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

Optimal. Leaf size=94 \[ \frac {13-6 x}{506 \left (2 x^2-x+3\right )}-\frac {13}{968} \log \left (2 x^2-x+3\right )+\frac {13}{968} \log \left (5 x^2+3 x+2\right )+\frac {241 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{484 \sqrt {31}} \]

[Out]

1/506*(13-6*x)/(2*x^2-x+3)-13/968*ln(2*x^2-x+3)+13/968*ln(5*x^2+3*x+2)+241/256036*arctan(1/23*(1-4*x)*23^(1/2)
)*23^(1/2)+69/15004*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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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} \[ \frac {13-6 x}{506 \left (2 x^2-x+3\right )}-\frac {13}{968} \log \left (2 x^2-x+3\right )+\frac {13}{968} \log \left (5 x^2+3 x+2\right )+\frac {241 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{484 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13 - 6*x)/(506*(3 - x + 2*x^2)) + (241*ArcTan[(1 - 4*x)/Sqrt[23]])/(11132*Sqrt[23]) + (69*ArcTan[(3 + 10*x)/S
qrt[31]])/(484*Sqrt[31]) - (13*Log[3 - x + 2*x^2])/968 + (13*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 )^2 \left (2+3 x+5 x^2\right )} \, dx &=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}-\frac {\int \frac {-1892-1067 x+330 x^2}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx}{5566}\\ &=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}-\frac {\int \frac {-3509+72358 x}{3-x+2 x^2} \, dx}{1346972}-\frac {\int \frac {-150282-180895 x}{2+3 x+5 x^2} \, dx}{1346972}\\ &=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}-\frac {241 \int \frac {1}{3-x+2 x^2} \, dx}{22264}-\frac {13}{968} \int \frac {-1+4 x}{3-x+2 x^2} \, dx+\frac {13}{968} \int \frac {3+10 x}{2+3 x+5 x^2} \, dx+\frac {69}{968} \int \frac {1}{2+3 x+5 x^2} \, dx\\ &=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}-\frac {13}{968} \log \left (3-x+2 x^2\right )+\frac {13}{968} \log \left (2+3 x+5 x^2\right )+\frac {241 \operatorname {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )}{11132}-\frac {69}{484} \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}+\frac {241 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{484 \sqrt {31}}-\frac {13}{968} \log \left (3-x+2 x^2\right )+\frac {13}{968} \log \left (2+3 x+5 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 94, normalized size = 1.00 \[ \frac {13-6 x}{506 \left (2 x^2-x+3\right )}-\frac {13}{968} \log \left (2 x^2-x+3\right )+\frac {13}{968} \log \left (5 x^2+3 x+2\right )-\frac {241 \tan ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{484 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13 - 6*x)/(506*(3 - x + 2*x^2)) - (241*ArcTan[(-1 + 4*x)/Sqrt[23]])/(11132*Sqrt[23]) + (69*ArcTan[(3 + 10*x)/
Sqrt[31]])/(484*Sqrt[31]) - (13*Log[3 - x + 2*x^2])/968 + (13*Log[2 + 3*x + 5*x^2])/968

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fricas [A]  time = 0.93, size = 117, normalized size = 1.24 \[ \frac {73002 \, \sqrt {31} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - 14942 \, \sqrt {23} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + 213187 \, {\left (2 \, x^{2} - x + 3\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) - 213187 \, {\left (2 \, x^{2} - x + 3\right )} \log \left (2 \, x^{2} - x + 3\right ) - 188232 \, x + 407836}{15874232 \, {\left (2 \, x^{2} - x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15874232*(73002*sqrt(31)*(2*x^2 - x + 3)*arctan(1/31*sqrt(31)*(10*x + 3)) - 14942*sqrt(23)*(2*x^2 - x + 3)*a
rctan(1/23*sqrt(23)*(4*x - 1)) + 213187*(2*x^2 - x + 3)*log(5*x^2 + 3*x + 2) - 213187*(2*x^2 - x + 3)*log(2*x^
2 - x + 3) - 188232*x + 407836)/(2*x^2 - x + 3)

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giac [A]  time = 0.19, size = 78, normalized size = 0.83 \[ \frac {69}{15004} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {241}{256036} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {6 \, x - 13}{506 \, {\left (2 \, x^{2} - x + 3\right )}} + \frac {13}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {13}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

69/15004*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 241/256036*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 1/5
06*(6*x - 13)/(2*x^2 - x + 3) + 13/968*log(5*x^2 + 3*x + 2) - 13/968*log(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 77, normalized size = 0.82 \[ \frac {69 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{15004}-\frac {241 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{256036}-\frac {13 \ln \left (2 x^{2}-x +3\right )}{968}+\frac {13 \ln \left (5 x^{2}+3 x +2\right )}{968}-\frac {\frac {66 x}{23}-\frac {143}{23}}{484 \left (x^{2}-\frac {1}{2} x +\frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/968*ln(5*x^2+3*x+2)+69/15004*31^(1/2)*arctan(1/31*(10*x+3)*31^(1/2))-1/484*(66/23*x-143/23)/(x^2-1/2*x+3/2)
-13/968*ln(2*x^2-x+3)-241/256036*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 \[ \frac {69}{15004} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {241}{256036} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {6 \, x - 13}{506 \, {\left (2 \, x^{2} - x + 3\right )}} + \frac {13}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {13}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

69/15004*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 241/256036*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 1/5
06*(6*x - 13)/(2*x^2 - x + 3) + 13/968*log(5*x^2 + 3*x + 2) - 13/968*log(2*x^2 - x + 3)

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mupad [B]  time = 3.58, size = 96, normalized size = 1.02 \[ -\frac {\frac {3\,x}{506}-\frac {13}{1012}}{x^2-\frac {x}{2}+\frac {3}{2}}-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {13}{968}+\frac {\sqrt {31}\,69{}\mathrm {i}}{30008}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {13}{968}+\frac {\sqrt {31}\,69{}\mathrm {i}}{30008}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {13}{968}+\frac {\sqrt {23}\,241{}\mathrm {i}}{512072}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {13}{968}+\frac {\sqrt {23}\,241{}\mathrm {i}}{512072}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*69i)/30008 + 13/968) - log(x - (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)
*69i)/30008 - 13/968) - ((3*x)/506 - 13/1012)/(x^2 - x/2 + 3/2) + log(x - (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*24
1i)/512072 - 13/968) - log(x + (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*241i)/512072 + 13/968)

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sympy [A]  time = 0.32, size = 102, normalized size = 1.09 \[ \frac {13 - 6 x}{1012 x^{2} - 506 x + 1518} - \frac {13 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{968} + \frac {13 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{968} - \frac {241 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{256036} + \frac {69 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{15004} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13 - 6*x)/(1012*x**2 - 506*x + 1518) - 13*log(x**2 - x/2 + 3/2)/968 + 13*log(x**2 + 3*x/5 + 2/5)/968 - 241*sq
rt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/256036 + 69*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/15004

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