3.82.4 \(\int \frac {1080+436 x+140 x^2+40 x^3+4 x^4+(436+88 x+40 x^2+8 x^3) \log (4)+(44+4 x^2) \log ^2(4)+(20+8 x+4 \log (4)) \log (x)}{75 x^2+30 x^3+3 x^4+(30 x^2+6 x^3) \log (4)+3 x^2 \log ^2(4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4 \left (-11+x^2+\frac {x-\log (x)}{5+x+\log (4)}\right )}{3 x} \]

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Rubi [B]  time = 0.80, antiderivative size = 273, normalized size of antiderivative = 10.50, number of steps used = 18, number of rules used = 10, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6688, 12, 6742, 43, 44, 2357, 2304, 2314, 31} \begin {gather*} \frac {4 x}{3}-\frac {4 \left (35+\log ^2(4)+10 \log (4)\right )}{3 (x+5+\log (4))}-\frac {4 x \log (x)}{3 (5+\log (4))^2 (x+5+\log (4))}-\frac {8 (270+(109+22 \log (2)) \log (4)) \log (x)}{3 (5+\log (4))^3}+\frac {4 (109+22 \log (4)) \log (x)}{3 (5+\log (4))^2}+\frac {8 (270+(109+22 \log (2)) \log (4)) \log (x+5+\log (4))}{3 (5+\log (4))^3}-\frac {4 (109+22 \log (4)) \log (x+5+\log (4))}{3 (5+\log (4))^2}+\frac {4 \log (x+5+\log (4))}{3 (5+\log (4))^2}+\frac {4 (5+\log (4))^2}{3 (x+5+\log (4))}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 (5+\log (4))^2 (x+5+\log (4))}+\frac {4 (109+22 \log (4))}{3 (5+\log (4)) (x+5+\log (4))}-\frac {4 \log (x)}{3 x (5+\log (4))}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 x (5+\log (4))^2}-\frac {4}{3 x (5+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1080 + 436*x + 140*x^2 + 40*x^3 + 4*x^4 + (436 + 88*x + 40*x^2 + 8*x^3)*Log[4] + (44 + 4*x^2)*Log[4]^2 +
(20 + 8*x + 4*Log[4])*Log[x])/(75*x^2 + 30*x^3 + 3*x^4 + (30*x^2 + 6*x^3)*Log[4] + 3*x^2*Log[4]^2),x]

[Out]

