3.41.58 $\int \frac{-10-2x+e^{-2-6x}(5+x{)}^2(10+30x+6x^2)}{-5x-x^2+e^{-2-6x}(5+x{)}^2(5x+x^2)}dx$

Optimal. Leaf size=26 $$\log \left(\frac{x^2}{-1+e^{-8+6(1-x)}(5+x{)}^2}\right)$$

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Rubi [F]  time = 0.75, 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 = {} $$\begin{array}{c}\int \frac{-10-2x+e^{-2-6x}(5+x{)}^2(10+30x+6x^2)}{-5x-x^2+e^{-2-6x}(5+x{)}^2(5x+x^2)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-10 - 2*x + E^(-2 - 6*x)*(5 + x)^2*(10 + 30*x + 6*x^2))/(-5*x - x^2 + E^(-2 - 6*x)*(5 + x)^2*(5*x + x^2))
,x]

[Out]

2*Log[x] - 14*Defer[Int][(-5 + E^(1 + 3*x) - x)^(-1), x] + 3*Defer[Int][x/(5 - E^(1 + 3*x) + x), x] + 14*Defer
[Int][(5 + E^(1 + 3*x) + x)^(-1), x] + 3*Defer[Int][x/(5 + E^(1 + 3*x) + x), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2(-25+e^{2+6x}-80x-30x^2-3x^3)}{x(e^{2+6x}-(5+x{)}^2)}dx\\ & =2\int \frac{-25+e^{2+6x}-80x-30x^2-3x^3}{x(e^{2+6x}-(5+x{)}^2)}dx\\ & =2\int (\frac{1}{x}+\frac{14+3x}{2(5-e^{1+3x}+x)}+\frac{14+3x}{2(5+e^{1+3x}+x)})dx\\ & =2\log (x)+\int \frac{14+3x}{5-e^{1+3x}+x}dx+\int \frac{14+3x}{5+e^{1+3x}+x}dx\\ & =2\log (x)+\int (-\frac{14}{-5+e^{1+3x}-x}+\frac{3x}{5-e^{1+3x}+x})dx+\int (\frac{14}{5+e^{1+3x}+x}+\frac{3x}{5+e^{1+3x}+x})dx\\ & =2\log (x)+3\int \frac{x}{5-e^{1+3x}+x}dx+3\int \frac{x}{5+e^{1+3x}+x}dx-14\int \frac{1}{-5+e^{1+3x}-x}dx+14\int \frac{1}{5+e^{1+3x}+x}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.23, size = 28, normalized size = 1.08 $$\begin{array}{c}6x+2\log (x)-\log (25-e^{2+6x}+10x+x^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 2*x + E^(-2 - 6*x)*(5 + x)^2*(10 + 30*x + 6*x^2))/(-5*x - x^2 + E^(-2 - 6*x)*(5 + x)^2*(5*x +
x^2)),x]

[Out]

6*x + 2*Log[x] - Log[25 - E^(2 + 6*x) + 10*x + x^2]

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fricas [A]  time = 0.93, size = 22, normalized size = 0.85 $$\begin{array}{c}2\log (x)-\log (e^{(-6x+2\log (x+5)-2)}-1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="fricas")

[Out]

2*log(x) - log(e^(-6*x + 2*log(x + 5) - 2) - 1)

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giac [A]  time = 0.31, size = 27, normalized size = 1.04 $$\begin{array}{c}6x-\log (-x^2-10x+e^{(6x+2)}-25)+2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="giac")

[Out]

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

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maple [A]  time = 0.11, size = 24, normalized size = 0.92

method	result	size
risch	$2\ln (x)-2-\ln ({(5+x)}^2{\mathrm{e}}^{-6x-2}-1)$	$24$
norman	$2\ln (x)-\ln ({\mathrm{e}}^{\ln (5+x)-3x-1}-1)-\ln ({\mathrm{e}}^{\ln (5+x)-3x-1}+1)$	$36$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2+30*x+10)*exp(ln(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(ln(5+x)-3*x-1)^2-x^2-5*x),x,method=_RETURNVER
BOSE)

[Out]

2*ln(x)-2-ln((5+x)^2*exp(-6*x-2)-1)

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maxima [A]  time = 0.45, size = 41, normalized size = 1.58 $$\begin{array}{c}6x-\log \left((x+e^{(3x+1)}+5)e^{(-1)}\right)-\log (-(x-e^{(3x+1)}+5)e^{(-1)})+2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+30*x+10)*exp(log(5+x)-3*x-1)^2-2*x-10)/((x^2+5*x)*exp(log(5+x)-3*x-1)^2-x^2-5*x),x, algorith
m="maxima")

[Out]

6*x - log((x + e^(3*x + 1) + 5)*e^(-1)) - log(-(x - e^(3*x + 1) + 5)*e^(-1)) + 2*log(x)

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mupad [B]  time = 0.24, size = 37, normalized size = 1.42 $$\begin{array}{c}2\ln (x)-\ln (25{\mathrm{e}}^{-6x}{\mathrm{e}}^{-2}+10x{\mathrm{e}}^{-6x}{\mathrm{e}}^{-2}+x^2{\mathrm{e}}^{-6x}{\mathrm{e}}^{-2}-1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(2*log(x + 5) - 6*x - 2)*(30*x + 6*x^2 + 10) + 10)/(5*x - exp(2*log(x + 5) - 6*x - 2)*(5*x + x^2
) + x^2),x)

[Out]

2*log(x) - log(25*exp(-6*x)*exp(-2) + 10*x*exp(-6*x)*exp(-2) + x^2*exp(-6*x)*exp(-2) - 1)

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sympy [A]  time = 0.36, size = 31, normalized size = 1.19 $$\begin{array}{c}2\log (x)-2\log (x+5)-\log (e^{-6x-2}-\frac{1}{x^2+10x+25})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2+30*x+10)*exp(ln(5+x)-3*x-1)**2-2*x-10)/((x**2+5*x)*exp(ln(5+x)-3*x-1)**2-x**2-5*x),x)

[Out]

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

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