\(\int \frac {(5-x) (3+2 x)^5}{(2+5 x+3 x^2)^3} \, dx\) [2396]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {32 x}{27}-\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}-1085 \log (1+x)+\frac {29375}{27} \log (2+3 x) \]

[Out]

-32/27*x+1/486*(-56041-57499*x)/(3*x^2+5*x+2)^2+1/486*(398585+502254*x)/(3*x^2+5*x+2)-1085*ln(1+x)+29375/27*ln
(2+3*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {830, 1674, 1671, 646, 31} \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {502254 x+398585}{486 \left (3 x^2+5 x+2\right )}-\frac {57499 x+56041}{486 \left (3 x^2+5 x+2\right )^2}-\frac {32 x}{27}-1085 \log (x+1)+\frac {29375}{27} \log (3 x+2) \]

[In]

Int[((5 - x)*(3 + 2*x)^5)/(2 + 5*x + 3*x^2)^3,x]

[Out]

(-32*x)/27 - (56041 + 57499*x)/(486*(2 + 5*x + 3*x^2)^2) + (398585 + 502254*x)/(486*(2 + 5*x + 3*x^2)) - 1085*
Log[1 + x] + (29375*Log[2 + 3*x])/27

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 830

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(a
+ b*x + c*x^2)^p*ExpandIntegrand[(d + e*x)^m*(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2
- 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[p, -1] && IGtQ[m, 0] && RationalQ[a, b, c, d, e, f, g]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1674

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {13}{2} (3+2 x)^5-\frac {1}{2} (3+2 x)^6}{\left (2+5 x+3 x^2\right )^3} \, dx \\ & = -\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}-\frac {1}{2} \int \frac {-\frac {72539}{243}-\frac {87920 x}{81}-\frac {9824 x^2}{27}+\frac {160 x^3}{9}+\frac {64 x^4}{3}}{\left (2+5 x+3 x^2\right )^2} \, dx \\ & = -\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}+\frac {1}{2} \int \frac {\frac {58942}{27}+\frac {160 x}{27}-\frac {64 x^2}{9}}{2+5 x+3 x^2} \, dx \\ & = -\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}+\frac {1}{2} \int \left (-\frac {64}{27}+\frac {10 (1969+16 x)}{9 \left (2+5 x+3 x^2\right )}\right ) \, dx \\ & = -\frac {32 x}{27}-\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}+\frac {5}{9} \int \frac {1969+16 x}{2+5 x+3 x^2} \, dx \\ & = -\frac {32 x}{27}-\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}-3255 \int \frac {1}{3+3 x} \, dx+\frac {29375}{9} \int \frac {1}{2+3 x} \, dx \\ & = -\frac {32 x}{27}-\frac {56041+57499 x}{486 \left (2+5 x+3 x^2\right )^2}+\frac {398585+502254 x}{486 \left (2+5 x+3 x^2\right )}-1085 \log (1+x)+\frac {29375}{27} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.27 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {245891+973450 x+1221179 x^2+486510 x^3-8352 x^4-1728 x^5+176250 \left (2+5 x+3 x^2\right )^2 \log (-4-6 x)-175770 \left (2+5 x+3 x^2\right )^2 \log (-2 (1+x))}{162 \left (2+5 x+3 x^2\right )^2} \]

[In]

Integrate[((5 - x)*(3 + 2*x)^5)/(2 + 5*x + 3*x^2)^3,x]

[Out]

(245891 + 973450*x + 1221179*x^2 + 486510*x^3 - 8352*x^4 - 1728*x^5 + 176250*(2 + 5*x + 3*x^2)^2*Log[-4 - 6*x]
- 175770*(2 + 5*x + 3*x^2)^2*Log[-2*(1 + x)])/(162*(2 + 5*x + 3*x^2)^2)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {32 x}{27}+\frac {\frac {9301}{3} x^{3}+\frac {1235675}{162} x^{2}+\frac {489989}{81} x +\frac {247043}{162}}{\left (3 x^{2}+5 x +2\right )^{2}}-1085 \ln \left (1+x \right )+\frac {29375 \ln \left (2+3 x \right )}{27}\)	\(48\)
norman	\(\frac {3175 x^{3}+\frac {165335}{27} x +\frac {418505}{54} x^{2}-\frac {32}{3} x^{5}+\frac {83201}{54}}{\left (3 x^{2}+5 x +2\right )^{2}}-1085 \ln \left (1+x \right )+\frac {29375 \ln \left (2+3 x \right )}{27}\)	\(49\)
default	\(-\frac {32 x}{27}-\frac {53125}{162 \left (2+3 x \right )^{2}}+\frac {6250}{9 \left (2+3 x \right )}+\frac {29375 \ln \left (2+3 x \right )}{27}+\frac {3}{\left (1+x \right )^{2}}+\frac {113}{1+x}-1085 \ln \left (1+x \right )\)	\(51\)
parallelrisch	\(-\frac {2109240 \ln \left (1+x \right ) x^{4}-2115000 \ln \left (x +\frac {2}{3}\right ) x^{4}+2304 x^{5}+7030800 \ln \left (1+x \right ) x^{3}-7050000 \ln \left (x +\frac {2}{3}\right ) x^{3}+748809 x^{4}+8671320 \ln \left (1+x \right ) x^{2}-8695000 \ln \left (x +\frac {2}{3}\right ) x^{2}+1810230 x^{3}+4687200 \ln \left (1+x \right ) x -4700000 \ln \left (x +\frac {2}{3}\right ) x +1404417 x^{2}+937440 \ln \left (1+x \right )-940000 \ln \left (x +\frac {2}{3}\right )+341340 x}{216 \left (3 x^{2}+5 x +2\right )^{2}}\)	\(119\)