(4*x)/3 - 4/(3*x*(5 + Log[4])) + (4*(5 + Log[4])^2)/(3*(5 + x + Log[4])) + (4*(109 + 22*Log[4]))/(3*(5 + Log[4
])*(5 + x + Log[4])) - (4*(270 + (109 + 22*Log[2])*Log[4]))/(3*x*(5 + Log[4])^2) - (4*(270 + (109 + 22*Log[2])
*Log[4]))/(3*(5 + Log[4])^2*(5 + x + Log[4])) - (4*(35 + 10*Log[4] + Log[4]^2))/(3*(5 + x + Log[4])) - (4*Log[
x])/(3*x*(5 + Log[4])) - (4*x*Log[x])/(3*(5 + Log[4])^2*(5 + x + Log[4])) + (4*(109 + 22*Log[4])*Log[x])/(3*(5
+ Log[4])^2) - (8*(270 + (109 + 22*Log[2])*Log[4])*Log[x])/(3*(5 + Log[4])^3) + (4*Log[5 + x + Log[4]])/(3*(5
+ Log[4])^2) - (4*(109 + 22*Log[4])*Log[5 + x + Log[4]])/(3*(5 + Log[4])^2) + (8*(270 + (109 + 22*Log[2])*Log
[4])*Log[5 + x + Log[4]])/(3*(5 + Log[4])^3)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1080+436 x+140 x^2+40 x^3+4 x^4+\left (436+88 x+40 x^2+8 x^3\right ) \log (4)+\left (44+4 x^2\right ) \log ^2(4)+(20+8 x+4 \log (4)) \log (x)}{30 x^3+3 x^4+\left (30 x^2+6 x^3\right ) \log (4)+x^2 \left (75+3 \log ^2(4)\right )} \, dx\\ &=\int \frac {4 \left (x^4+2 x^3 (5+\log (4))+x (109+22 \log (4))+270 \left (1+\frac {1}{270} (109+22 \log (2)) \log (4)\right )+x^2 \left (35+10 \log (4)+\log ^2(4)\right )+(5+2 x+\log (4)) \log (x)\right )}{3 x^2 (5+x+\log (4))^2} \, dx\\ &=\frac {4}{3} \int \frac {x^4+2 x^3 (5+\log (4))+x (109+22 \log (4))+270 \left (1+\frac {1}{270} (109+22 \log (2)) \log (4)\right )+x^2 \left (35+10 \log (4)+\log ^2(4)\right )+(5+2 x+\log (4)) \log (x)}{x^2 (5+x+\log (4))^2} \, dx\\ &=\frac {4}{3} \int \left (\frac {x^2}{(5+x+\log (4))^2}+\frac {2 x (5+\log (4))}{(5+x+\log (4))^2}+\frac {109+22 \log (4)}{x (5+x+\log (4))^2}+\frac {270+(109+22 \log (2)) \log (4)}{x^2 (5+x+\log (4))^2}+\frac {35+10 \log (4)+\log ^2(4)}{(5+x+\log (4))^2}+\frac {(5+2 x+\log (4)) \log (x)}{x^2 (5+x+\log (4))^2}\right ) \, dx\\ &=-\frac {4 \left (35+10 \log (4)+\log ^2(4)\right )}{3 (5+x+\log (4))}+\frac {4}{3} \int \frac {x^2}{(5+x+\log (4))^2} \, dx+\frac {4}{3} \int \frac {(5+2 x+\log (4)) \log (x)}{x^2 (5+x+\log (4))^2} \, dx+\frac {1}{3} (8 (5+\log (4))) \int \frac {x}{(5+x+\log (4))^2} \, dx+\frac {1}{3} (4 (109+22 \log (4))) \int \frac {1}{x (5+x+\log (4))^2} \, dx+\frac {1}{3} (4 (270+(109+22 \log (2)) \log (4))) \int \frac {1}{x^2 (5+x+\log (4))^2} \, dx\\ &=-\frac {4 \left (35+10 \log (4)+\log ^2(4)\right )}{3 (5+x+\log (4))}+\frac {4}{3} \int \left (1+\frac {(5+\log (4))^2}{(5+x+\log (4))^2}-\frac {2 (5+\log (4))}{5+x+\log (4)}\right ) \, dx+\frac {4}{3} \int \left (\frac {\log (x)}{x^2 (5+\log (4))}-\frac {\log (x)}{(5+\log (4)) (5+x+\log (4))^2}\right ) \, dx+\frac {1}{3} (8 (5+\log (4))) \int \left (\frac {-5-\log (4)}{(5+x+\log (4))^2}+\frac {1}{5+x+\log (4)}\right ) \, dx+\frac {1}{3} (4 (109+22 \log (4))) \int \left (\frac {1}{x (5+\log (4))^2}-\frac {1}{(5+\log (4)) (5+x+\log (4))^2}-\frac {1}{(5+\log (4))^2 (5+x+\log (4))}\right ) \, dx+\frac {1}{3} (4 (270+(109+22 \log (2)) \log (4))) \int \left (-\frac {2}{x (5+\log (4))^3}+\frac {1}{x^2 (5+\log (4))^2}+\frac {1}{(5+\log (4))^2 (5+x+\log (4))^2}+\frac {2}{(5+\log (4))^3 (5+x+\log (4))}\right ) \, dx\\ &=\frac {4 x}{3}+\frac {4 (5+\log (4))^2}{3 (5+x+\log (4))}+\frac {4 (109+22 \log (4))}{3 (5+\log (4)) (5+x+\log (4))}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 x (5+\log (4))^2}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 (5+\log (4))^2 (5+x+\log (4))}-\frac {4 \left (35+10 \log (4)+\log ^2(4)\right )}{3 (5+x+\log (4))}+\frac {4 (109+22 \log (4)) \log (x)}{3 (5+\log (4))^2}-\frac {8 (270+(109+22 \log (2)) \log (4)) \log (x)}{3 (5+\log (4))^3}-\frac {4 (109+22 \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^2}+\frac {8 (270+(109+22 \log (2)) \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^3}+\frac {4 \int \frac {\log (x)}{x^2} \, dx}{3 (5+\log (4))}-\frac {4 \int \frac {\log (x)}{(5+x+\log (4))^2} \, dx}{3 (5+\log (4))}\\ &=\frac {4 x}{3}-\frac {4}{3 x (5+\log (4))}+\frac {4 (5+\log (4))^2}{3 (5+x+\log (4))}+\frac {4 (109+22 \log (4))}{3 (5+\log (4)) (5+x+\log (4))}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 x (5+\log (4))^2}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 (5+\log (4))^2 (5+x+\log (4))}-\frac {4 \left (35+10 \log (4)+\log ^2(4)\right )}{3 (5+x+\log (4))}-\frac {4 \log (x)}{3 x (5+\log (4))}-\frac {4 x \log (x)}{3 (5+\log (4))^2 (5+x+\log (4))}+\frac {4 (109+22 \log (4)) \log (x)}{3 (5+\log (4))^2}-\frac {8 (270+(109+22 \log (2)) \log (4)) \log (x)}{3 (5+\log (4))^3}-\frac {4 (109+22 \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^2}+\frac {8 (270+(109+22 \log (2)) \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^3}+\frac {4 \int \frac {1}{5+x+\log (4)} \, dx}{3 (5+\log (4))^2}\\ &=\frac {4 x}{3}-\frac {4}{3 x (5+\log (4))}+\frac {4 (5+\log (4))^2}{3 (5+x+\log (4))}+\frac {4 (109+22 \log (4))}{3 (5+\log (4)) (5+x+\log (4))}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 x (5+\log (4))^2}-\frac {4 (270+(109+22 \log (2)) \log (4))}{3 (5+\log (4))^2 (5+x+\log (4))}-\frac {4 \left (35+10 \log (4)+\log ^2(4)\right )}{3 (5+x+\log (4))}-\frac {4 \log (x)}{3 x (5+\log (4))}-\frac {4 x \log (x)}{3 (5+\log (4))^2 (5+x+\log (4))}+\frac {4 (109+22 \log (4)) \log (x)}{3 (5+\log (4))^2}-\frac {8 (270+(109+22 \log (2)) \log (4)) \log (x)}{3 (5+\log (4))^3}+\frac {4 \log (5+x+\log (4))}{3 (5+\log (4))^2}-\frac {4 (109+22 \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^2}+\frac {8 (270+(109+22 \log (2)) \log (4)) \log (5+x+\log (4))}{3 (5+\log (4))^3}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.36, size = 75, normalized size = 2.88 \begin {gather*} \frac {4 \left (x^3 (5+\log (4))^2-11 (5+\log (4))^3+x^2 (5+\log (4))^3+x \left (-250+\log ^2(4)-2 \log (4) (50+\log (2048))\right )-(5+\log (4))^2 \log (x)\right )}{3 x (5+\log (4))^2 (5+x+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1080 + 436*x + 140*x^2 + 40*x^3 + 4*x^4 + (436 + 88*x + 40*x^2 + 8*x^3)*Log[4] + (44 + 4*x^2)*Log[4
]^2 + (20 + 8*x + 4*Log[4])*Log[x])/(75*x^2 + 30*x^3 + 3*x^4 + (30*x^2 + 6*x^3)*Log[4] + 3*x^2*Log[4]^2),x]

