3.47.96 \(\int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log (x^2)}{25 x} \, dx\)

Optimal. Leaf size=17 \[ \left (1+\frac {4 x}{5}+\log (4)+5 \log \left (x^2\right )\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.06, antiderivative size = 43, normalized size of antiderivative = 2.53, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 14, 76, 2346, 2301, 2295} \begin {gather*} \frac {16 x^2}{25}+25 \log ^2\left (x^2\right )+8 x \log \left (x^2\right )-16 x+\frac {8}{5} x (11+\log (4))+20 (1+\log (4)) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(500 + 440*x + 32*x^2 + (500 + 40*x)*Log[4] + (2500 + 200*x)*Log[x^2])/(25*x),x]

[Out]

-16*x + (16*x^2)/25 + (8*x*(11 + Log[4]))/5 + 20*(1 + Log[4])*Log[x] + 8*x*Log[x^2] + 25*Log[x^2]^2

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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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} &=\frac {1}{25} \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{x} \, dx\\ &=\frac {1}{25} \int \left (\frac {4 (25+2 x) (5+4 x+5 \log (4))}{x}+\frac {100 (25+2 x) \log \left (x^2\right )}{x}\right ) \, dx\\ &=\frac {4}{25} \int \frac {(25+2 x) (5+4 x+5 \log (4))}{x} \, dx+4 \int \frac {(25+2 x) \log \left (x^2\right )}{x} \, dx\\ &=\frac {4}{25} \int \left (8 x+\frac {125 (1+\log (4))}{x}+10 (11+\log (4))\right ) \, dx+8 \int \log \left (x^2\right ) \, dx+100 \int \frac {\log \left (x^2\right )}{x} \, dx\\ &=-16 x+\frac {16 x^2}{25}+\frac {8}{5} x (11+\log (4))+20 (1+\log (4)) \log (x)+8 x \log \left (x^2\right )+25 \log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.01, size = 45, normalized size = 2.65 \begin {gather*} \frac {8 x}{5}+\frac {16 x^2}{25}+\frac {8}{5} x \log (4)+20 \log (x)+20 \log (4) \log (x)+8 x \log \left (x^2\right )+25 \log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(500 + 440*x + 32*x^2 + (500 + 40*x)*Log[4] + (2500 + 200*x)*Log[x^2])/(25*x),x]

[Out]

(8*x)/5 + (16*x^2)/25 + (8*x*Log[4])/5 + 20*Log[x] + 20*Log[4]*Log[x] + 8*x*Log[x^2] + 25*Log[x^2]^2

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 37, normalized size = 2.18 \begin {gather*} \frac {16}{25} \, x^{2} + \frac {16}{5} \, x \log \relax (2) + 2 \, {\left (4 \, x + 10 \, \log \relax (2) + 5\right )} \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + \frac {8}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((200*x+2500)*log(x^2)+2*(40*x+500)*log(2)+32*x^2+440*x+500)/x,x, algorithm="fricas")

[Out]

16/25*x^2 + 16/5*x*log(2) + 2*(4*x + 10*log(2) + 5)*log(x^2) + 25*log(x^2)^2 + 8/5*x

________________________________________________________________________________________

giac [B]  time = 0.20, size = 40, normalized size = 2.35 \begin {gather*} \frac {16}{25} \, x^{2} + \frac {8}{5} \, x {\left (2 \, \log \relax (2) + 1\right )} + 8 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + 20 \, {\left (2 \, \log \relax (2) + 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((200*x+2500)*log(x^2)+2*(40*x+500)*log(2)+32*x^2+440*x+500)/x,x, algorithm="giac")

[Out]

16/25*x^2 + 8/5*x*(2*log(2) + 1) + 8*x*log(x^2) + 25*log(x^2)^2 + 20*(2*log(2) + 1)*log(x)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 41, normalized size = 2.41

method	result	size
norman	\(\left (\frac {8}{5}+\frac {16 \ln \relax (2)}{5}\right ) x +\left (10+20 \ln \relax (2)\right ) \ln \left (x^{2}\right )+\frac {16 x^{2}}{25}+25 \ln \left (x^{2}\right )^{2}+8 x \ln \left (x^{2}\right )\)	\(41\)
default	\(\frac {16 x^{2}}{25}+\frac {8 x}{5}+20 \ln \relax (x )+40 \ln \relax (2) \ln \relax (x )+\frac {16 x \ln \relax (2)}{5}+8 x \ln \left (x^{2}\right )+100 \ln \relax (x ) \ln \left (x^{2}\right )-100 \ln \relax (x )^{2}\)	\(46\)
risch	\(\frac {16 x^{2}}{25}+\frac {8 x}{5}+20 \ln \relax (x )+40 \ln \relax (2) \ln \relax (x )+\frac {16 x \ln \relax (2)}{5}+8 x \ln \left (x^{2}\right )+100 \ln \relax (x ) \ln \left (x^{2}\right )-100 \ln \relax (x )^{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((200*x+2500)*ln(x^2)+2*(40*x+500)*ln(2)+32*x^2+440*x+500)/x,x,method=_RETURNVERBOSE)

[Out]

(8/5+16/5*ln(2))*x+(10+20*ln(2))*ln(x^2)+16/25*x^2+25*ln(x^2)^2+8*x*ln(x^2)

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 39, normalized size = 2.29 \begin {gather*} \frac {16}{25} \, x^{2} + \frac {16}{5} \, x \log \relax (2) + 8 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + 40 \, \log \relax (2) \log \relax (x) + \frac {8}{5} \, x + 20 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((200*x+2500)*log(x^2)+2*(40*x+500)*log(2)+32*x^2+440*x+500)/x,x, algorithm="maxima")

[Out]

16/25*x^2 + 16/5*x*log(2) + 8*x*log(x^2) + 25*log(x^2)^2 + 40*log(2)*log(x) + 8/5*x + 20*log(x)

________________________________________________________________________________________

mupad [B]  time = 3.17, size = 27, normalized size = 1.59 \begin {gather*} \frac {\left (4\,x+25\,\ln \left (x^2\right )\right )\,\left (4\,x+25\,\ln \left (x^2\right )+20\,\ln \relax (2)+10\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((88*x)/5 + (2*log(2)*(40*x + 500))/25 + (32*x^2)/25 + (log(x^2)*(200*x + 2500))/25 + 20)/x,x)

[Out]

((4*x + 25*log(x^2))*(4*x + 25*log(x^2) + 20*log(2) + 10))/25

________________________________________________________________________________________

sympy [B]  time = 0.17, size = 44, normalized size = 2.59 \begin {gather*} \frac {16 x^{2}}{25} + 8 x \log {\left (x^{2} \right )} + \frac {x \left (40 + 80 \log {\relax (2 )}\right )}{25} + 20 \left (1 + 2 \log {\relax (2 )}\right ) \log {\relax (x )} + 25 \log {\left (x^{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((200*x+2500)*ln(x**2)+2*(40*x+500)*ln(2)+32*x**2+440*x+500)/x,x)

[Out]

16*x**2/25 + 8*x*log(x**2) + x*(40 + 80*log(2))/25 + 20*(1 + 2*log(2))*log(x) + 25*log(x**2)**2

________________________________________________________________________________________