3.17.40 $\int \frac{-875-350x+100x^2+68x^3+13x^4+(-25-10x+5x^2+2x^3)\log (5-x^2)}{e^3(-6125-2800x+1255x^2+640x^3-11x^4-16x^5+x^6)+e^3(-350-80x+80x^2+16x^3-2x^4)\log (5-x^2)+e^3(-5+x^2){\log }^2(5-x^2)}dx$

Optimal. Leaf size=32 $$1+\frac{x}{e^3(7+\frac{x-x^2+\log (5-x^2)}{5+x})}$$

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Rubi [F]  time = 1.69, 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 = {} $$\begin{array}{c}\int \frac{-875-350x+100x^2+68x^3+13x^4+(-25-10x+5x^2+2x^3)\log (5-x^2)}{e^3(-6125-2800x+1255x^2+640x^3-11x^4-16x^5+x^6)+e^3(-350-80x+80x^2+16x^3-2x^4)\log (5-x^2)+e^3(-5+x^2){\log }^2(5-x^2)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-875 - 350*x + 100*x^2 + 68*x^3 + 13*x^4 + (-25 - 10*x + 5*x^2 + 2*x^3)*Log[5 - x^2])/(E^3*(-6125 - 2800*
x + 1255*x^2 + 640*x^3 - 11*x^4 - 16*x^5 + x^6) + E^3*(-350 - 80*x + 80*x^2 + 16*x^3 - 2*x^4)*Log[5 - x^2] + E
^3*(-5 + x^2)*Log[5 - x^2]^2),x]

[Out]

(-10*Defer[Int][(-35 - 8*x + x^2 - Log[5 - x^2])^(-2), x])/E^3 + (5*Defer[Int][1/((Sqrt[5] - x)*(-35 - 8*x + x
^2 - Log[5 - x^2])^2), x])/E^3 + (5*Sqrt[5]*Defer[Int][1/((Sqrt[5] - x)*(-35 - 8*x + x^2 - Log[5 - x^2])^2), x
])/E^3 - (42*Defer[Int][x/(-35 - 8*x + x^2 - Log[5 - x^2])^2, x])/E^3 + (2*Defer[Int][x^2/(-35 - 8*x + x^2 - L
og[5 - x^2])^2, x])/E^3 + (2*Defer[Int][x^3/(-35 - 8*x + x^2 - Log[5 - x^2])^2, x])/E^3 - (5*Defer[Int][1/((Sq
rt[5] + x)*(-35 - 8*x + x^2 - Log[5 - x^2])^2), x])/E^3 + (5*Sqrt[5]*Defer[Int][1/((Sqrt[5] + x)*(-35 - 8*x +
x^2 - Log[5 - x^2])^2), x])/E^3 - (5*Defer[Int][(-35 - 8*x + x^2 - Log[5 - x^2])^(-1), x])/E^3 - (2*Defer[Int]
[x/(-35 - 8*x + x^2 - Log[5 - x^2]), x])/E^3

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{875+350x-100x^2-68x^3-13x^4-(-25-10x+5x^2+2x^3)\log (5-x^2)}{e^3(5-x^2){(35+8x-x^2+\log (5-x^2))}^2}dx\\ & =\frac{\int \frac{875+350x-100x^2-68x^3-13x^4-(-25-10x+5x^2+2x^3)\log (5-x^2)}{(5-x^2){(35+8x-x^2+\log (5-x^2))}^2}dx}{e^3}\\ & =\frac{\int (\frac{2x(100-10x-26x^2+x^3+x^4)}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}+\frac{-5-2x}{-35-8x+x^2-\log (5-x^2)})dx}{e^3}\\ & =\frac{\int \frac{-5-2x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}+\frac{2\int \frac{x(100-10x-26x^2+x^3+x^4)}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}\\ & =\frac{\int (-\frac{5}{-35-8x+x^2-\log (5-x^2)}-\frac{2x}{-35-8x+x^2-\log (5-x^2)})dx}{e^3}+\frac{2\int (-\frac{5}{{(-35-8x+x^2-\log (5-x^2))}^2}-\frac{21x}{{(-35-8x+x^2-\log (5-x^2))}^2}+\frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}+\frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}-\frac{5(5+x)}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2})dx}{e^3}\\ & =\frac{2\int \frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{2\int \frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{2\int \frac{x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{5\int \frac{1}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{10\int \frac{1}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{10\int \frac{5+x}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{42\int \frac{x}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}\\ & =\frac{2\int \frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{2\int \frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{2\int \frac{x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{5\int \frac{1}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{10\int (\frac{5}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}+\frac{x}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2})dx}{e^3}-\frac{10\int \frac{1}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{42\int \frac{x}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}\\ & =\frac{2\int \frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{2\int \frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{2\int \frac{x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{5\int \frac{1}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{10\int \frac{1}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{10\int \frac{x}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{42\int \frac{x}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{50\int \frac{1}{(-5+x^2){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}\\ & =\frac{2\int \frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{2\int \frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{2\int \frac{x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{5\int \frac{1}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{10\int (-\frac{1}{2(\sqrt{5}-x){(-35-8x+x^2-\log (5-x^2))}^2}+\frac{1}{2(\sqrt{5}+x){(-35-8x+x^2-\log (5-x^2))}^2})dx}{e^3}-\frac{10\int \frac{1}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{42\int \frac{x}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{50\int (-\frac{1}{2\sqrt{5}(\sqrt{5}-x){(-35-8x+x^2-\log (5-x^2))}^2}-\frac{1}{2\sqrt{5}(\sqrt{5}+x){(-35-8x+x^2-\log (5-x^2))}^2})dx}{e^3}\\ & =\frac{2\int \frac{x^2}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{2\int \frac{x^3}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{2\int \frac{x}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}+\frac{5\int \frac{1}{(\sqrt{5}-x){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{5\int \frac{1}{(\sqrt{5}+x){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{5\int \frac{1}{-35-8x+x^2-\log (5-x^2)}dx}{e^3}-\frac{10\int \frac{1}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}-\frac{42\int \frac{x}{{(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{\left(5\sqrt{5}\right)\int \frac{1}{(\sqrt{5}-x){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}+\frac{\left(5\sqrt{5}\right)\int \frac{1}{(\sqrt{5}+x){(-35-8x+x^2-\log (5-x^2))}^2}dx}{e^3}\end{array}\end{array}$$

