3.38.88 $\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x+10\log (25x))}{20-20x+5x^2}dx$

Optimal. Leaf size=26 $$e^{\frac{x(e^{10}-\frac{6}{5x}-\log (25x))}{-2+x}}$$

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Rubi [F]  time = 1.93, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x+10\log (25x))}{20-20x+5x^2}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5*x^2),
x]

[Out]

(2*(3 - 5*E^10)*Defer[Int][E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))/(-2 + x)^2, x])/5 - Defer[Int][E^((
-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))/(-2 + x), x] + 2*Defer[Int][(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-
10 + 5*x))*Log[25*x])/(-2 + x)^2, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x+10\log (25x))}{5(-2+x{)}^2}dx\\ & =\frac{1}{5}\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x+10\log (25x))}{(-2+x{)}^2}dx\\ & =\frac{1}{5}\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16(1-\frac{5e^{10}}{8})-5x+10\log (25x))}{(2-x{)}^2}dx\\ & =\frac{1}{5}\int (\frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x)}{(-2+x{)}^2}+\frac{10e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}\log (25x)}{(-2+x{)}^2})dx\\ & =\frac{1}{5}\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(16-10e^{10}-5x)}{(-2+x{)}^2}dx+2\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}\log (25x)}{(-2+x{)}^2}dx\\ & =\frac{1}{5}\int (-\frac{2e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}(-3+5e^{10})}{(-2+x{)}^2}-\frac{5e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}}{-2+x})dx+2\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}\log (25x)}{(-2+x{)}^2}dx\\ & =2\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}\log (25x)}{(-2+x{)}^2}dx+\frac{1}{5}\left(2(3-5e^{10})\right)\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}}{(-2+x{)}^2}dx-\int \frac{e^{\frac{-6+5e^{10}x-5x\log (25x)}{-10+5x}}}{-2+x}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.37, size = 44, normalized size = 1.69 $$\begin{array}{c}{5}^{-1+\frac{2+x}{2-x}}{e}^{\frac{6-5e^{10}x}{10-5x}}{x}^{-\frac{x}{-2+x}}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5
*x^2),x]

[Out]

(5^(-1 + (2 + x)/(2 - x))*E^((6 - 5*E^10*x)/(10 - 5*x)))/x^(x/(-2 + x))

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fricas [A]  time = 0.63, size = 22, normalized size = 0.85 $$\begin{array}{c}{e}^{\left(\frac{5x{e}^{10}-5x\log \left(25x\right)-6}{5(x-2)}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="fricas")

[Out]

e^(1/5*(5*x*e^10 - 5*x*log(25*x) - 6)/(x - 2))

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giac [A]  time = 0.25, size = 30, normalized size = 1.15 $$\begin{array}{c}{e}^{(\frac{x{e}^{10}}{x-2}-\frac{x\log \left(25x\right)}{x-2}-\frac{6}{5(x-2)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="giac")

[Out]

e^(x*e^10/(x - 2) - x*log(25*x)/(x - 2) - 6/5/(x - 2))

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maple [A]  time = 0.14, size = 23, normalized size = 0.88

method	result	size
risch	${\mathrm{e}}^{\frac{-5x\ln \left(25x\right)+5x{\mathrm{e}}^{10}-6}{5x-10}}$	$23$
norman	$\frac{x{\mathrm{e}}^{\frac{-5x\ln \left(25x\right)+5x{\mathrm{e}}^{10}-6}{5x-10}}-2{\mathrm{e}}^{\frac{-5x\ln \left(25x\right)+5x{\mathrm{e}}^{10}-6}{5x-10}}}{x-2}$	$62$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x,method=_RE
TURNVERBOSE)

[Out]

exp(1/5*(-5*x*ln(25*x)+5*x*exp(10)-6)/(x-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}-\frac{1}{5}\int \frac{(5x+10e^{10}-10\log \left(25x\right)-16)e^{\left(\frac{5x{e}^{10}-5x\log \left(25x\right)-6}{5(x-2)}\right)}}{x^2-4x+4}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="maxima")

[Out]

-1/5*integrate((5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x*log(25*x) - 6)/(x - 2))/(x^2 - 4*x +
4), x)

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mupad [B]  time = 2.50, size = 52, normalized size = 2.00 $$\begin{array}{c}\frac{{\mathrm{e}}^{\frac{5x{\mathrm{e}}^{10}}{5x-10}}{\mathrm{e}}^{-\frac{6}{5x-10}}}{5^{\frac{10x}{5x-10}}{x}^{\frac{5x}{5x-10}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x*log(25*x) - 5*x*exp(10) + 6)/(5*x - 10))*(5*x - 10*log(25*x) + 10*exp(10) - 16))/(5*x^2 - 20*x
+ 20),x)

[Out]

(exp((5*x*exp(10))/(5*x - 10))*exp(-6/(5*x - 10)))/(5^((10*x)/(5*x - 10))*x^((5*x)/(5*x - 10)))

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sympy [A]  time = 0.58, size = 22, normalized size = 0.85 $$\begin{array}{c}{e}^{\frac{-5x\log \left(25x\right)+5x{e}^{10}-6}{5x-10}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(25*x)-10*exp(5)**2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)**2-6)/(5*x-10))/(5*x**2-20*x+20),x)

[Out]

exp((-5*x*log(25*x) + 5*x*exp(10) - 6)/(5*x - 10))

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