3.90.22 \(\int \frac {15625000 x+78125000 x^2+(2500 x+25000 x^2) \log (9)+2 x^2 \log ^2(9)+(3125000+15625000 x+2500 x \log (9)) \log (x)}{15625 x} \, dx\)

Optimal. Leaf size=20 \[ 25 \left (\frac {1}{625} x \log (9)+2 (5 x+\log (x))\right )^2 \]

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Rubi [B]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 7, number of rules used = 6, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6, 12, 14, 2346, 2301, 2295} \begin {gather*} 100 \log ^2(x)+\frac {(x (6250+\log (9))+1250)^2}{15625}+\frac {4}{25} x (6250+\log (9)) \log (x)-\frac {4}{25} x (6250+\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15625000*x + 78125000*x^2 + (2500*x + 25000*x^2)*Log[9] + 2*x^2*Log[9]^2 + (3125000 + 15625000*x + 2500*x
*Log[9])*Log[x])/(15625*x),x]

[Out]

(-4*x*(6250 + Log[9]))/25 + (1250 + x*(6250 + Log[9]))^2/15625 + (4*x*(6250 + Log[9])*Log[x])/25 + 100*Log[x]^
2

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15625000 x+\left (2500 x+25000 x^2\right ) \log (9)+x^2 \left (78125000+2 \log ^2(9)\right )+(3125000+15625000 x+2500 x \log (9)) \log (x)}{15625 x} \, dx\\ &=\frac {\int \frac {15625000 x+\left (2500 x+25000 x^2\right ) \log (9)+x^2 \left (78125000+2 \log ^2(9)\right )+(3125000+15625000 x+2500 x \log (9)) \log (x)}{x} \, dx}{15625}\\ &=\frac {\int \left (2 (6250+\log (9)) (1250+x (6250+\log (9)))+\frac {2500 (1250+x (6250+\log (9))) \log (x)}{x}\right ) \, dx}{15625}\\ &=\frac {(1250+x (6250+\log (9)))^2}{15625}+\frac {4}{25} \int \frac {(1250+x (6250+\log (9))) \log (x)}{x} \, dx\\ &=\frac {(1250+x (6250+\log (9)))^2}{15625}+200 \int \frac {\log (x)}{x} \, dx+\frac {1}{25} (4 (6250+\log (9))) \int \log (x) \, dx\\ &=-\frac {4}{25} x (6250+\log (9))+\frac {(1250+x (6250+\log (9)))^2}{15625}+\frac {4}{25} x (6250+\log (9)) \log (x)+100 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 46, normalized size = 2.30 \begin {gather*} 2500 x^2+\frac {4}{5} x^2 \log (9)+\frac {x^2 \log ^2(9)}{15625}+1000 x \log (x)+\frac {4}{25} x \log (9) \log (x)+100 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15625000*x + 78125000*x^2 + (2500*x + 25000*x^2)*Log[9] + 2*x^2*Log[9]^2 + (3125000 + 15625000*x +
2500*x*Log[9])*Log[x])/(15625*x),x]

[Out]

2500*x^2 + (4*x^2*Log[9])/5 + (x^2*Log[9]^2)/15625 + 1000*x*Log[x] + (4*x*Log[9]*Log[x])/25 + 100*Log[x]^2

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fricas [B]  time = 0.50, size = 40, normalized size = 2.00 \begin {gather*} \frac {4}{15625} \, x^{2} \log \relax (3)^{2} + \frac {8}{5} \, x^{2} \log \relax (3) + 2500 \, x^{2} + \frac {8}{25} \, {\left (x \log \relax (3) + 3125 \, x\right )} \log \relax (x) + 100 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15625*((5000*x*log(3)[phone]*x+3125000)*log(x)+8*x^2*log(3)^2+2*(25000*x^2+2500*x)*log(3)+781250
00*x^2+15625000*x)/x,x, algorithm="fricas")

[Out]

4/15625*x^2*log(3)^2 + 8/5*x^2*log(3) + 2500*x^2 + 8/25*(x*log(3) + 3125*x)*log(x) + 100*log(x)^2

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giac [A]  time = 0.15, size = 31, normalized size = 1.55 \begin {gather*} \frac {4}{15625} \, {\left (\log \relax (3)^{2} + 6250 \, \log \relax (3) + 9765625\right )} x^{2} + \frac {8}{25} \, x {\left (\log \relax (3) + 3125\right )} \log \relax (x) + 100 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15625*((5000*x*log(3)[phone]*x+3125000)*log(x)+8*x^2*log(3)^2+2*(25000*x^2+2500*x)*log(3)+781250
00*x^2+15625000*x)/x,x, algorithm="giac")

