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

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} $$\begin{array}{c}{13}^{\frac{x^2}{10\log (50-x)}}{e}^{\frac{x^2}{5\log (50-x)}}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{4x^2+x^2\log (169)}{20\log (50-x)}}(x^2(-4-\log (169))+(-400x+8x^2+(-100x+2x^2)\log (169))\log (50-x))}{(-1000+20x){\log }^2(50-x)}dx\\ & =\int \frac{e^{\frac{x^2(4+\log (169))}{20\log (50-x)}}x(4+\log (169))(x+100\log (50-x)-2x\log (50-x))}{(1000-20x){\log }^2(50-x)}dx\\ & =(4+\log (169))\int \frac{e^{\frac{x^2(4+\log (169))}{20\log (50-x)}}x(x+100\log (50-x)-2x\log (50-x))}{(1000-20x){\log }^2(50-x)}dx\\ & ={13}^{\frac{x^2}{10\log (50-x)}}{e}^{\frac{x^2}{5\log (50-x)}}\end{array}\end{array}$$

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

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 $$\begin{array}{c}{e}^{\left(\frac{x^2\log \left(13\right)+2x^2}{10\log (-x+50)}\right)}\end{array}$$

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 $$\begin{array}{c}{e}^{(\frac{x^2\log \left(13\right)}{10\log (-x+50)}+\frac{x^2}{5\log (-x+50)})}\end{array}$$

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

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 $$\begin{array}{c}\text{Exception raised: RuntimeError}\end{array}$$

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 $$\begin{array}{c}{\mathrm{e}}^{\frac{x^2\ln \left(13\right)+2x^2}{10\ln (50-x)}}\end{array}$$

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 $$\begin{array}{c}{e}^{\frac{\frac{x^2}{5}+\frac{x^2\log \left(13\right)}{10}}{\log (50-x)}}\end{array}$$

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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