[Out]

(4*(x^3*(5 + Log[4])^2 - 11*(5 + Log[4])^3 + x^2*(5 + Log[4])^3 + x*(-250 + Log[4]^2 - 2*Log[4]*(50 + Log[2048
])) - (5 + Log[4])^2*Log[x]))/(3*x*(5 + Log[4])^2*(5 + x + Log[4]))

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fricas [A]  time = 0.81, size = 42, normalized size = 1.62 \begin {gather*} \frac {4 \, {\left (x^{3} + 5 \, x^{2} + 2 \, {\left (x^{2} - 11\right )} \log \relax (2) - 10 \, x - \log \relax (x) - 55\right )}}{3 \, {\left (x^{2} + 2 \, x \log \relax (2) + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)+8*x+20)*log(x)+4*(4*x^2+44)*log(2)^2+2*(8*x^3+40*x^2+88*x+436)*log(2)+4*x^4+40*x^3+140*x^
2+436*x+1080)/(12*x^2*log(2)^2+2*(6*x^3+30*x^2)*log(2)+3*x^4+30*x^3+75*x^2),x, algorithm="fricas")

[Out]

4/3*(x^3 + 5*x^2 + 2*(x^2 - 11)*log(2) - 10*x - log(x) - 55)/(x^2 + 2*x*log(2) + 5*x)

