3.46.49 \(\int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+(240 x^2+128 x^3-160 x^4) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+(60 x+32 x^2-40 x^3) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx\)

Optimal. Leaf size=27 \[ 2 x^2+\frac {1}{4+5 \left (\frac {3}{2 x}-x\right )+\log (5 x)} \]

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Rubi [F]  time = 0.68, 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 {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(30 + 896*x + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + (240*x^2 + 128*x^3 - 160*x^4)*Log[5*x] + 16*x^3*Log[
5*x]^2)/(225 + 240*x - 236*x^2 - 160*x^3 + 100*x^4 + (60*x + 32*x^2 - 40*x^3)*Log[5*x] + 4*x^2*Log[5*x]^2),x]

[Out]

2*x^2 + 30*Defer[Int][(-15 - 8*x + 10*x^2 - 2*x*Log[5*x])^(-2), x] - 4*Defer[Int][x/(-15 - 8*x + 10*x^2 - 2*x*
Log[5*x])^2, x] + 20*Defer[Int][x^2/(-15 - 8*x + 10*x^2 - 2*x*Log[5*x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (15+448 x+490 x^2-472 x^3-320 x^4+200 x^5+8 x^2 \left (15+8 x-10 x^2\right ) \log (5 x)+8 x^3 \log ^2(5 x)\right )}{\left (15+8 x-10 x^2+2 x \log (5 x)\right )^2} \, dx\\ &=2 \int \frac {15+448 x+490 x^2-472 x^3-320 x^4+200 x^5+8 x^2 \left (15+8 x-10 x^2\right ) \log (5 x)+8 x^3 \log ^2(5 x)}{\left (15+8 x-10 x^2+2 x \log (5 x)\right )^2} \, dx\\ &=2 \int \left (2 x+\frac {15-2 x+10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}\right ) \, dx\\ &=2 x^2+2 \int \frac {15-2 x+10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx\\ &=2 x^2+2 \int \left (\frac {15}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}-\frac {2 x}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}+\frac {10 x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2}\right ) \, dx\\ &=2 x^2-4 \int \frac {x}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx+20 \int \frac {x^2}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx+30 \int \frac {1}{\left (-15-8 x+10 x^2-2 x \log (5 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 27, normalized size = 1.00 \begin {gather*} 2 \left (x^2+\frac {x}{15+8 x-10 x^2+2 x \log (5 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 + 896*x + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + (240*x^2 + 128*x^3 - 160*x^4)*Log[5*x] + 16*x^
3*Log[5*x]^2)/(225 + 240*x - 236*x^2 - 160*x^3 + 100*x^4 + (60*x + 32*x^2 - 40*x^3)*Log[5*x] + 4*x^2*Log[5*x]^
2),x]

[Out]

2*(x^2 + x/(15 + 8*x - 10*x^2 + 2*x*Log[5*x]))

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fricas [A]  time = 0.60, size = 49, normalized size = 1.81 \begin {gather*} \frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} \log \left (5 \, x\right ) - 8 \, x^{3} - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-640*x^4-944*x^3+980*x^2+896*x+30)/(4*
x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*log(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="fricas")

[Out]

2*(10*x^4 - 2*x^3*log(5*x) - 8*x^3 - 15*x^2 - x)/(10*x^2 - 2*x*log(5*x) - 8*x - 15)

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giac [A]  time = 0.25, size = 28, normalized size = 1.04 \begin {gather*} 2 \, x^{2} - \frac {2 \, x}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-640*x^4-944*x^3+980*x^2+896*x+30)/(4*
x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*log(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="giac")

[Out]

2*x^2 - 2*x/(10*x^2 - 2*x*log(5*x) - 8*x - 15)

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maple [A]  time = 0.06, size = 29, normalized size = 1.07

method	result	size
risch	\(2 x^{2}-\frac {2 x}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\)	\(29\)
norman	\(\frac {-30 x^{2}-2 x -16 x^{3}+20 x^{4}-4 x^{3} \ln \left (5 x \right )}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^3*ln(5*x)^2+(-160*x^4+128*x^3+240*x^2)*ln(5*x)+400*x^5-640*x^4-944*x^3+980*x^2+896*x+30)/(4*x^2*ln(5
*x)^2+(-40*x^3+32*x^2+60*x)*ln(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x,method=_RETURNVERBOSE)

[Out]

2*x^2-2*x/(10*x^2-2*x*ln(5*x)-8*x-15)

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maxima [B]  time = 0.61, size = 53, normalized size = 1.96 \begin {gather*} \frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} {\left (\log \relax (5) + 4\right )} - 2 \, x^{3} \log \relax (x) - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x {\left (\log \relax (5) + 4\right )} - 2 \, x \log \relax (x) - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-640*x^4-944*x^3+980*x^2+896*x+30)/(4*
x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*log(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="maxima")

[Out]

2*(10*x^4 - 2*x^3*(log(5) + 4) - 2*x^3*log(x) - 15*x^2 - x)/(10*x^2 - 2*x*(log(5) + 4) - 2*x*log(x) - 15)

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mupad [B]  time = 3.43, size = 28, normalized size = 1.04 \begin {gather*} \frac {2\,x}{8\,x+2\,x\,\ln \left (5\,x\right )-10\,x^2+15}+2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((896*x + log(5*x)*(240*x^2 + 128*x^3 - 160*x^4) + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + 16*x^3*log(5*x)^
2 + 30)/(240*x + log(5*x)*(60*x + 32*x^2 - 40*x^3) - 236*x^2 - 160*x^3 + 100*x^4 + 4*x^2*log(5*x)^2 + 225),x)

[Out]

(2*x)/(8*x + 2*x*log(5*x) - 10*x^2 + 15) + 2*x^2

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sympy [A]  time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} 2 x^{2} + \frac {2 x}{- 10 x^{2} + 2 x \log {\left (5 x \right )} + 8 x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**3*ln(5*x)**2+(-160*x**4+128*x**3+240*x**2)*ln(5*x)+400*x**5-640*x**4-944*x**3+980*x**2+896*x+
30)/(4*x**2*ln(5*x)**2+(-40*x**3+32*x**2+60*x)*ln(5*x)+100*x**4-160*x**3-236*x**2+240*x+225),x)

[Out]

2*x**2 + 2*x/(-10*x**2 + 2*x*log(5*x) + 8*x + 15)

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