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

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

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

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

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

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 {gather*} -\frac {x^{2} + 5 \, x}{{\left (x^{2} - 8 \, x - 35\right )} e^{3} - e^{3} \log \left (-x^{2} + 5\right )} \end {gather*}

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 {gather*} -\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \end {gather*}

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	\(-\frac {\left (5+x \right ) x \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\)	\(29\)
norman	\(\frac {-13 x \,{\mathrm e}^{-3}-{\mathrm e}^{-3} \ln \left (-x^{2}+5\right )-35 \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\)	\(50\)

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 {gather*} -\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \end {gather*}

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 {gather*} \frac {{\mathrm {e}}^{-3}\,\left (13\,x+\ln \left (5-x^2\right )+35\right )}{8\,x+\ln \left (5-x^2\right )-x^2+35} \end {gather*}

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 {gather*} \frac {x^{2} + 5 x}{- x^{2} e^{3} + 8 x e^{3} + e^{3} \log {\left (5 - x^{2} \right )} + 35 e^{3}} \end {gather*}

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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