3.252 $$\int \frac{(1+2 x)^3 (1+3 x+4 x^2)}{(2-x+3 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=86 $\frac{2 (12839-3871 x)}{5589 \left (3 x^2-x+2\right )^{3/2}}+\frac{32}{27} \sqrt{3 x^2-x+2}-\frac{28 (42240 x+35809)}{128547 \sqrt{3 x^2-x+2}}-\frac{296 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{27 \sqrt{3}}$

[Out]

(2*(12839 - 3871*x))/(5589*(2 - x + 3*x^2)^(3/2)) - (28*(35809 + 42240*x))/(128547*Sqrt[2 - x + 3*x^2]) + (32*
Sqrt[2 - x + 3*x^2])/27 - (296*ArcSinh[(1 - 6*x)/Sqrt[23]])/(27*Sqrt[3])

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Rubi [A]  time = 0.112569, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {1660, 640, 619, 215} $\frac{2 (12839-3871 x)}{5589 \left (3 x^2-x+2\right )^{3/2}}+\frac{32}{27} \sqrt{3 x^2-x+2}-\frac{28 (42240 x+35809)}{128547 \sqrt{3 x^2-x+2}}-\frac{296 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{27 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/(2 - x + 3*x^2)^(5/2),x]

[Out]

(2*(12839 - 3871*x))/(5589*(2 - x + 3*x^2)^(3/2)) - (28*(35809 + 42240*x))/(128547*Sqrt[2 - x + 3*x^2]) + (32*
Sqrt[2 - x + 3*x^2])/27 - (296*ArcSinh[(1 - 6*x)/Sqrt[23]])/(27*Sqrt[3])

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 640

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

Rule 619

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

Rule 215

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

Rubi steps

\begin{align*} \int \frac{(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2-x+3 x^2\right )^{5/2}} \, dx &=\frac{2 (12839-3871 x)}{5589 \left (2-x+3 x^2\right )^{3/2}}+\frac{2}{69} \int \frac{-\frac{4361}{81}+\frac{7682 x}{9}+\frac{2852 x^2}{3}+368 x^3}{\left (2-x+3 x^2\right )^{3/2}} \, dx\\ &=\frac{2 (12839-3871 x)}{5589 \left (2-x+3 x^2\right )^{3/2}}-\frac{28 (35809+42240 x)}{128547 \sqrt{2-x+3 x^2}}+\frac{4 \int \frac{\frac{37030}{9}+\frac{4232 x}{3}}{\sqrt{2-x+3 x^2}} \, dx}{1587}\\ &=\frac{2 (12839-3871 x)}{5589 \left (2-x+3 x^2\right )^{3/2}}-\frac{28 (35809+42240 x)}{128547 \sqrt{2-x+3 x^2}}+\frac{32}{27} \sqrt{2-x+3 x^2}+\frac{296}{27} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2 (12839-3871 x)}{5589 \left (2-x+3 x^2\right )^{3/2}}-\frac{28 (35809+42240 x)}{128547 \sqrt{2-x+3 x^2}}+\frac{32}{27} \sqrt{2-x+3 x^2}+\frac{296 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{27 \sqrt{69}}\\ &=\frac{2 (12839-3871 x)}{5589 \left (2-x+3 x^2\right )^{3/2}}-\frac{28 (35809+42240 x)}{128547 \sqrt{2-x+3 x^2}}+\frac{32}{27} \sqrt{2-x+3 x^2}-\frac{296 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{27 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0628087, size = 71, normalized size = 0.83 $\frac{2 \left (228528 x^4-743712 x^3+25890 x^2+78292 \sqrt{3} \left (3 x^2-x+2\right )^{3/2} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )-358377 x-134217\right )}{42849 \left (3 x^2-x+2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/(2 - x + 3*x^2)^(5/2),x]

[Out]

(2*(-134217 - 358377*x + 25890*x^2 - 743712*x^3 + 228528*x^4 + 78292*Sqrt[3]*(2 - x + 3*x^2)^(3/2)*ArcSinh[(-1
+ 6*x)/Sqrt[23]]))/(42849*(2 - x + 3*x^2)^(3/2))

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Maple [B]  time = 0.056, size = 163, normalized size = 1.9 \begin{align*}{\frac{32\,{x}^{4}}{3} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{296\,{x}^{3}}{27} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,{x}^{2}}{27} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{296\,\sqrt{3}}{81}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }-{\frac{461\,x}{81} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{296\,x}{27}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{-65264+391584\,x}{128547}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{-13763+82578\,x}{33534} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{1727}{1458} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{148}{81}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x)

[Out]

