3.2230 \(\int \frac {1}{(1+2 x)^2 (2+3 x+5 x^2)^3} \, dx\)

Optimal. Leaf size=114 \[ \frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}+\frac {5820 x+6427}{47089 (2 x+1) \left (5 x^2+3 x+2\right )}-\frac {192 \log \left (5 x^2+3 x+2\right )}{2401}-\frac {51516}{329623 (2 x+1)}+\frac {384 \log (2 x+1)}{2401}-\frac {1065012 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{2307361 \sqrt {31}} \]

[Out]

-51516/329623/(1+2*x)+1/434*(37+20*x)/(1+2*x)/(5*x^2+3*x+2)^2+1/47089*(6427+5820*x)/(1+2*x)/(5*x^2+3*x+2)+384/
2401*ln(1+2*x)-192/2401*ln(5*x^2+3*x+2)-1065012/71528191*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {740, 822, 800, 634, 618, 204, 628} \[ \frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}+\frac {5820 x+6427}{47089 (2 x+1) \left (5 x^2+3 x+2\right )}-\frac {192 \log \left (5 x^2+3 x+2\right )}{2401}-\frac {51516}{329623 (2 x+1)}+\frac {384 \log (2 x+1)}{2401}-\frac {1065012 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{2307361 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + 2*x)^2*(2 + 3*x + 5*x^2)^3),x]

[Out]

-51516/(329623*(1 + 2*x)) + (37 + 20*x)/(434*(1 + 2*x)*(2 + 3*x + 5*x^2)^2) + (6427 + 5820*x)/(47089*(1 + 2*x)
*(2 + 3*x + 5*x^2)) - (1065012*ArcTan[(3 + 10*x)/Sqrt[31]])/(2307361*Sqrt[31]) + (384*Log[1 + 2*x])/2401 - (19
2*Log[2 + 3*x + 5*x^2])/2401

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), 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] && IntegerQ[m]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx &=\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {1}{434} \int \frac {382+160 x}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}+\frac {\int \frac {74796+46560 x}{(1+2 x)^2 \left (2+3 x+5 x^2\right )} \, dx}{94178}\\ &=\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}+\frac {\int \left (\frac {206064}{7 (1+2 x)^2}+\frac {1476096}{49 (1+2 x)}-\frac {12 (181007+307520 x)}{49 \left (2+3 x+5 x^2\right )}\right ) \, dx}{94178}\\ &=-\frac {51516}{329623 (1+2 x)}+\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}+\frac {384 \log (1+2 x)}{2401}-\frac {6 \int \frac {181007+307520 x}{2+3 x+5 x^2} \, dx}{2307361}\\ &=-\frac {51516}{329623 (1+2 x)}+\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}+\frac {384 \log (1+2 x)}{2401}-\frac {192 \int \frac {3+10 x}{2+3 x+5 x^2} \, dx}{2401}-\frac {532506 \int \frac {1}{2+3 x+5 x^2} \, dx}{2307361}\\ &=-\frac {51516}{329623 (1+2 x)}+\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}+\frac {384 \log (1+2 x)}{2401}-\frac {192 \log \left (2+3 x+5 x^2\right )}{2401}+\frac {1065012 \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )}{2307361}\\ &=-\frac {51516}{329623 (1+2 x)}+\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}-\frac {1065012 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{2307361 \sqrt {31}}+\frac {384 \log (1+2 x)}{2401}-\frac {192 \log \left (2+3 x+5 x^2\right )}{2401}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 98, normalized size = 0.86 \[ \frac {4 \left (-\frac {47089 (270 x-43)}{8 \left (5 x^2+3 x+2\right )^2}-\frac {217 (51910 x-15179)}{4 \left (5 x^2+3 x+2\right )}-1429968 \log \left (4 \left (5 x^2+3 x+2\right )\right )-\frac {1668296}{2 x+1}+2859936 \log (2 x+1)-266253 \sqrt {31} \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )\right )}{71528191} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 + 2*x)^2*(2 + 3*x + 5*x^2)^3),x]

[Out]

(4*(-1668296/(1 + 2*x) - (47089*(-43 + 270*x))/(8*(2 + 3*x + 5*x^2)^2) - (217*(-15179 + 51910*x))/(4*(2 + 3*x
+ 5*x^2)) - 266253*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]] + 2859936*Log[1 + 2*x] - 1429968*Log[4*(2 + 3*x + 5*x^
2)]))/71528191

________________________________________________________________________________________

fricas [A]  time = 1.47, size = 161, normalized size = 1.41 \[ -\frac {558948600 \, x^{4} + 582332520 \, x^{3} + 2130024 \, \sqrt {31} {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 497710766 \, x^{2} + 11439744 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) - 22879488 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \log \left (2 \, x + 1\right ) + 167764870 \, x + 38185273}{143056382 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+2*x)^2/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

