3.2229 \(\int \frac {1}{(1+2 x) (2+3 x+5 x^2)^3} \, dx\)

Optimal. Leaf size=89 \[ \frac {20 x+37}{434 \left (5 x^2+3 x+2\right )^2}+\frac {2 (2290 x+2609)}{47089 \left (5 x^2+3 x+2\right )}-\frac {16}{343} \log \left (5 x^2+3 x+2\right )+\frac {32}{343} \log (2 x+1)+\frac {125624 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{329623 \sqrt {31}} \]

[Out]

1/434*(37+20*x)/(5*x^2+3*x+2)^2+2/47089*(2609+2290*x)/(5*x^2+3*x+2)+32/343*ln(1+2*x)-16/343*ln(5*x^2+3*x+2)+12
5624/10218313*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 89, 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} \[ \frac {20 x+37}{434 \left (5 x^2+3 x+2\right )^2}+\frac {2 (2290 x+2609)}{47089 \left (5 x^2+3 x+2\right )}-\frac {16}{343} \log \left (5 x^2+3 x+2\right )+\frac {32}{343} \log (2 x+1)+\frac {125624 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{329623 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(37 + 20*x)/(434*(2 + 3*x + 5*x^2)^2) + (2*(2609 + 2290*x))/(47089*(2 + 3*x + 5*x^2)) + (125624*ArcTan[(3 + 10
*x)/Sqrt[31]])/(329623*Sqrt[31]) + (32*Log[1 + 2*x])/343 - (16*Log[2 + 3*x + 5*x^2])/343

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) \left (2+3 x+5 x^2\right )^3} \, dx &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {1}{434} \int \frac {308+120 x}{(1+2 x) \left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {\int \frac {39912+18320 x}{(1+2 x) \left (2+3 x+5 x^2\right )} \, dx}{94178}\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {\int \left (\frac {123008}{7 (1+2 x)}-\frac {8 (-4171+38440 x)}{7 \left (2+3 x+5 x^2\right )}\right ) \, dx}{94178}\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {32}{343} \log (1+2 x)-\frac {4 \int \frac {-4171+38440 x}{2+3 x+5 x^2} \, dx}{329623}\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {32}{343} \log (1+2 x)-\frac {16}{343} \int \frac {3+10 x}{2+3 x+5 x^2} \, dx+\frac {62812 \int \frac {1}{2+3 x+5 x^2} \, dx}{329623}\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {32}{343} \log (1+2 x)-\frac {16}{343} \log \left (2+3 x+5 x^2\right )-\frac {125624 \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )}{329623}\\ &=\frac {37+20 x}{434 \left (2+3 x+5 x^2\right )^2}+\frac {2 (2609+2290 x)}{47089 \left (2+3 x+5 x^2\right )}+\frac {125624 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{329623 \sqrt {31}}+\frac {32}{343} \log (1+2 x)-\frac {16}{343} \log \left (2+3 x+5 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 78, normalized size = 0.88 \[ \frac {8 \left (-59582 \log \left (4 \left (5 x^2+3 x+2\right )\right )+\frac {217 \left (45800 x^3+79660 x^2+53968 x+28901\right )}{16 \left (5 x^2+3 x+2\right )^2}+119164 \log (2 x+1)+15703 \sqrt {31} \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )\right )}{10218313} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*((217*(28901 + 53968*x + 79660*x^2 + 45800*x^3))/(16*(2 + 3*x + 5*x^2)^2) + 15703*Sqrt[31]*ArcTan[(3 + 10*x
)/Sqrt[31]] + 119164*Log[1 + 2*x] - 59582*Log[4*(2 + 3*x + 5*x^2)]))/10218313

