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

Optimal. Leaf size=94 \[ \frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}+\frac {3}{968} \log \left (2 x^2-x+3\right )-\frac {3}{968} \log \left (5 x^2+3 x+2\right )+\frac {7 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{15004 \sqrt {31}} \]

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Rubi [A]  time = 0.09, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {974, 1072, 634, 618, 204, 628} \begin {gather*} \frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}+\frac {3}{968} \log \left (2 x^2-x+3\right )-\frac {3}{968} \log \left (5 x^2+3 x+2\right )+\frac {7 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{15004 \sqrt {31}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + 65*x)/(682*(2 + 3*x + 5*x^2)) + (7*ArcTan[(1 - 4*x)/Sqrt[23]])/(484*Sqrt[23]) + (2891*ArcTan[(3 + 10*x)/S
qrt[31]])/(15004*Sqrt[31]) + (3*Log[3 - x + 2*x^2])/968 - (3*Log[2 + 3*x + 5*x^2])/968

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 974

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a
*c^2*e - b^2*c*e + b^3*f + b*c*(c*d - 3*a*f) + c*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*x)*(a + b*x + c*x^2)^(p +
1)*(d + e*x + f*x^2)^(q + 1))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), x] - Dist[1/
((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*
x^2)^q*Simp[2*c*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) - (2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(a*f*(
p + 1) - c*d*(p + 2)) - e*(b^2*c*e - 2*a*c^2*e - b^3*f - b*c*(c*d - 3*a*f))*(p + q + 2) + (2*f*(2*a*c^2*e - b^
2*c*e + b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(b*f*(p + 1) - c*e*(2*p +
q + 4)))*x + c*f*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e,
f, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 - (b*d - a*e)*(c*e
- b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1072

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)
), x_Symbol] :> With[{q = c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2}, Dist[1/q, In
t[(A*c^2*d - a*c*C*d - A*b*c*e + a*B*c*e + A*b^2*f - a*b*B*f - a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d - A*c*e +
a*C*e + A*b*f - a*B*f)*x)/(a + b*x + c*x^2), x], x] + Dist[1/q, Int[(c*C*d^2 - B*c*d*e + A*c*e^2 + b*B*d*f - A
*c*d*f - a*C*d*f - A*b*e*f + a*A*f^2 - f*(B*c*d - b*C*d - A*c*e + a*C*e + A*b*f - a*B*f)*x)/(d + e*x + f*x^2),
x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx &=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}-\frac {\int \frac {-1804+1397 x-1430 x^2}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx}{7502}\\ &=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}-\frac {\int \frac {18755-22506 x}{3-x+2 x^2} \, dx}{1815484}-\frac {\int \frac {-158026+56265 x}{2+3 x+5 x^2} \, dx}{1815484}\\ &=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}+\frac {3}{968} \int \frac {-1+4 x}{3-x+2 x^2} \, dx-\frac {3}{968} \int \frac {3+10 x}{2+3 x+5 x^2} \, dx-\frac {7}{968} \int \frac {1}{3-x+2 x^2} \, dx+\frac {2891 \int \frac {1}{2+3 x+5 x^2} \, dx}{30008}\\ &=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}+\frac {3}{968} \log \left (3-x+2 x^2\right )-\frac {3}{968} \log \left (2+3 x+5 x^2\right )+\frac {7}{484} \operatorname {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )-\frac {2891 \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )}{15004}\\ &=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}+\frac {7 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{15004 \sqrt {31}}+\frac {3}{968} \log \left (3-x+2 x^2\right )-\frac {3}{968} \log \left (2+3 x+5 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 94, normalized size = 1.00 \begin {gather*} \frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}+\frac {3}{968} \log \left (2 x^2-x+3\right )-\frac {3}{968} \log \left (5 x^2+3 x+2\right )-\frac {7 \tan ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{15004 \sqrt {31}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + 65*x)/(682*(2 + 3*x + 5*x^2)) - (7*ArcTan[(-1 + 4*x)/Sqrt[23]])/(484*Sqrt[23]) + (2891*ArcTan[(3 + 10*x)/
Sqrt[31]])/(15004*Sqrt[31]) + (3*Log[3 - x + 2*x^2])/968 - (3*Log[2 + 3*x + 5*x^2])/968

