3.2.69 \(\int \frac {384 x^2+288 x^3+112 x^4-84 x^5+40 x^6+10 x^7}{128+96 x+24 x^2+2 x^3+(64+48 x+12 x^2+x^3) \log (3)} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 x^2 \left (x+\frac {x \left (-x+x^2\right )^2}{(4+x)^2}\right )}{2+\log (3)} \]

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Rubi [B]  time = 0.12, antiderivative size = 80, normalized size of antiderivative = 2.76, number of steps used = 2, number of rules used = 1, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2074} \begin {gather*} \frac {2 x^5}{2+\log (3)}-\frac {20 x^4}{2+\log (3)}+\frac {132 x^3}{2+\log (3)}-\frac {720 x^2}{2+\log (3)}+\frac {3680 x}{2+\log (3)}+\frac {84480}{(x+4) (2+\log (3))}-\frac {51200}{(x+4)^2 (2+\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(384*x^2 + 288*x^3 + 112*x^4 - 84*x^5 + 40*x^6 + 10*x^7)/(128 + 96*x + 24*x^2 + 2*x^3 + (64 + 48*x + 12*x^
2 + x^3)*Log[3]),x]

[Out]

(3680*x)/(2 + Log[3]) - (720*x^2)/(2 + Log[3]) + (132*x^3)/(2 + Log[3]) - (20*x^4)/(2 + Log[3]) + (2*x^5)/(2 +
Log[3]) - 51200/((4 + x)^2*(2 + Log[3])) + 84480/((4 + x)*(2 + Log[3]))

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {3680}{2+\log (3)}-\frac {1440 x}{2+\log (3)}+\frac {396 x^2}{2+\log (3)}-\frac {80 x^3}{2+\log (3)}+\frac {10 x^4}{2+\log (3)}+\frac {102400}{(4+x)^3 (2+\log (3))}-\frac {84480}{(4+x)^2 (2+\log (3))}\right ) \, dx\\ &=\frac {3680 x}{2+\log (3)}-\frac {720 x^2}{2+\log (3)}+\frac {132 x^3}{2+\log (3)}-\frac {20 x^4}{2+\log (3)}+\frac {2 x^5}{2+\log (3)}-\frac {51200}{(4+x)^2 (2+\log (3))}+\frac {84480}{(4+x) (2+\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.59 \begin {gather*} \frac {2 \left (478208+239104 x+29888 x^2+16 x^3+8 x^4+2 x^5-2 x^6+x^7\right )}{(4+x)^2 (2+\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(384*x^2 + 288*x^3 + 112*x^4 - 84*x^5 + 40*x^6 + 10*x^7)/(128 + 96*x + 24*x^2 + 2*x^3 + (64 + 48*x +
12*x^2 + x^3)*Log[3]),x]

[Out]

(2*(478208 + 239104*x + 29888*x^2 + 16*x^3 + 8*x^4 + 2*x^5 - 2*x^6 + x^7))/((4 + x)^2*(2 + Log[3]))

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fricas [A]  time = 0.66, size = 58, normalized size = 2.00 \begin {gather*} \frac {2 \, {\left (x^{7} - 2 \, x^{6} + 2 \, x^{5} + 8 \, x^{4} + 16 \, x^{3} + 8960 \, x^{2} + 71680 \, x + 143360\right )}}{2 \, x^{2} + {\left (x^{2} + 8 \, x + 16\right )} \log \relax (3) + 16 \, x + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^7+40*x^6-84*x^5+112*x^4+288*x^3+384*x^2)/((x^3+12*x^2+48*x+64)*log(3)+2*x^3+24*x^2+96*x+128),x
, algorithm="fricas")

[Out]

2*(x^7 - 2*x^6 + 2*x^5 + 8*x^4 + 16*x^3 + 8960*x^2 + 71680*x + 143360)/(2*x^2 + (x^2 + 8*x + 16)*log(3) + 16*x
+ 32)

