, normalized size = 2.29 \begin {gather*} \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} (4+\log (169))}{2 (2+\log (13))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4*x^2 + x^2*Log[169])/(20*Log[50 - x]))*(-4*x^2 - x^2*Log[169] + (-400*x + 8*x^2 + (-100*x + 2*
x^2)*Log[169])*Log[50 - x]))/((-1000 + 20*x)*Log[50 - x]^2),x]

[Out]

(13^(x^2/(10*Log[50 - x]))*E^(x^2/(5*Log[50 - x]))*(4 + Log[169]))/(2*(2 + Log[13]))

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fricas [A]  time = 0.94, size = 23, normalized size = 1.10 \begin {gather*} e^{\left (\frac {x^{2} \log \left (13\right ) + 2 \, x^{2}}{10 \, \log \left (-x + 50\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)-4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2
)/log(-x+50))/(20*x-1000)/log(-x+50)^2,x, algorithm="fricas")

[Out]

e^(1/10*(x^2*log(13) + 2*x^2)/log(-x + 50))

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giac [A]  time = 0.15, size = 30, normalized size = 1.43 \begin {gather*} e^{\left (\frac {x^{2} \log \left (13\right )}{10 \, \log \left (-x + 50\right )} + \frac {x^{2}}{5 \, \log \left (-x + 50\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)-4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2
)/log(-x+50))/(20*x-1000)/log(-x+50)^2,x, algorithm="giac")

[Out]

e^(1/10*x^2*log(13)/log(-x + 50) + 1/5*x^2/log(-x + 50))

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maple [A]  time = 0.11, size = 19, normalized size = 0.90

method	result	size
risch	\({\mathrm e}^{\frac {x^{2} \left (\ln \left (13\right )+2\right )}{10 \ln \left (-x +50\right )}}\)	\(19\)
norman	\({\mathrm e}^{\frac {2 x^{2} \ln \left (13\right )+4 x^{2}}{20 \ln \left (-x +50\right )}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(2*x^2-100*x)*ln(13)+8*x^2-400*x)*ln(-x+50)-2*x^2*ln(13)-4*x^2)*exp(1/20*(2*x^2*ln(13)+4*x^2)/ln(-x+50
))/(20*x-1000)/ln(-x+50)^2,x,method=_RETURNVERBOSE)

[Out]

exp(1/10*x^2*(ln(13)+2)/ln(-x+50))

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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(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)-4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2
)/log(-x+50))/(20*x-1000)/log(-x+50)^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 = 3.52, size = 23, normalized size = 1.10 \begin {gather*} {\mathrm {e}}^{\frac {x^2\,\ln \left (13\right )+2\,x^2}{10\,\ln \left (50-x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(((x^2*log(13))/10 + x^2/5)/log(50 - x))*(2*x^2*log(13) + log(50 - x)*(400*x + 2*log(13)*(100*x - 2*x
^2) - 8*x^2) + 4*x^2))/(log(50 - x)^2*(20*x - 1000)),x)

[Out]

exp((x^2*log(13) + 2*x^2)/(10*log(50 - x)))

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sympy [A]  time = 0.51, size = 19, normalized size = 0.90 \begin {gather*} e^{\frac {\frac {x^{2}}{5} + \frac {x^{2} \log {\left (13 \right )}}{10}}{\log {\left (50 - x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x**2-100*x)*ln(13)+8*x**2-400*x)*ln(-x+50)-2*x**2*ln(13)-4*x**2)*exp(1/20*(2*x**2*ln(13)+4*x*
*2)/ln(-x+50))/(20*x-1000)/ln(-x+50)**2,x)

[Out]

exp((x**2/5 + x**2*log(13)/10)/log(50 - x))

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