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giac [B]  time = 0.23, size = 60, normalized size = 2.31 \begin {gather*} \frac {4}{3} \, {\left (\frac {1}{2 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} + 5 \, x + 20 \, \log \relax (2) + 25} - \frac {1}{2 \, x \log \relax (2) + 5 \, x}\right )} \log \relax (x) + \frac {4}{3} \, x + \frac {4}{3 \, {\left (x + 2 \, \log \relax (2) + 5\right )}} - \frac {44}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)+8*x+20)*log(x)+4*(4*x^2+44)*log(2)^2+2*(8*x^3+40*x^2+88*x+436)*log(2)+4*x^4+40*x^3+140*x^
2+436*x+1080)/(12*x^2*log(2)^2+2*(6*x^3+30*x^2)*log(2)+3*x^4+30*x^3+75*x^2),x, algorithm="giac")

[Out]

4/3*(1/(2*x*log(2) + 4*log(2)^2 + 5*x + 20*log(2) + 25) - 1/(2*x*log(2) + 5*x))*log(x) + 4/3*x + 4/3/(x + 2*lo
g(2) + 5) - 44/3/x

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maple [A]  time = 0.15, size = 43, normalized size = 1.65

method	result	size
norman	\(\frac {\left (-\frac {16 \ln \relax (2)^{2}}{3}-\frac {80 \ln \relax (2)}{3}-\frac {140}{3}\right ) x +\frac {4 x^{3}}{3}-\frac {220}{3}-\frac {4 \ln \relax (x )}{3}-\frac {88 \ln \relax (2)}{3}}{x \left (2 \ln \relax (2)+5+x \right )}\)	\(43\)
risch	\(-\frac {4 \ln \relax (x )}{3 \left (2 \ln \relax (2)+5+x \right ) x}+\frac {\frac {8 x^{2} \ln \relax (2)}{3}+\frac {4 x^{3}}{3}+\frac {20 x^{2}}{3}-\frac {88 \ln \relax (2)}{3}-\frac {40 x}{3}-\frac {220}{3}}{x \left (2 \ln \relax (2)+5+x \right )}\)	\(56\)
default	\(\frac {4 x}{3}+\frac {4}{3 \left (2 \ln \relax (2)+5+x \right )}+\frac {4 \ln \relax (x )}{3 \left (2 \ln \relax (2)+5\right )^{2}}-\frac {88 \ln \relax (2)}{3 \left (2 \ln \relax (2)+5\right ) x}-\frac {220}{3 \left (2 \ln \relax (2)+5\right ) x}-\frac {4 \ln \relax (x )}{3 \left (2 \ln \relax (2)+5\right ) x}-\frac {4 \ln \relax (x ) x}{3 \left (2 \ln \relax (2)+5\right )^{2} \left (2 \ln \relax (2)+5+x \right )}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*ln(2)+8*x+20)*ln(x)+4*(4*x^2+44)*ln(2)^2+2*(8*x^3+40*x^2+88*x+436)*ln(2)+4*x^4+40*x^3+140*x^2+436*x+10
80)/(12*x^2*ln(2)^2+2*(6*x^3+30*x^2)*ln(2)+3*x^4+30*x^3+75*x^2),x,method=_RETURNVERBOSE)

[Out]

((-16/3*ln(2)^2-80/3*ln(2)-140/3)*x+4/3*x^3-220/3-4/3*ln(x)-88/3*ln(2))/x/(2*ln(2)+5+x)

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maxima [B]  time = 0.54, size = 683, normalized size = 26.27 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)+8*x+20)*log(x)+4*(4*x^2+44)*log(2)^2+2*(8*x^3+40*x^2+88*x+436)*log(2)+4*x^4+40*x^3+140*x^
2+436*x+1080)/(12*x^2*log(2)^2+2*(6*x^3+30*x^2)*log(2)+3*x^4+30*x^3+75*x^2),x, algorithm="maxima")

[Out]