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giac [B]  time = 0.36, size = 236, normalized size = 8.14 \begin {gather*} \frac {2 \, {\left (x^{5} \log \relax (3)^{4} + 8 \, x^{5} \log \relax (3)^{3} - 10 \, x^{4} \log \relax (3)^{4} + 24 \, x^{5} \log \relax (3)^{2} - 80 \, x^{4} \log \relax (3)^{3} + 66 \, x^{3} \log \relax (3)^{4} + 32 \, x^{5} \log \relax (3) - 240 \, x^{4} \log \relax (3)^{2} + 528 \, x^{3} \log \relax (3)^{3} - 360 \, x^{2} \log \relax (3)^{4} + 16 \, x^{5} - 320 \, x^{4} \log \relax (3) + 1584 \, x^{3} \log \relax (3)^{2} - 2880 \, x^{2} \log \relax (3)^{3} + 1840 \, x \log \relax (3)^{4} - 160 \, x^{4} + 2112 \, x^{3} \log \relax (3) - 8640 \, x^{2} \log \relax (3)^{2} + 14720 \, x \log \relax (3)^{3} + 1056 \, x^{3} - 11520 \, x^{2} \log \relax (3) + 44160 \, x \log \relax (3)^{2} - 5760 \, x^{2} + 58880 \, x \log \relax (3) + 29440 \, x\right )}}{\log \relax (3)^{5} + 10 \, \log \relax (3)^{4} + 40 \, \log \relax (3)^{3} + 80 \, \log \relax (3)^{2} + 80 \, \log \relax (3) + 32} + \frac {2560 \, {\left (33 \, x + 112\right )}}{{\left (x + 4\right )}^{2} {\left (\log \relax (3) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^7+40*x^6-84*x^5+112*x^4+288*x^3+384*x^2)/((x^3+12*x^2+48*x+64)*log(3)+2*x^3+24*x^2+96*x+128),x
, algorithm="giac")

[Out]

2*(x^5*log(3)^4 + 8*x^5*log(3)^3 - 10*x^4*log(3)^4 + 24*x^5*log(3)^2 - 80*x^4*log(3)^3 + 66*x^3*log(3)^4 + 32*
x^5*log(3) - 240*x^4*log(3)^2 + 528*x^3*log(3)^3 - 360*x^2*log(3)^4 + 16*x^5 - 320*x^4*log(3) + 1584*x^3*log(3
)^2 - 2880*x^2*log(3)^3 + 1840*x*log(3)^4 - 160*x^4 + 2112*x^3*log(3) - 8640*x^2*log(3)^2 + 14720*x*log(3)^3 +
1056*x^3 - 11520*x^2*log(3) + 44160*x*log(3)^2 - 5760*x^2 + 58880*x*log(3) + 29440*x)/(log(3)^5 + 10*log(3)^4
+ 40*log(3)^3 + 80*log(3)^2 + 80*log(3) + 32) + 2560*(33*x + 112)/((x + 4)^2*(log(3) + 2))

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maple [A]  time = 0.06, size = 45, normalized size = 1.55

method	result	size
default	\(\frac {2 x^{5}-20 x^{4}+132 x^{3}-720 x^{2}+3680 x -\frac {51200}{\left (4+x \right )^{2}}+\frac {84480}{4+x}}{2+\ln \relax (3)}\)	\(45\)
gosper	\(\frac {2 x^{3} \left (x^{4}-2 x^{3}+2 x^{2}+8 x +16\right )}{x^{2} \ln \relax (3)+8 x \ln \relax (3)+2 x^{2}+16 \ln \relax (3)+16 x +32}\)	\(51\)
norman	\(\frac {\frac {32 x^{3}}{2+\ln \relax (3)}+\frac {16 x^{4}}{2+\ln \relax (3)}+\frac {4 x^{5}}{2+\ln \relax (3)}-\frac {4 x^{6}}{2+\ln \relax (3)}+\frac {2 x^{7}}{2+\ln \relax (3)}}{\left (4+x \right )^{2}}\)	\(63\)
risch	\(\frac {2 x^{5}}{2+\ln \relax (3)}-\frac {20 x^{4}}{2+\ln \relax (3)}+\frac {132 x^{3}}{2+\ln \relax (3)}-\frac {720 x^{2}}{2+\ln \relax (3)}+\frac {3680 x}{2+\ln \relax (3)}+\frac {\left (84480 \ln \relax (3)+168960\right ) x +286720 \ln \relax (3)+573440}{\left (2+\ln \relax (3)\right ) \left (x^{2} \ln \relax (3)+8 x \ln \relax (3)+2 x^{2}+16 \ln \relax (3)+16 x +32\right )}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^7+40*x^6-84*x^5+112*x^4+288*x^3+384*x^2)/((x^3+12*x^2+48*x+64)*ln(3)+2*x^3+24*x^2+96*x+128),x,method
=_RETURNVERBOSE)

