\(\int \frac {-50 x^2-x^3+(75 x^2+3 x^3) \log (\frac {1}{25} (25 x+x^2))}{800+32 x} \, dx\) [2427]

Optimal result

Integrand size = 43, antiderivative size = 19 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {1}{32} x^3 \left (-1+\log \left (x+\frac {x^2}{25}\right )\right ) \]

[Out]

1/32*x^3*(ln(1/25*x^2+x)-1)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6820, 12, 6874, 78, 2581, 30, 45} \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {1}{32} x^3 \log \left (\frac {1}{25} x (x+25)\right )-\frac {x^3}{32} \]

[In]

Int[(-50*x^2 - x^3 + (75*x^2 + 3*x^3)*Log[(25*x + x^2)/25])/(800 + 32*x),x]

[Out]

-1/32*x^3 + (x^3*Log[(x*(25 + x))/25])/32

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

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 78

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 2581

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(h*(m + 1))), x] + (-Dist[b*p*(r/(h
*(m + 1))), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[d*q*(r/(h*(m + 1))), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (-50-x+3 (25+x) \log \left (\frac {1}{25} x (25+x)\right )\right )}{32 (25+x)} \, dx \\ & = \frac {1}{32} \int \frac {x^2 \left (-50-x+3 (25+x) \log \left (\frac {1}{25} x (25+x)\right )\right )}{25+x} \, dx \\ & = \frac {1}{32} \int \left (-\frac {x^2 (50+x)}{25+x}+3 x^2 \log \left (\frac {1}{25} x (25+x)\right )\right ) \, dx \\ & = -\left (\frac {1}{32} \int \frac {x^2 (50+x)}{25+x} \, dx\right )+\frac {3}{32} \int x^2 \log \left (\frac {1}{25} x (25+x)\right ) \, dx \\ & = \frac {1}{32} x^3 \log \left (\frac {1}{25} x (25+x)\right )-\frac {\int x^2 \, dx}{32}-\frac {1}{32} \int \frac {x^3}{25+x} \, dx-\frac {1}{32} \int \left (-625+25 x+x^2+\frac {15625}{25+x}\right ) \, dx \\ & = \frac {625 x}{32}-\frac {25 x^2}{64}-\frac {x^3}{48}-\frac {15625}{32} \log (25+x)+\frac {1}{32} x^3 \log \left (\frac {1}{25} x (25+x)\right )-\frac {1}{32} \int \left (625-25 x+x^2-\frac {15625}{25+x}\right ) \, dx \\ & = -\frac {x^3}{32}+\frac {1}{32} x^3 \log \left (\frac {1}{25} x (25+x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {1}{32} \left (-x^3+x^3 \log \left (\frac {1}{25} x (25+x)\right )\right ) \]

[In]

Integrate[(-50*x^2 - x^3 + (75*x^2 + 3*x^3)*Log[(25*x + x^2)/25])/(800 + 32*x),x]

[Out]

(-x^3 + x^3*Log[(x*(25 + x))/25])/32

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
norman	\(-\frac {x^{3}}{32}+\frac {x^{3} \ln \left (\frac {1}{25} x^{2}+x \right )}{32}\)	\(20\)
risch	\(-\frac {x^{3}}{32}+\frac {x^{3} \ln \left (\frac {1}{25} x^{2}+x \right )}{32}\)	\(20\)
parallelrisch	\(-\frac {x^{3}}{32}+\frac {x^{3} \ln \left (\frac {1}{25} x^{2}+x \right )}{32}\)	\(20\)
default	\(-\frac {x^{3} \ln \left (5\right )}{16}+\frac {x^{3} \ln \left (x^{2}+25 x \right )}{32}-\frac {x^{3}}{32}\)	\(27\)
parts	\(-\frac {x^{3} \ln \left (5\right )}{16}+\frac {x^{3} \ln \left (x^{2}+25 x \right )}{32}-\frac {x^{3}}{32}\)	\(27\)

[In]

int(((3*x^3+75*x^2)*ln(1/25*x^2+x)-x^3-50*x^2)/(32*x+800),x,method=_RETURNVERBOSE)

[Out]

-1/32*x^3+1/32*x^3*ln(1/25*x^2+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {1}{32} \, x^{3} \log \left (\frac {1}{25} \, x^{2} + x\right ) - \frac {1}{32} \, x^{3} \]

[In]

integrate(((3*x^3+75*x^2)*log(1/25*x^2+x)-x^3-50*x^2)/(32*x+800),x, algorithm="fricas")

[Out]

1/32*x^3*log(1/25*x^2 + x) - 1/32*x^3

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {x^{3} \log {\left (\frac {x^{2}}{25} + x \right )}}{32} - \frac {x^{3}}{32} \]

[In]

integrate(((3*x**3+75*x**2)*ln(1/25*x**2+x)-x**3-50*x**2)/(32*x+800),x)

[Out]

x**3*log(x**2/25 + x)/32 - x**3/32

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (15) = 30\).

Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=-\frac {1}{48} \, x^{3} {\left (3 \, \log \left (5\right ) + 1\right )} + \frac {1}{32} \, x^{3} \log \left (x\right ) - \frac {1}{96} \, x^{3} + \frac {1}{32} \, {\left (x^{3} + 15625\right )} \log \left (x + 25\right ) - \frac {15625}{32} \, \log \left (x + 25\right ) \]

[In]

integrate(((3*x^3+75*x^2)*log(1/25*x^2+x)-x^3-50*x^2)/(32*x+800),x, algorithm="maxima")

[Out]

-1/48*x^3*(3*log(5) + 1) + 1/32*x^3*log(x) - 1/96*x^3 + 1/32*(x^3 + 15625)*log(x + 25) - 15625/32*log(x + 25)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {1}{32} \, x^{3} \log \left (\frac {1}{25} \, x^{2} + x\right ) - \frac {1}{32} \, x^{3} \]

[In]

integrate(((3*x^3+75*x^2)*log(1/25*x^2+x)-x^3-50*x^2)/(32*x+800),x, algorithm="giac")

[Out]

1/32*x^3*log(1/25*x^2 + x) - 1/32*x^3

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-50 x^2-x^3+\left (75 x^2+3 x^3\right ) \log \left (\frac {1}{25} \left (25 x+x^2\right )\right )}{800+32 x} \, dx=\frac {x^3\,\left (\ln \left (\frac {x^2}{25}+x\right )-1\right )}{32} \]

[In]

int(-(50*x^2 - log(x + x^2/25)*(75*x^2 + 3*x^3) + x^3)/(32*x + 800),x)

[Out]

(x^3*(log(x + x^2/25) - 1))/32