3.306 \(\int \frac{(d+e x) (2+x+3 x^2-5 x^3+4 x^4)}{3+2 x+5 x^2} \, dx\)

Optimal. Leaf size=99 \[ \frac{1}{75} x^3 (20 d-33 e)-\frac{3}{250} x^2 (55 d-27 e)+\frac{(2290 d-881 e) \log \left (5 x^2+2 x+3\right )}{6250}+\frac{1}{625} x (405 d+458 e)-\frac{(2115 d+5989 e) \tan ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{3125 \sqrt{14}}+\frac{e x^4}{5} \]

[Out]

((405*d + 458*e)*x)/625 - (3*(55*d - 27*e)*x^2)/250 + ((20*d - 33*e)*x^3)/75 + (e*x^4)/5 - ((2115*d + 5989*e)*
ArcTan[(1 + 5*x)/Sqrt[14]])/(3125*Sqrt[14]) + ((2290*d - 881*e)*Log[3 + 2*x + 5*x^2])/6250

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Rubi [A]  time = 0.108266, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {1628, 634, 618, 204, 628} \[ \frac{1}{75} x^3 (20 d-33 e)-\frac{3}{250} x^2 (55 d-27 e)+\frac{(2290 d-881 e) \log \left (5 x^2+2 x+3\right )}{6250}+\frac{1}{625} x (405 d+458 e)-\frac{(2115 d+5989 e) \tan ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{3125 \sqrt{14}}+\frac{e x^4}{5} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2),x]

[Out]

((405*d + 458*e)*x)/625 - (3*(55*d - 27*e)*x^2)/250 + ((20*d - 33*e)*x^3)/75 + (e*x^4)/5 - ((2115*d + 5989*e)*
ArcTan[(1 + 5*x)/Sqrt[14]])/(3125*Sqrt[14]) + ((2290*d - 881*e)*Log[3 + 2*x + 5*x^2])/6250

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+e x) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx &=\int \left (\frac{1}{625} (405 d+458 e)-\frac{3}{125} (55 d-27 e) x+\frac{1}{25} (20 d-33 e) x^2+\frac{4 e x^3}{5}+\frac{35 d-1374 e+(2290 d-881 e) x}{625 \left (3+2 x+5 x^2\right )}\right ) \, dx\\ &=\frac{1}{625} (405 d+458 e) x-\frac{3}{250} (55 d-27 e) x^2+\frac{1}{75} (20 d-33 e) x^3+\frac{e x^4}{5}+\frac{1}{625} \int \frac{35 d-1374 e+(2290 d-881 e) x}{3+2 x+5 x^2} \, dx\\ &=\frac{1}{625} (405 d+458 e) x-\frac{3}{250} (55 d-27 e) x^2+\frac{1}{75} (20 d-33 e) x^3+\frac{e x^4}{5}+\frac{(-2115 d-5989 e) \int \frac{1}{3+2 x+5 x^2} \, dx}{3125}+\frac{(2290 d-881 e) \int \frac{2+10 x}{3+2 x+5 x^2} \, dx}{6250}\\ &=\frac{1}{625} (405 d+458 e) x-\frac{3}{250} (55 d-27 e) x^2+\frac{1}{75} (20 d-33 e) x^3+\frac{e x^4}{5}+\frac{(2290 d-881 e) \log \left (3+2 x+5 x^2\right )}{6250}+\frac{(2 (2115 d+5989 e)) \operatorname{Subst}\left (\int \frac{1}{-56-x^2} \, dx,x,2+10 x\right )}{3125}\\ &=\frac{1}{625} (405 d+458 e) x-\frac{3}{250} (55 d-27 e) x^2+\frac{1}{75} (20 d-33 e) x^3+\frac{e x^4}{5}-\frac{(2115 d+5989 e) \tan ^{-1}\left (\frac{1+5 x}{\sqrt{14}}\right )}{3125 \sqrt{14}}+\frac{(2290 d-881 e) \log \left (3+2 x+5 x^2\right )}{6250}\\ \end{align*}

Mathematica [A]  time = 0.0503886, size = 86, normalized size = 0.87 \[ \frac{35 x \left (5 d \left (200 x^2-495 x+486\right )+3 e \left (250 x^3-550 x^2+405 x+916\right )\right )+21 (2290 d-881 e) \log \left (5 x^2+2 x+3\right )-3 \sqrt{14} (2115 d+5989 e) \tan ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{131250} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2),x]

[Out]

(35*x*(5*d*(486 - 495*x + 200*x^2) + 3*e*(916 + 405*x - 550*x^2 + 250*x^3)) - 3*Sqrt[14]*(2115*d + 5989*e)*Arc
Tan[(1 + 5*x)/Sqrt[14]] + 21*(2290*d - 881*e)*Log[3 + 2*x + 5*x^2])/131250

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Maple [A]  time = 0.046, size = 102, normalized size = 1. \begin{align*}{\frac{e{x}^{4}}{5}}+{\frac{4\,d{x}^{3}}{15}}-{\frac{11\,{x}^{3}e}{25}}-{\frac{33\,d{x}^{2}}{50}}+{\frac{81\,e{x}^{2}}{250}}+{\frac{81\,dx}{125}}+{\frac{458\,ex}{625}}+{\frac{229\,\ln \left ( 5\,{x}^{2}+2\,x+3 \right ) d}{625}}-{\frac{881\,e\ln \left ( 5\,{x}^{2}+2\,x+3 \right ) }{6250}}-{\frac{423\,\sqrt{14}d}{8750}\arctan \left ({\frac{ \left ( 10\,x+2 \right ) \sqrt{14}}{28}} \right ) }-{\frac{5989\,\sqrt{14}e}{43750}\arctan \left ({\frac{ \left ( 10\,x+2 \right ) \sqrt{14}}{28}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)