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fricas [A]  time = 1.14, size = 136, normalized size = 1.53 \[ \frac {9938600 \, x^{3} + 251248 \, \sqrt {31} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 17286220 \, x^{2} - 953312 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) + 1906624 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \log \left (2 \, x + 1\right ) + 11711056 \, x + 6271517}{20436626 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20436626*(9938600*x^3 + 251248*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*arctan(1/31*sqrt(31)*(10*x + 3
)) + 17286220*x^2 - 953312*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*log(5*x^2 + 3*x + 2) + 1906624*(25*x^4 + 30*x
^3 + 29*x^2 + 12*x + 4)*log(2*x + 1) + 11711056*x + 6271517)/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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giac [A]  time = 0.16, size = 68, normalized size = 0.76 \[ \frac {125624}{10218313} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {45800 \, x^{3} + 79660 \, x^{2} + 53968 \, x + 28901}{94178 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}} - \frac {16}{343} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {32}{343} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125624/10218313*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/94178*(45800*x^3 + 79660*x^2 + 53968*x + 28901)/
(5*x^2 + 3*x + 2)^2 - 16/343*log(5*x^2 + 3*x + 2) + 32/343*log(abs(2*x + 1))

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maple [A]  time = 0.06, size = 68, normalized size = 0.76 \[ \frac {125624 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{10218313}+\frac {32 \ln \left (2 x +1\right )}{343}-\frac {16 \ln \left (5 x^{2}+3 x +2\right )}{343}-\frac {25 \left (-\frac {6412}{961} x^{3}-\frac {55762}{4805} x^{2}-\frac {188888}{24025} x -\frac {202307}{48050}\right )}{343 \left (5 x^{2}+3 x +2\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/343*ln(2*x+1)-25/343*(-6412/961*x^3-55762/4805*x^2-188888/24025*x-202307/48050)/(5*x^2+3*x+2)^2-16/343*ln(5
*x^2+3*x+2)+125624/10218313*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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maxima [A]  time = 2.00, size = 77, normalized size = 0.87 \[ \frac {125624}{10218313} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {45800 \, x^{3} + 79660 \, x^{2} + 53968 \, x + 28901}{94178 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} - \frac {16}{343} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {32}{343} \, \log \left (2 \, x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125624/10218313*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/94178*(45800*x^3 + 79660*x^2 + 53968*x + 28901)/
(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 16/343*log(5*x^2 + 3*x + 2) + 32/343*log(2*x + 1)

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mupad [B]  time = 0.13, size = 82, normalized size = 0.92 \[ \frac {32\,\ln \left (x+\frac {1}{2}\right )}{343}-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {16}{343}+\frac {\sqrt {31}\,62812{}\mathrm {i}}{10218313}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {16}{343}+\frac {\sqrt {31}\,62812{}\mathrm {i}}{10218313}\right )+\frac {\frac {916\,x^3}{47089}+\frac {1138\,x^2}{33635}+\frac {26984\,x}{1177225}+\frac {28901}{2354450}}{x^4+\frac {6\,x^3}{5}+\frac {29\,x^2}{25}+\frac {12\,x}{25}+\frac {4}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(32*log(x + 1/2))/343 - log(x - (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*62812i)/10218313 + 16/343) + log(x + (31^(
1/2)*1i)/10 + 3/10)*((31^(1/2)*62812i)/10218313 - 16/343) + ((26984*x)/1177225 + (1138*x^2)/33635 + (916*x^3)/
47089 + 28901/2354450)/((12*x)/25 + (29*x^2)/25 + (6*x^3)/5 + x^4 + 4/25)

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sympy [A]  time = 0.25, size = 90, normalized size = 1.01 \[ \frac {45800 x^{3} + 79660 x^{2} + 53968 x + 28901}{2354450 x^{4} + 2825340 x^{3} + 2731162 x^{2} + 1130136 x + 376712} + \frac {32 \log {\left (x + \frac {1}{2} \right )}}{343} - \frac {16 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{343} + \frac {125624 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{10218313} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(45800*x**3 + 79660*x**2 + 53968*x + 28901)/(2354450*x**4 + 2825340*x**3 + 2731162*x**2 + 1130136*x + 376712)
+ 32*log(x + 1/2)/343 - 16*log(x**2 + 3*x/5 + 2/5)/343 + 125624*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31
)/10218313

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