-176/3*((2*x + 2*log(2) + 5)/((4*log(2)^2 + 20*log(2) + 25)*x^2 + (8*log(2)^3 + 60*log(2)^2 + 150*log(2) + 125
)*x) - 2*log(x + 2*log(2) + 5)/(8*log(2)^3 + 60*log(2)^2 + 150*log(2) + 125) + 2*log(x)/(8*log(2)^3 + 60*log(2
)^2 + 150*log(2) + 125))*log(2)^2 - 872/3*((2*x + 2*log(2) + 5)/((4*log(2)^2 + 20*log(2) + 25)*x^2 + (8*log(2)
^3 + 60*log(2)^2 + 150*log(2) + 125)*x) - 2*log(x + 2*log(2) + 5)/(8*log(2)^3 + 60*log(2)^2 + 150*log(2) + 125
) + 2*log(x)/(8*log(2)^3 + 60*log(2)^2 + 150*log(2) + 125))*log(2) + 16/3*((2*log(2) + 5)/(x + 2*log(2) + 5) +
log(x + 2*log(2) + 5))*log(2) - 176/3*(log(x + 2*log(2) + 5)/(4*log(2)^2 + 20*log(2) + 25) - log(x)/(4*log(2)
^2 + 20*log(2) + 25) - 1/(x*(2*log(2) + 5) + 4*log(2)^2 + 20*log(2) + 25))*log(2) - 8/3*(2*log(2) + 5)*log(x +
2*log(2) + 5) + 4/3*x - 16/3*log(2)^2/(x + 2*log(2) + 5) - 4/3*(x*(2*log(2) + 5) + 4*log(2)^2 + (x^2 + x*(2*l
og(2) + 5) + 4*log(2)^2 + 20*log(2) + 25)*log(x) + 20*log(2) + 25)/((4*log(2)^2 + 20*log(2) + 25)*x^2 + (8*log
(2)^3 + 60*log(2)^2 + 150*log(2) + 125)*x) - 360*(2*x + 2*log(2) + 5)/((4*log(2)^2 + 20*log(2) + 25)*x^2 + (8*
log(2)^3 + 60*log(2)^2 + 150*log(2) + 125)*x) - 4/3*(4*log(2)^2 + 20*log(2) + 25)/(x + 2*log(2) + 5) + 40/3*(2
*log(2) + 5)/(x + 2*log(2) + 5) - 80/3*log(2)/(x + 2*log(2) + 5) + 720*log(x + 2*log(2) + 5)/(8*log(2)^3 + 60*
log(2)^2 + 150*log(2) + 125) - 144*log(x + 2*log(2) + 5)/(4*log(2)^2 + 20*log(2) + 25) - 720*log(x)/(8*log(2)^
3 + 60*log(2)^2 + 150*log(2) + 125) + 436/3*log(x)/(4*log(2)^2 + 20*log(2) + 25) + 436/3/(x*(2*log(2) + 5) + 4
*log(2)^2 + 20*log(2) + 25) - 140/3/(x + 2*log(2) + 5) + 40/3*log(x + 2*log(2) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {436\,x+\ln \relax (x)\,\left (8\,x+8\,\ln \relax (2)+20\right )+2\,\ln \relax (2)\,\left (8\,x^3+40\,x^2+88\,x+436\right )+4\,{\ln \relax (2)}^2\,\left (4\,x^2+44\right )+140\,x^2+40\,x^3+4\,x^4+1080}{12\,x^2\,{\ln \relax (2)}^2+2\,\ln \relax (2)\,\left (6\,x^3+30\,x^2\right )+75\,x^2+30\,x^3+3\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((436*x + log(x)*(8*x + 8*log(2) + 20) + 2*log(2)*(88*x + 40*x^2 + 8*x^3 + 436) + 4*log(2)^2*(4*x^2 + 44) +
140*x^2 + 40*x^3 + 4*x^4 + 1080)/(12*x^2*log(2)^2 + 2*log(2)*(30*x^2 + 6*x^3) + 75*x^2 + 30*x^3 + 3*x^4),x)

[Out]

int((436*x + log(x)*(8*x + 8*log(2) + 20) + 2*log(2)*(88*x + 40*x^2 + 8*x^3 + 436) + 4*log(2)^2*(4*x^2 + 44) +
140*x^2 + 40*x^3 + 4*x^4 + 1080)/(12*x^2*log(2)^2 + 2*log(2)*(30*x^2 + 6*x^3) + 75*x^2 + 30*x^3 + 3*x^4), x)

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sympy [A]  time = 0.78, size = 49, normalized size = 1.88 \begin {gather*} \frac {4 x}{3} - \frac {4 \log {\relax (x )}}{3 x^{2} + 6 x \log {\relax (2 )} + 15 x} + \frac {- 40 x - 220 - 88 \log {\relax (2 )}}{3 x^{2} + x \left (6 \log {\relax (2 )} + 15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*ln(2)+8*x+20)*ln(x)+4*(4*x**2+44)*ln(2)**2+2*(8*x**3+40*x**2+88*x+436)*ln(2)+4*x**4+40*x**3+140*
x**2+436*x+1080)/(12*x**2*ln(2)**2+2*(6*x**3+30*x**2)*ln(2)+3*x**4+30*x**3+75*x**2),x)

[Out]

4*x/3 - 4*log(x)/(3*x**2 + 6*x*log(2) + 15*x) + (-40*x - 220 - 88*log(2))/(3*x**2 + x*(6*log(2) + 15))

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