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Mathematica [A]  time = 1.32, size = 28, normalized size = 0.88 $$\begin{array}{c}\frac{x(5+x)}{e^3(35+8x-x^2+\log (5-x^2))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-875 - 350*x + 100*x^2 + 68*x^3 + 13*x^4 + (-25 - 10*x + 5*x^2 + 2*x^3)*Log[5 - x^2])/(E^3*(-6125 -
2800*x + 1255*x^2 + 640*x^3 - 11*x^4 - 16*x^5 + x^6) + E^3*(-350 - 80*x + 80*x^2 + 16*x^3 - 2*x^4)*Log[5 - x^
2] + E^3*(-5 + x^2)*Log[5 - x^2]^2),x]

[Out]

(x*(5 + x))/(E^3*(35 + 8*x - x^2 + Log[5 - x^2]))

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fricas [A]  time = 0.66, size = 35, normalized size = 1.09 $$\begin{array}{c}-\frac{x^2+5x}{(x^2-8x-35)e^3-e^3\log (-x^2+5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-875)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-
2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*log(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, a
lgorithm="fricas")

[Out]

-(x^2 + 5*x)/((x^2 - 8*x - 35)*e^3 - e^3*log(-x^2 + 5))

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giac [A]  time = 0.38, size = 39, normalized size = 1.22 $$\begin{array}{c}-\frac{x^2+5x}{x^2{e}^3-8x{e}^3-e^3\log (-x^2+5)-35e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-875)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-
2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*log(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, a
lgorithm="giac")

[Out]

-(x^2 + 5*x)/(x^2*e^3 - 8*x*e^3 - e^3*log(-x^2 + 5) - 35*e^3)

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maple [A]  time = 0.23, size = 29, normalized size = 0.91

method	result	size
risch	$-\frac{(5+x)x{\mathrm{e}}^{-3}}{x^2-8x-\ln (-x^2+5)-35}$	$29$
norman	$\frac{-13x{\mathrm{e}}^{-3}-{\mathrm{e}}^{-3}\ln (-x^2+5)-35{\mathrm{e}}^{-3}}{x^2-8x-\ln (-x^2+5)-35}$	$50$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+5*x^2-10*x-25)*ln(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-875)/((x^2-5)*exp(3)*ln(-x^2+5)^2+(-2*x^4+16
*x^3+80*x^2-80*x-350)*exp(3)*ln(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x,method=_RET
URNVERBOSE)

[Out]

-(5+x)*x*exp(-3)/(x^2-8*x-ln(-x^2+5)-35)

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maxima [A]  time = 0.89, size = 39, normalized size = 1.22 $$\begin{array}{c}-\frac{x^2+5x}{x^2{e}^3-8x{e}^3-e^3\log (-x^2+5)-35e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+5*x^2-10*x-25)*log(-x^2+5)+13*x^4+68*x^3+100*x^2-350*x-875)/((x^2-5)*exp(3)*log(-x^2+5)^2+(-
2*x^4+16*x^3+80*x^2-80*x-350)*exp(3)*log(-x^2+5)+(x^6-16*x^5-11*x^4+640*x^3+1255*x^2-2800*x-6125)*exp(3)),x, a
lgorithm="maxima")

[Out]

-(x^2 + 5*x)/(x^2*e^3 - 8*x*e^3 - e^3*log(-x^2 + 5) - 35*e^3)

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mupad [B]  time = 1.65, size = 36, normalized size = 1.12 $$\begin{array}{c}\frac{{\mathrm{e}}^{-3}(13x+\ln (5-x^2)+35)}{8x+\ln (5-x^2)-x^2+35}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((350*x + log(5 - x^2)*(10*x - 5*x^2 - 2*x^3 + 25) - 100*x^2 - 68*x^3 - 13*x^4 + 875)/(exp(3)*(2800*x - 125
5*x^2 - 640*x^3 + 11*x^4 + 16*x^5 - x^6 + 6125) - log(5 - x^2)^2*exp(3)*(x^2 - 5) + log(5 - x^2)*exp(3)*(80*x
- 80*x^2 - 16*x^3 + 2*x^4 + 350)),x)

[Out]

(exp(-3)*(13*x + log(5 - x^2) + 35))/(8*x + log(5 - x^2) - x^2 + 35)

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sympy [A]  time = 0.19, size = 34, normalized size = 1.06 $$\begin{array}{c}\frac{x^2+5x}{-x^2{e}^3+8x{e}^3+e^3\log (5-x^2)+35e^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+5*x**2-10*x-25)*ln(-x**2+5)+13*x**4+68*x**3+100*x**2-350*x-875)/((x**2-5)*exp(3)*ln(-x**2+5
)**2+(-2*x**4+16*x**3+80*x**2-80*x-350)*exp(3)*ln(-x**2+5)+(x**6-16*x**5-11*x**4+640*x**3+1255*x**2-2800*x-612
5)*exp(3)),x)

[Out]

(x**2 + 5*x)/(-x**2*exp(3) + 8*x*exp(3) + exp(3)*log(5 - x**2) + 35*exp(3))

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