[Out]

2/(2+ln(3))*(x^5-10*x^4+66*x^3-360*x^2+1840*x-25600/(4+x)^2+42240/(4+x))

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maxima [B]  time = 0.44, size = 61, normalized size = 2.10 \begin {gather*} \frac {2560 \, {\left (33 \, x + 112\right )}}{x^{2} {\left (\log \relax (3) + 2\right )} + 8 \, x {\left (\log \relax (3) + 2\right )} + 16 \, \log \relax (3) + 32} + \frac {2 \, {\left (x^{5} - 10 \, x^{4} + 66 \, x^{3} - 360 \, x^{2} + 1840 \, x\right )}}{\log \relax (3) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^7+40*x^6-84*x^5+112*x^4+288*x^3+384*x^2)/((x^3+12*x^2+48*x+64)*log(3)+2*x^3+24*x^2+96*x+128),x
, algorithm="maxima")

[Out]

2560*(33*x + 112)/(x^2*(log(3) + 2) + 8*x*(log(3) + 2) + 16*log(3) + 32) + 2*(x^5 - 10*x^4 + 66*x^3 - 360*x^2
+ 1840*x)/(log(3) + 2)

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mupad [B]  time = 0.70, size = 541, normalized size = 18.66 \begin {gather*} x\,\left (\frac {288}{\ln \relax (3)+2}-\frac {\left (64\,\ln \relax (3)+128\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{\ln \relax (3)+2}-\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {112}{\ln \relax (3)+2}-\frac {10\,\left (64\,\ln \relax (3)+128\right )}{{\left (\ln \relax (3)+2\right )}^2}-\frac {\left (48\,\ln \relax (3)+96\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{\ln \relax (3)+2}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {84}{\ln \relax (3)+2}+\frac {10\,\left (48\,\ln \relax (3)+96\right )}{{\left (\ln \relax (3)+2\right )}^2}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{\ln \relax (3)+2}\right )}{\ln \relax (3)+2}\right )}{\ln \relax (3)+2}+\frac {\left (48\,\ln \relax (3)+96\right )\,\left (\frac {84}{\ln \relax (3)+2}+\frac {10\,\left (48\,\ln \relax (3)+96\right )}{{\left (\ln \relax (3)+2\right )}^2}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{\ln \relax (3)+2}\right )}{\ln \relax (3)+2}\right )+\frac {84480\,x+286720}{\left (\ln \relax (3)+2\right )\,x^2+\left (8\,\ln \relax (3)+16\right )\,x+16\,\ln \relax (3)+32}-x^3\,\left (\frac {28}{\ln \relax (3)+2}+\frac {10\,\left (48\,\ln \relax (3)+96\right )}{3\,{\left (\ln \relax (3)+2\right )}^2}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{3\,\left (\ln \relax (3)+2\right )}\right )+\frac {2\,x^5}{\ln \relax (3)+2}+x^4\,\left (\frac {10}{\ln \relax (3)+2}-\frac {5\,\left (12\,\ln \relax (3)+24\right )}{2\,{\left (\ln \relax (3)+2\right )}^2}\right )+x^2\,\left (\frac {56}{\ln \relax (3)+2}-\frac {5\,\left (64\,\ln \relax (3)+128\right )}{{\left (\ln \relax (3)+2\right )}^2}-\frac {\left (48\,\ln \relax (3)+96\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{2\,\left (\ln \relax (3)+2\right )}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {84}{\ln \relax (3)+2}+\frac {10\,\left (48\,\ln \relax (3)+96\right )}{{\left (\ln \relax (3)+2\right )}^2}+\frac {\left (12\,\ln \relax (3)+24\right )\,\left (\frac {40}{\ln \relax (3)+2}-\frac {10\,\left (12\,\ln \relax (3)+24\right )}{{\left (\ln \relax (3)+2\right )}^2}\right )}{\ln \relax (3)+2}\right )}{2\,\left (\ln \relax (3)+2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((384*x^2 + 288*x^3 + 112*x^4 - 84*x^5 + 40*x^6 + 10*x^7)/(96*x + log(3)*(48*x + 12*x^2 + x^3 + 64) + 24*x^
2 + 2*x^3 + 128),x)