[Out]

1/5*e*x^4+4/15*d*x^3-11/25*x^3*e-33/50*d*x^2+81/250*e*x^2+81/125*d*x+458/625*e*x+229/625*ln(5*x^2+2*x+3)*d-881
/6250*e*ln(5*x^2+2*x+3)-423/8750*14^(1/2)*arctan(1/28*(10*x+2)*14^(1/2))*d-5989/43750*14^(1/2)*arctan(1/28*(10
*x+2)*14^(1/2))*e

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Maxima [A]  time = 1.56544, size = 113, normalized size = 1.14 \begin{align*} \frac{1}{5} \, e x^{4} + \frac{1}{75} \,{\left (20 \, d - 33 \, e\right )} x^{3} - \frac{3}{250} \,{\left (55 \, d - 27 \, e\right )} x^{2} - \frac{1}{43750} \, \sqrt{14}{\left (2115 \, d + 5989 \, e\right )} \arctan \left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) + \frac{1}{625} \,{\left (405 \, d + 458 \, e\right )} x + \frac{1}{6250} \,{\left (2290 \, d - 881 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="maxima")

[Out]

1/5*e*x^4 + 1/75*(20*d - 33*e)*x^3 - 3/250*(55*d - 27*e)*x^2 - 1/43750*sqrt(14)*(2115*d + 5989*e)*arctan(1/14*
sqrt(14)*(5*x + 1)) + 1/625*(405*d + 458*e)*x + 1/6250*(2290*d - 881*e)*log(5*x^2 + 2*x + 3)

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Fricas [A]  time = 1.31163, size = 275, normalized size = 2.78 \begin{align*} \frac{1}{5} \, e x^{4} + \frac{1}{75} \,{\left (20 \, d - 33 \, e\right )} x^{3} - \frac{3}{250} \,{\left (55 \, d - 27 \, e\right )} x^{2} - \frac{1}{43750} \, \sqrt{14}{\left (2115 \, d + 5989 \, e\right )} \arctan \left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) + \frac{1}{625} \,{\left (405 \, d + 458 \, e\right )} x + \frac{1}{6250} \,{\left (2290 \, d - 881 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="fricas")

[Out]

1/5*e*x^4 + 1/75*(20*d - 33*e)*x^3 - 3/250*(55*d - 27*e)*x^2 - 1/43750*sqrt(14)*(2115*d + 5989*e)*arctan(1/14*
sqrt(14)*(5*x + 1)) + 1/625*(405*d + 458*e)*x + 1/6250*(2290*d - 881*e)*log(5*x^2 + 2*x + 3)

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Sympy [C]  time = 0.686389, size = 163, normalized size = 1.65 \begin{align*} \frac{e x^{4}}{5} + x^{3} \left (\frac{4 d}{15} - \frac{11 e}{25}\right ) + x^{2} \left (- \frac{33 d}{50} + \frac{81 e}{250}\right ) + x \left (\frac{81 d}{125} + \frac{458 e}{625}\right ) + \left (\frac{229 d}{625} - \frac{881 e}{6250} - \frac{\sqrt{14} i \left (2115 d + 5989 e\right )}{87500}\right ) \log{\left (x + \frac{423 d + \frac{5989 e}{5} + \frac{\sqrt{14} i \left (2115 d + 5989 e\right )}{5}}{2115 d + 5989 e} \right )} + \left (\frac{229 d}{625} - \frac{881 e}{6250} + \frac{\sqrt{14} i \left (2115 d + 5989 e\right )}{87500}\right ) \log{\left (x + \frac{423 d + \frac{5989 e}{5} - \frac{\sqrt{14} i \left (2115 d + 5989 e\right )}{5}}{2115 d + 5989 e} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x**4-5*x**3+3*x**2+x+2)/(5*x**2+2*x+3),x)

[Out]

e*x**4/5 + x**3*(4*d/15 - 11*e/25) + x**2*(-33*d/50 + 81*e/250) + x*(81*d/125 + 458*e/625) + (229*d/625 - 881*
e/6250 - sqrt(14)*I*(2115*d + 5989*e)/87500)*log(x + (423*d + 5989*e/5 + sqrt(14)*I*(2115*d + 5989*e)/5)/(2115
*d + 5989*e)) + (229*d/625 - 881*e/6250 + sqrt(14)*I*(2115*d + 5989*e)/87500)*log(x + (423*d + 5989*e/5 - sqrt
(14)*I*(2115*d + 5989*e)/5)/(2115*d + 5989*e))

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Giac [A]  time = 1.1224, size = 119, normalized size = 1.2 \begin{align*} \frac{1}{5} \, x^{4} e + \frac{4}{15} \, d x^{3} - \frac{11}{25} \, x^{3} e - \frac{33}{50} \, d x^{2} + \frac{81}{250} \, x^{2} e - \frac{1}{43750} \, \sqrt{14}{\left (2115 \, d + 5989 \, e\right )} \arctan \left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) + \frac{81}{125} \, d x + \frac{458}{625} \, x e + \frac{1}{6250} \,{\left (2290 \, d - 881 \, e\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="giac")

[Out]

1/5*x^4*e + 4/15*d*x^3 - 11/25*x^3*e - 33/50*d*x^2 + 81/250*x^2*e - 1/43750*sqrt(14)*(2115*d + 5989*e)*arctan(
1/14*sqrt(14)*(5*x + 1)) + 81/125*d*x + 458/625*x*e + 1/6250*(2290*d - 881*e)*log(5*x^2 + 2*x + 3)