\(\int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log (x^2)}{25 x} \, dx\) [4696]

Optimal result

Integrand size = 35, antiderivative size = 17 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\left (1+\frac {4 x}{5}+\log (4)+5 \log \left (x^2\right )\right )^2 \]

[Out]

(1+5*ln(x^2)+4/5*x+2*ln(2))^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(43\) vs. \(2(17)=34\).

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 14, 77, 2388, 2338, 2332} \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {16 x^2}{25}+25 \log ^2\left (x^2\right )+8 x \log \left (x^2\right )-16 x+\frac {8}{5} x (11+\log (4))+20 (1+\log (4)) \log (x) \]

[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 77

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 2332

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

Rule 2338

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 2388

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{align*} \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{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(45\) vs. \(2(17)=34\).

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.65 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {8 x}{5}+\frac {16 x^2}{25}+\frac {8}{5} x \log (4)+20 \log (x)+20 \log (4) \log (x)+8 x \log \left (x^2\right )+25 \log ^2\left (x^2\right ) \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(40\) vs. \(2(17)=34\).

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.41

method	result	size
norman	\(\left (\frac {8}{5}+\frac {16 \ln \left (2\right )}{5}\right ) x +\left (10+20 \ln \left (2\right )\right ) \ln \left (x^{2}\right )+\frac {16 x^{2}}{25}+25 \ln \left (x^{2}\right )^{2}+8 x \ln \left (x^{2}\right )\)	\(41\)
parallelrisch	\(\frac {16 x \ln \left (2\right )}{5}+20 \ln \left (2\right ) \ln \left (x^{2}\right )+\frac {16 x^{2}}{25}+8 x \ln \left (x^{2}\right )+25 \ln \left (x^{2}\right )^{2}+\frac {8 x}{5}+10 \ln \left (x^{2}\right )\)	\(44\)
risch	\(\frac {16 x \ln \left (2\right )}{5}+\frac {16 x^{2}}{25}+\frac {8 x}{5}+40 \ln \left (2\right ) \ln \left (x \right )+20 \ln \left (x \right )+8 x \ln \left (x^{2}\right )+100 \ln \left (x \right ) \ln \left (x^{2}\right )-100 \ln \left (x \right )^{2}\)	\(46\)
parts	\(\frac {16 x \ln \left (2\right )}{5}+\frac {16 x^{2}}{25}+\frac {8 x}{5}+40 \ln \left (2\right ) \ln \left (x \right )+20 \ln \left (x \right )+8 x \ln \left (x^{2}\right )+100 \ln \left (x \right ) \ln \left (x^{2}\right )-100 \ln \left (x \right )^{2}\)	\(46\)
default	\(\frac {16 x^{2}}{25}+\frac {8 x}{5}+20 \ln \left (x \right )+\frac {8 \ln \left (2\right ) \left (2 x +25 \ln \left (x \right )\right )}{5}+8 x \ln \left (x^{2}\right )+100 \ln \left (x \right ) \ln \left (x^{2}\right )-100 \ln \left (x \right )^{2}\)	\(47\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {16}{25} \, x^{2} + \frac {16}{5} \, x \log \left (2\right ) + 2 \, {\left (4 \, x + 10 \, \log \left (2\right ) + 5\right )} \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + \frac {8}{5} \, x \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.59 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {16 x^{2}}{25} + 8 x \log {\left (x^{2} \right )} + \frac {x \left (40 + 80 \log {\left (2 \right )}\right )}{25} + 20 \cdot \left (1 + 2 \log {\left (2 \right )}\right ) \log {\left (x \right )} + 25 \log {\left (x^{2} \right )}^{2} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.29 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {16}{25} \, x^{2} + \frac {16}{5} \, x \log \left (2\right ) + 8 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + 40 \, \log \left (2\right ) \log \left (x\right ) + \frac {8}{5} \, x + 20 \, \log \left (x\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.35 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {16}{25} \, x^{2} + \frac {8}{5} \, x {\left (2 \, \log \left (2\right ) + 1\right )} + 8 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )^{2} + 20 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 10.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {500+440 x+32 x^2+(500+40 x) \log (4)+(2500+200 x) \log \left (x^2\right )}{25 x} \, dx=\frac {\left (4\,x+25\,\ln \left (x^2\right )\right )\,\left (4\,x+25\,\ln \left (x^2\right )+20\,\ln \left (2\right )+10\right )}{25} \]

[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