[Out]

x*(288/(log(3) + 2) - ((64*log(3) + 128)*(40/(log(3) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2))/(log(3) + 2
) - ((12*log(3) + 24)*(112/(log(3) + 2) - (10*(64*log(3) + 128))/(log(3) + 2)^2 - ((48*log(3) + 96)*(40/(log(3
) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2))/(log(3) + 2) + ((12*log(3) + 24)*(84/(log(3) + 2) + (10*(48*lo
g(3) + 96))/(log(3) + 2)^2 + ((12*log(3) + 24)*(40/(log(3) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2))/(log(
3) + 2)))/(log(3) + 2)))/(log(3) + 2) + ((48*log(3) + 96)*(84/(log(3) + 2) + (10*(48*log(3) + 96))/(log(3) + 2
)^2 + ((12*log(3) + 24)*(40/(log(3) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2))/(log(3) + 2)))/(log(3) + 2))
+ (84480*x + 286720)/(16*log(3) + x*(8*log(3) + 16) + x^2*(log(3) + 2) + 32) - x^3*(28/(log(3) + 2) + (10*(48
*log(3) + 96))/(3*(log(3) + 2)^2) + ((12*log(3) + 24)*(40/(log(3) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2)
)/(3*(log(3) + 2))) + (2*x^5)/(log(3) + 2) + x^4*(10/(log(3) + 2) - (5*(12*log(3) + 24))/(2*(log(3) + 2)^2)) +
x^2*(56/(log(3) + 2) - (5*(64*log(3) + 128))/(log(3) + 2)^2 - ((48*log(3) + 96)*(40/(log(3) + 2) - (10*(12*lo
g(3) + 24))/(log(3) + 2)^2))/(2*(log(3) + 2)) + ((12*log(3) + 24)*(84/(log(3) + 2) + (10*(48*log(3) + 96))/(lo
g(3) + 2)^2 + ((12*log(3) + 24)*(40/(log(3) + 2) - (10*(12*log(3) + 24))/(log(3) + 2)^2))/(log(3) + 2)))/(2*(l
og(3) + 2)))

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sympy [B]  time = 0.68, size = 76, normalized size = 2.62 \begin {gather*} \frac {2 x^{5}}{\log {\relax (3 )} + 2} - \frac {20 x^{4}}{\log {\relax (3 )} + 2} + \frac {132 x^{3}}{\log {\relax (3 )} + 2} - \frac {720 x^{2}}{\log {\relax (3 )} + 2} + \frac {3680 x}{\log {\relax (3 )} + 2} + \frac {84480 x + 286720}{x^{2} \left (\log {\relax (3 )} + 2\right ) + x \left (8 \log {\relax (3 )} + 16\right ) + 16 \log {\relax (3 )} + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x**7+40*x**6-84*x**5+112*x**4+288*x**3+384*x**2)/((x**3+12*x**2+48*x+64)*ln(3)+2*x**3+24*x**2+96
*x+128),x)

[Out]

2*x**5/(log(3) + 2) - 20*x**4/(log(3) + 2) + 132*x**3/(log(3) + 2) - 720*x**2/(log(3) + 2) + 3680*x/(log(3) +
2) + (84480*x + 286720)/(x**2*(log(3) + 2) + x*(8*log(3) + 16) + 16*log(3) + 32)

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