3.54.85 \(\int \frac {-45 x+x^2+(-250-50 x) \log (\frac {5+x}{5})}{10+2 x} \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{4} x \left (x-100 \log \left (\frac {5+x}{5}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 27, normalized size of antiderivative = 1.59, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {6742, 77, 2389, 2295} \begin {gather*} \frac {x^2}{4}-25 (x+5) \log \left (\frac {x+5}{5}\right )+125 \log (x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45*x + x^2 + (-250 - 50*x)*Log[(5 + x)/5])/(10 + 2*x),x]

[Out]

x^2/4 - 25*(5 + x)*Log[(5 + x)/5] + 125*Log[5 + x]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, 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 \left (\frac {(-45+x) x}{2 (5+x)}-25 \log \left (1+\frac {x}{5}\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {(-45+x) x}{5+x} \, dx-25 \int \log \left (1+\frac {x}{5}\right ) \, dx\\ &=\frac {1}{2} \int \left (-50+x+\frac {250}{5+x}\right ) \, dx-125 \operatorname {Subst}\left (\int \log (x) \, dx,x,1+\frac {x}{5}\right )\\ &=\frac {x^2}{4}-25 (5+x) \log \left (\frac {5+x}{5}\right )+125 \log (5+x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.01, size = 39, normalized size = 2.29 \begin {gather*} 25 x-\frac {55 (5+x)}{2}+\frac {1}{4} (5+x)^2-25 (5+x) \log \left (\frac {5+x}{5}\right )+125 \log (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45*x + x^2 + (-250 - 50*x)*Log[(5 + x)/5])/(10 + 2*x),x]

[Out]

25*x - (55*(5 + x))/2 + (5 + x)^2/4 - 25*(5 + x)*Log[(5 + x)/5] + 125*Log[5 + x]

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 15, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, x^{2} - 25 \, x \log \left (\frac {1}{5} \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x-250)*log(1+1/5*x)+x^2-45*x)/(2*x+10),x, algorithm="fricas")

[Out]

1/4*x^2 - 25*x*log(1/5*x + 1)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 15, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, x^{2} - 25 \, x \log \left (\frac {1}{5} \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x-250)*log(1+1/5*x)+x^2-45*x)/(2*x+10),x, algorithm="giac")

[Out]

1/4*x^2 - 25*x*log(1/5*x + 1)

________________________________________________________________________________________

maple [A]  time = 0.19, size = 16, normalized size = 0.94

method	result	size
norman	\(\frac {x^{2}}{4}-25 \ln \left (1+\frac {x}{5}\right ) x\)	\(16\)
risch	\(\frac {x^{2}}{4}-25 \ln \left (1+\frac {x}{5}\right ) x\)	\(16\)
derivativedivides	\(-125 \left (1+\frac {x}{5}\right ) \ln \left (1+\frac {x}{5}\right )-\frac {25}{2}-\frac {5 x}{2}+\frac {25 \left (1+\frac {x}{5}\right )^{2}}{4}+125 \ln \left (1+\frac {x}{5}\right )\)	\(36\)
default	\(-125 \left (1+\frac {x}{5}\right ) \ln \left (1+\frac {x}{5}\right )-\frac {25}{2}-\frac {5 x}{2}+\frac {25 \left (1+\frac {x}{5}\right )^{2}}{4}+125 \ln \left (1+\frac {x}{5}\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-50*x-250)*ln(1+1/5*x)+x^2-45*x)/(2*x+10),x,method=_RETURNVERBOSE)

[Out]

1/4*x^2-25*ln(1+1/5*x)*x

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 38, normalized size = 2.24 \begin {gather*} \frac {1}{4} \, x^{2} + 125 \, \log \relax (5) \log \left (x + 5\right ) - 125 \, \log \left (x + 5\right )^{2} - 25 \, {\left (x - 5 \, \log \left (x + 5\right )\right )} \log \left (\frac {1}{5} \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x-250)*log(1+1/5*x)+x^2-45*x)/(2*x+10),x, algorithm="maxima")

[Out]

1/4*x^2 + 125*log(5)*log(x + 5) - 125*log(x + 5)^2 - 25*(x - 5*log(x + 5))*log(1/5*x + 1)

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 15, normalized size = 0.88 \begin {gather*} \frac {x^2}{4}-25\,x\,\ln \left (\frac {x}{5}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(45*x + log(x/5 + 1)*(50*x + 250) - x^2)/(2*x + 10),x)

[Out]

x^2/4 - 25*x*log(x/5 + 1)

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 14, normalized size = 0.82 \begin {gather*} \frac {x^{2}}{4} - 25 x \log {\left (\frac {x}{5} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x-250)*ln(1+1/5*x)+x**2-45*x)/(2*x+10),x)

[Out]

x**2/4 - 25*x*log(x/5 + 1)

________________________________________________________________________________________