[In]

int((5-x)*(3+2*x)^5/(3*x^2+5*x+2)^3,x,method=_RETURNVERBOSE)

[Out]

-32/27*x+9*(9301/27*x^3+1235675/1458*x^2+489989/729*x+247043/1458)/(3*x^2+5*x+2)^2-1085*ln(1+x)+29375/27*ln(2+
3*x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.61 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {1728 \, x^{5} + 5760 \, x^{4} - 495150 \, x^{3} - 1231835 \, x^{2} - 176250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 175770 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) - 979210 \, x - 247043}{162 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]

[In]

integrate((5-x)*(3+2*x)^5/(3*x^2+5*x+2)^3,x, algorithm="fricas")

[Out]

-1/162*(1728*x^5 + 5760*x^4 - 495150*x^3 - 1231835*x^2 - 176250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x +
2) + 175770*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(x + 1) - 979210*x - 247043)/(9*x^4 + 30*x^3 + 37*x^2 + 2
0*x + 4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=- \frac {32 x}{27} - \frac {- 502254 x^{3} - 1235675 x^{2} - 979978 x - 247043}{1458 x^{4} + 4860 x^{3} + 5994 x^{2} + 3240 x + 648} + \frac {29375 \log {\left (x + \frac {2}{3} \right )}}{27} - 1085 \log {\left (x + 1 \right )} \]

[In]

integrate((5-x)*(3+2*x)**5/(3*x**2+5*x+2)**3,x)

[Out]

-32*x/27 - (-502254*x**3 - 1235675*x**2 - 979978*x - 247043)/(1458*x**4 + 4860*x**3 + 5994*x**2 + 3240*x + 648
) + 29375*log(x + 2/3)/27 - 1085*log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {32}{27} \, x + \frac {502254 \, x^{3} + 1235675 \, x^{2} + 979978 \, x + 247043}{162 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + \frac {29375}{27} \, \log \left (3 \, x + 2\right ) - 1085 \, \log \left (x + 1\right ) \]

[In]

integrate((5-x)*(3+2*x)^5/(3*x^2+5*x+2)^3,x, algorithm="maxima")

[Out]

-32/27*x + 1/162*(502254*x^3 + 1235675*x^2 + 979978*x + 247043)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 29375/2
7*log(3*x + 2) - 1085*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.77 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {32}{27} \, x + \frac {502254 \, x^{3} + 1235675 \, x^{2} + 979978 \, x + 247043}{162 \, {\left (3 \, x + 2\right )}^{2} {\left (x + 1\right )}^{2}} + \frac {29375}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 1085 \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate((5-x)*(3+2*x)^5/(3*x^2+5*x+2)^3,x, algorithm="giac")

[Out]

-32/27*x + 1/162*(502254*x^3 + 1235675*x^2 + 979978*x + 247043)/((3*x + 2)^2*(x + 1)^2) + 29375/27*log(abs(3*x
+ 2)) - 1085*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {(5-x) (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {29375\,\ln \left (x+\frac {2}{3}\right )}{27}-1085\,\ln \left (x+1\right )-\frac {32\,x}{27}+\frac {\frac {9301\,x^3}{27}+\frac {1235675\,x^2}{1458}+\frac {489989\,x}{729}+\frac {247043}{1458}}{x^4+\frac {10\,x^3}{3}+\frac {37\,x^2}{9}+\frac {20\,x}{9}+\frac {4}{9}} \]

[In]

int(-((2*x + 3)^5*(x - 5))/(5*x + 3*x^2 + 2)^3,x)

[Out]

(29375*log(x + 2/3))/27 - 1085*log(x + 1) - (32*x)/27 + ((489989*x)/729 + (1235675*x^2)/1458 + (9301*x^3)/27 +
247043/1458)/((20*x)/9 + (37*x^2)/9 + (10*x^3)/3 + x^4 + 4/9)