-1/143056382*(558948600*x^4 + 582332520*x^3 + 2130024*sqrt(31)*(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)*
arctan(1/31*sqrt(31)*(10*x + 3)) + 497710766*x^2 + 11439744*(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)*log
(5*x^2 + 3*x + 2) - 22879488*(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)*log(2*x + 1) + 167764870*x + 38185
273)/(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 108, normalized size = 0.95 \[ -\frac {1065012}{71528191} \, \sqrt {31} \arctan \left (-\frac {1}{31} \, \sqrt {31} {\left (\frac {7}{2 \, x + 1} - 2\right )}\right ) - \frac {32}{343 \, {\left (2 \, x + 1\right )}} + \frac {4 \, {\left (\frac {1178375}{2 \, x + 1} - \frac {2320190}{{\left (2 \, x + 1\right )}^{2}} + \frac {87843}{{\left (2 \, x + 1\right )}^{3}} - 1304250\right )}}{2307361 \, {\left (\frac {4}{2 \, x + 1} - \frac {7}{{\left (2 \, x + 1\right )}^{2}} - 5\right )}^{2}} - \frac {192}{2401} \, \log \left (-\frac {4}{2 \, x + 1} + \frac {7}{{\left (2 \, x + 1\right )}^{2}} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+2*x)^2/(5*x^2+3*x+2)^3,x, algorithm="giac")

[Out]

-1065012/71528191*sqrt(31)*arctan(-1/31*sqrt(31)*(7/(2*x + 1) - 2)) - 32/343/(2*x + 1) + 4/2307361*(1178375/(2
*x + 1) - 2320190/(2*x + 1)^2 + 87843/(2*x + 1)^3 - 1304250)/(4/(2*x + 1) - 7/(2*x + 1)^2 - 5)^2 - 192/2401*lo
g(-4/(2*x + 1) + 7/(2*x + 1)^2 + 5)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 77, normalized size = 0.68 \[ -\frac {1065012 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{71528191}+\frac {384 \ln \left (2 x +1\right )}{2401}-\frac {192 \ln \left (5 x^{2}+3 x +2\right )}{2401}-\frac {32}{343 \left (2 x +1\right )}-\frac {25 \left (\frac {72674}{961} x^{3}+\frac {111769}{4805} x^{2}+\frac {613046}{24025} x -\frac {490329}{48050}\right )}{2401 \left (5 x^{2}+3 x +2\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x+1)^2/(5*x^2+3*x+2)^3,x)

[Out]

-32/343/(2*x+1)+384/2401*ln(2*x+1)-25/2401*(72674/961*x^3+111769/4805*x^2+613046/24025*x-490329/48050)/(5*x^2+
3*x+2)^2-192/2401*ln(5*x^2+3*x+2)-1065012/71528191*31^(1/2)*arctan(1/31*(10*x+3)*31^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.82, size = 87, normalized size = 0.76 \[ -\frac {1065012}{71528191} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {2575800 \, x^{4} + 2683560 \, x^{3} + 2293598 \, x^{2} + 773110 \, x + 175969}{659246 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )}} - \frac {192}{2401} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {384}{2401} \, \log \left (2 \, x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+2*x)^2/(5*x^2+3*x+2)^3,x, algorithm="maxima")

[Out]

-1065012/71528191*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 1/659246*(2575800*x^4 + 2683560*x^3 + 2293598*x^
2 + 773110*x + 175969)/(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4) - 192/2401*log(5*x^2 + 3*x + 2) + 384/24
01*log(2*x + 1)

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 93, normalized size = 0.82 \[ \frac {384\,\ln \left (x+\frac {1}{2}\right )}{2401}-\frac {\frac {25758\,x^4}{329623}+\frac {134178\,x^3}{1648115}+\frac {1146799\,x^2}{16481150}+\frac {77311\,x}{3296230}+\frac {175969}{32962300}}{x^5+\frac {17\,x^4}{10}+\frac {44\,x^3}{25}+\frac {53\,x^2}{50}+\frac {2\,x}{5}+\frac {2}{25}}+\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {192}{2401}+\frac {\sqrt {31}\,532506{}\mathrm {i}}{71528191}\right )-\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {192}{2401}+\frac {\sqrt {31}\,532506{}\mathrm {i}}{71528191}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x + 1)^2*(3*x + 5*x^2 + 2)^3),x)

[Out]

(384*log(x + 1/2))/2401 - ((77311*x)/3296230 + (1146799*x^2)/16481150 + (134178*x^3)/1648115 + (25758*x^4)/329
623 + 175969/32962300)/((2*x)/5 + (53*x^2)/50 + (44*x^3)/25 + (17*x^4)/10 + x^5 + 2/25) + log(x - (31^(1/2)*1i
)/10 + 3/10)*((31^(1/2)*532506i)/71528191 - 192/2401) - log(x + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*532506i)/7
1528191 + 192/2401)

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 102, normalized size = 0.89 \[ \frac {- 2575800 x^{4} - 2683560 x^{3} - 2293598 x^{2} - 773110 x - 175969}{32962300 x^{5} + 56035910 x^{4} + 58013648 x^{3} + 34940038 x^{2} + 13184920 x + 2636984} + \frac {384 \log {\left (x + \frac {1}{2} \right )}}{2401} - \frac {192 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{2401} - \frac {1065012 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{71528191} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+2*x)**2/(5*x**2+3*x+2)**3,x)

[Out]

(-2575800*x**4 - 2683560*x**3 - 2293598*x**2 - 773110*x - 175969)/(32962300*x**5 + 56035910*x**4 + 58013648*x*
*3 + 34940038*x**2 + 13184920*x + 2636984) + 384*log(x + 1/2)/2401 - 192*log(x**2 + 3*x/5 + 2/5)/2401 - 106501
2*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/71528191

________________________________________________________________________________________