32/3*x^4/(3*x^2-x+2)^(3/2)-296/27*x^3/(3*x^2-x+2)^(3/2)+8/27*x^2/(3*x^2-x+2)^(3/2)+296/81*3^(1/2)*arcsinh(6/23
*23^(1/2)*(x-1/6))-461/81*x/(3*x^2-x+2)^(3/2)-296/27*x/(3*x^2-x+2)^(1/2)+65264/128547*(-1+6*x)/(3*x^2-x+2)^(1/
2)+13763/33534*(-1+6*x)/(3*x^2-x+2)^(3/2)-1727/1458/(3*x^2-x+2)^(3/2)-148/81/(3*x^2-x+2)^(1/2)

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Maxima [B]  time = 1.52883, size = 273, normalized size = 3.17 \begin{align*} \frac{32 \, x^{4}}{3 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} + \frac{296}{42849} \, x{\left (\frac{426 \, x}{\sqrt{3 \, x^{2} - x + 2}} - \frac{4761 \, x^{2}}{{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} - \frac{71}{\sqrt{3 \, x^{2} - x + 2}} + \frac{805 \, x}{{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} - \frac{2162}{{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}}\right )} + \frac{296}{81} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (6 \, x - 1\right )}\right ) - \frac{42032}{42849} \, \sqrt{3 \, x^{2} - x + 2} - \frac{47072 \, x}{42849 \, \sqrt{3 \, x^{2} - x + 2}} + \frac{52 \, x^{2}}{9 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} - \frac{23104}{14283 \, \sqrt{3 \, x^{2} - x + 2}} - \frac{7742 \, x}{1863 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} + \frac{1666}{1863 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="maxima")

[Out]

32/3*x^4/(3*x^2 - x + 2)^(3/2) + 296/42849*x*(426*x/sqrt(3*x^2 - x + 2) - 4761*x^2/(3*x^2 - x + 2)^(3/2) - 71/
sqrt(3*x^2 - x + 2) + 805*x/(3*x^2 - x + 2)^(3/2) - 2162/(3*x^2 - x + 2)^(3/2)) + 296/81*sqrt(3)*arcsinh(1/23*
sqrt(23)*(6*x - 1)) - 42032/42849*sqrt(3*x^2 - x + 2) - 47072/42849*x/sqrt(3*x^2 - x + 2) + 52/9*x^2/(3*x^2 -
x + 2)^(3/2) - 23104/14283/sqrt(3*x^2 - x + 2) - 7742/1863*x/(3*x^2 - x + 2)^(3/2) + 1666/1863/(3*x^2 - x + 2)
^(3/2)

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Fricas [A]  time = 1.14331, size = 325, normalized size = 3.78 \begin{align*} \frac{2 \,{\left (39146 \, \sqrt{3}{\left (9 \, x^{4} - 6 \, x^{3} + 13 \, x^{2} - 4 \, x + 4\right )} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + 3 \,{\left (76176 \, x^{4} - 247904 \, x^{3} + 8630 \, x^{2} - 119459 \, x - 44739\right )} \sqrt{3 \, x^{2} - x + 2}\right )}}{42849 \,{\left (9 \, x^{4} - 6 \, x^{3} + 13 \, x^{2} - 4 \, x + 4\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="fricas")

[Out]

2/42849*(39146*sqrt(3)*(9*x^4 - 6*x^3 + 13*x^2 - 4*x + 4)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^
2 + 24*x - 25) + 3*(76176*x^4 - 247904*x^3 + 8630*x^2 - 119459*x - 44739)*sqrt(3*x^2 - x + 2))/(9*x^4 - 6*x^3
+ 13*x^2 - 4*x + 4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (2 x + 1\right )^{3} \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} - x + 2\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**3*(4*x**2+3*x+1)/(3*x**2-x+2)**(5/2),x)

[Out]

Integral((2*x + 1)**3*(4*x**2 + 3*x + 1)/(3*x**2 - x + 2)**(5/2), x)

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Giac [A]  time = 1.15634, size = 90, normalized size = 1.05 \begin{align*} -\frac{296}{81} \, \sqrt{3} \log \left (-2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )} + 1\right ) + \frac{2 \,{\left ({\left (2 \,{\left (8 \,{\left (4761 \, x - 15494\right )} x + 4315\right )} x - 119459\right )} x - 44739\right )}}{14283 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2-x+2)^(5/2),x, algorithm="giac")

[Out]

-296/81*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1) + 2/14283*((2*(8*(4761*x - 15494)*x + 43
15)*x - 119459)*x - 44739)/(3*x^2 - x + 2)^(3/2)