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 117, normalized size = 1.24 \begin {gather*} \frac {132986 \, \sqrt {31} {\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - 13454 \, \sqrt {23} {\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - 66309 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) + 66309 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \log \left (2 \, x^{2} - x + 3\right ) + 2039180 \, x + 125488}{21395704 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21395704*(132986*sqrt(31)*(5*x^2 + 3*x + 2)*arctan(1/31*sqrt(31)*(10*x + 3)) - 13454*sqrt(23)*(5*x^2 + 3*x +
2)*arctan(1/23*sqrt(23)*(4*x - 1)) - 66309*(5*x^2 + 3*x + 2)*log(5*x^2 + 3*x + 2) + 66309*(5*x^2 + 3*x + 2)*l
og(2*x^2 - x + 3) + 2039180*x + 125488)/(5*x^2 + 3*x + 2)

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giac [A]  time = 0.20, size = 78, normalized size = 0.83 \begin {gather*} \frac {2891}{465124} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {7}{11132} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {65 \, x + 4}{682 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac {3}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {3}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2891/465124*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 7/11132*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/6
82*(65*x + 4)/(5*x^2 + 3*x + 2) - 3/968*log(5*x^2 + 3*x + 2) + 3/968*log(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 77, normalized size = 0.82 \begin {gather*} \frac {2891 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{465124}-\frac {7 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{11132}+\frac {3 \ln \left (2 x^{2}-x +3\right )}{968}-\frac {3 \ln \left (5 x^{2}+3 x +2\right )}{968}-\frac {-\frac {286 x}{31}-\frac {88}{155}}{484 \left (x^{2}+\frac {3}{5} x +\frac {2}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/484*(-286/31*x-88/155)/(x^2+3/5*x+2/5)-3/968*ln(5*x^2+3*x+2)+2891/465124*31^(1/2)*arctan(1/31*(10*x+3)*31^(
1/2))+3/968*ln(2*x^2-x+3)-7/11132*23^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))

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maxima [A]  time = 0.96, size = 78, normalized size = 0.83 \begin {gather*} \frac {2891}{465124} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {7}{11132} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {65 \, x + 4}{682 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac {3}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {3}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2891/465124*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 7/11132*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/6
82*(65*x + 4)/(5*x^2 + 3*x + 2) - 3/968*log(5*x^2 + 3*x + 2) + 3/968*log(2*x^2 - x + 3)

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mupad [B]  time = 3.57, size = 95, normalized size = 1.01 \begin {gather*} \frac {\frac {13\,x}{682}+\frac {2}{1705}}{x^2+\frac {3\,x}{5}+\frac {2}{5}}+\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {3}{968}+\frac {\sqrt {23}\,7{}\mathrm {i}}{22264}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {3}{968}+\frac {\sqrt {23}\,7{}\mathrm {i}}{22264}\right )-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {3}{968}+\frac {\sqrt {31}\,2891{}\mathrm {i}}{930248}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {3}{968}+\frac {\sqrt {31}\,2891{}\mathrm {i}}{930248}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((13*x)/682 + 2/1705)/((3*x)/5 + x^2 + 2/5) + log(x - (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*7i)/22264 + 3/968) - l
og(x + (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*7i)/22264 - 3/968) - log(x - (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*2891
i)/930248 + 3/968) + log(x + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*2891i)/930248 - 3/968)

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sympy [A]  time = 0.32, size = 102, normalized size = 1.09 \begin {gather*} \frac {65 x + 4}{3410 x^{2} + 2046 x + 1364} + \frac {3 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{968} - \frac {3 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{968} - \frac {7 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{11132} + \frac {2891 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{465124} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(65*x + 4)/(3410*x**2 + 2046*x + 1364) + 3*log(x**2 - x/2 + 3/2)/968 - 3*log(x**2 + 3*x/5 + 2/5)/968 - 7*sqrt(
23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/11132 + 2891*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/465124

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