[Out]

4/15625*(log(3)^2 + 6250*log(3) + 9765625)*x^2 + 8/25*x*(log(3) + 3125)*log(x) + 100*log(x)^2

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maple [A]  time = 0.03, size = 34, normalized size = 1.70

method	result	size
norman	\(\left (\frac {4 \ln \relax (3)^{2}}{15625}+\frac {8 \ln \relax (3)}{5}+2500\right ) x^{2}+\left (\frac {8 \ln \relax (3)}{25}+1000\right ) x \ln \relax (x )+100 \ln \relax (x )^{2}\)	\(34\)
risch	\(100 \ln \relax (x )^{2}+\frac {8 \left (\ln \relax (3)+3125\right ) x \ln \relax (x )}{25}+\frac {4 x^{2} \ln \relax (3)^{2}}{15625}+\frac {8 x^{2} \ln \relax (3)}{5}+2500 x^{2}\)	\(38\)
default	\(\frac {4 x^{2} \ln \relax (3)^{2}}{15625}+\frac {8 \ln \relax (3) \left (x \ln \relax (x )-x \right )}{25}+\frac {8 x^{2} \ln \relax (3)}{5}+1000 x \ln \relax (x )+\frac {8 x \ln \relax (3)}{25}+2500 x^{2}+100 \ln \relax (x )^{2}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15625*((5000*x*ln(3)[phone]*x+3125000)*ln(x)+8*x^2*ln(3)^2+2*(25000*x^2+2500*x)*ln(3)[phone]*x^2+156
25000*x)/x,x,method=_RETURNVERBOSE)

[Out]

(4/15625*ln(3)^2+8/5*ln(3)+2500)*x^2+(8/25*ln(3)+1000)*x*ln(x)+100*ln(x)^2

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maxima [B]  time = 0.36, size = 50, normalized size = 2.50 \begin {gather*} \frac {4}{15625} \, x^{2} \log \relax (3)^{2} + \frac {8}{5} \, x^{2} \log \relax (3) + 2500 \, x^{2} + \frac {8}{25} \, {\left (x \log \relax (x) - x\right )} \log \relax (3) + \frac {8}{25} \, x \log \relax (3) + 1000 \, x \log \relax (x) + 100 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15625*((5000*x*log(3)[phone]*x+3125000)*log(x)+8*x^2*log(3)^2+2*(25000*x^2+2500*x)*log(3)+781250
00*x^2+15625000*x)/x,x, algorithm="maxima")

[Out]

4/15625*x^2*log(3)^2 + 8/5*x^2*log(3) + 2500*x^2 + 8/25*(x*log(x) - x)*log(3) + 8/25*x*log(3) + 1000*x*log(x)
+ 100*log(x)^2

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mupad [B]  time = 7.19, size = 16, normalized size = 0.80 \begin {gather*} \frac {4\,{\left (3125\,x+625\,\ln \relax (x)+x\,\ln \relax (3)\right )}^2}{15625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1000*x + (8*x^2*log(3)^2)/15625 + (2*log(3)*(2500*x + 25000*x^2))/15625 + (log(x)*(15625000*x + 5000*x*lo
g(3) + 3125000))/15625 + 5000*x^2)/x,x)

[Out]

(4*(3125*x + 625*log(x) + x*log(3))^2)/15625

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sympy [B]  time = 0.15, size = 41, normalized size = 2.05 \begin {gather*} x^{2} \left (\frac {4 \log {\relax (3 )}^{2}}{15625} + \frac {8 \log {\relax (3 )}}{5} + 2500\right ) + \left (\frac {8 x \log {\relax (3 )}}{25} + 1000 x\right ) \log {\relax (x )} + 100 \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15625*((5000*x*ln(3)[phone]*x+3125000)*ln(x)+8*x**2*ln(3)**2+2*(25000*x**2+2500*x)*ln(3)+7812500
0*x**2+15625000*x)/x,x)

[Out]

x**2*(4*log(3)**2/15625 + 8*log(3)/5 + 2500) + (8*x*log(3)/25 + 1000*x)*log(x) + 100*log(x)**2

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