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

Optimal. Leaf size=73 \[ \frac{279 x+398}{162 \left (3 x^2+2\right )^{3/2}}+\frac{32}{27} \sqrt{3 x^2+2}-\frac{465 x+152}{54 \sqrt{3 x^2+2}}+\frac{8 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}} \]

[Out]

(398 + 279*x)/(162*(2 + 3*x^2)^(3/2)) - (152 + 465*x)/(54*Sqrt[2 + 3*x^2]) + (32*Sqrt[2 + 3*x^2])/27 + (8*ArcS
inh[Sqrt[3/2]*x])/Sqrt[3]

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Rubi [A]  time = 0.0849701, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {1814, 641, 215} \[ \frac{279 x+398}{162 \left (3 x^2+2\right )^{3/2}}+\frac{32}{27} \sqrt{3 x^2+2}-\frac{465 x+152}{54 \sqrt{3 x^2+2}}+\frac{8 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(398 + 279*x)/(162*(2 + 3*x^2)^(3/2)) - (152 + 465*x)/(54*Sqrt[2 + 3*x^2]) + (32*Sqrt[2 + 3*x^2])/27 + (8*ArcS
inh[Sqrt[3/2]*x])/Sqrt[3]

Rule 1814

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

Rule 641

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

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+3 x^2\right )^{5/2}} \, dx &=\frac{398+279 x}{162 \left (2+3 x^2\right )^{3/2}}-\frac{1}{6} \int \frac{\frac{22}{3}-\frac{280 x}{3}-144 x^2-64 x^3}{\left (2+3 x^2\right )^{3/2}} \, dx\\ &=\frac{398+279 x}{162 \left (2+3 x^2\right )^{3/2}}-\frac{152+465 x}{54 \sqrt{2+3 x^2}}+\frac{1}{12} \int \frac{96+\frac{128 x}{3}}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{162 \left (2+3 x^2\right )^{3/2}}-\frac{152+465 x}{54 \sqrt{2+3 x^2}}+\frac{32}{27} \sqrt{2+3 x^2}+8 \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{162 \left (2+3 x^2\right )^{3/2}}-\frac{152+465 x}{54 \sqrt{2+3 x^2}}+\frac{32}{27} \sqrt{2+3 x^2}+\frac{8 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0738641, size = 63, normalized size = 0.86 \[ \frac{1728 x^4-4185 x^3+936 x^2+432 \sqrt{3} \left (3 x^2+2\right )^{3/2} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )-2511 x+254}{162 \left (3 x^2+2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(254 - 2511*x + 936*x^2 - 4185*x^3 + 1728*x^4 + 432*Sqrt[3]*(2 + 3*x^2)^(3/2)*ArcSinh[Sqrt[3/2]*x])/(162*(2 +
3*x^2)^(3/2))

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Maple [A]  time = 0.062, size = 91, normalized size = 1.3 \begin{align*}{\frac{32\,{x}^{4}}{3} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{52\,{x}^{2}}{9} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{127}{81} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}-8\,{\frac{{x}^{3}}{ \left ( 3\,{x}^{2}+2 \right ) ^{3/2}}}-{\frac{107\,x}{18}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{8\,\sqrt{3}}{3}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }-{\frac{65\,x}{18} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{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+2)^(5/2),x)

[Out]

32/3*x^4/(3*x^2+2)^(3/2)+52/9*x^2/(3*x^2+2)^(3/2)+127/81/(3*x^2+2)^(3/2)-8*x^3/(3*x^2+2)^(3/2)-107/18*x/(3*x^2
+2)^(1/2)+8/3*arcsinh(1/2*x*6^(1/2))*3^(1/2)-65/18*x/(3*x^2+2)^(3/2)

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Maxima [A]  time = 1.50444, size = 142, normalized size = 1.95 \begin{align*} \frac{32 \, x^{4}}{3 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{8}{3} \, x{\left (\frac{9 \, x^{2}}{{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} + \frac{4}{{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}}\right )} + \frac{8}{3} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) - \frac{11 \, x}{18 \, \sqrt{3 \, x^{2} + 2}} + \frac{52 \, x^{2}}{9 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{65 \, x}{18 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} + \frac{127}{81 \,{\left (3 \, x^{2} + 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+2)^(5/2),x, algorithm="maxima")

[Out]

32/3*x^4/(3*x^2 + 2)^(3/2) - 8/3*x*(9*x^2/(3*x^2 + 2)^(3/2) + 4/(3*x^2 + 2)^(3/2)) + 8/3*sqrt(3)*arcsinh(1/2*s
qrt(6)*x) - 11/18*x/sqrt(3*x^2 + 2) + 52/9*x^2/(3*x^2 + 2)^(3/2) - 65/18*x/(3*x^2 + 2)^(3/2) + 127/81/(3*x^2 +
2)^(3/2)

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Fricas [A]  time = 1.54365, size = 232, normalized size = 3.18 \begin{align*} \frac{216 \, \sqrt{3}{\left (9 \, x^{4} + 12 \, x^{2} + 4\right )} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) +{\left (1728 \, x^{4} - 4185 \, x^{3} + 936 \, x^{2} - 2511 \, x + 254\right )} \sqrt{3 \, x^{2} + 2}}{162 \,{\left (9 \, x^{4} + 12 \, x^{2} + 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+2)^(5/2),x, algorithm="fricas")

[Out]

1/162*(216*sqrt(3)*(9*x^4 + 12*x^2 + 4)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + (1728*x^4 - 4185*x^3 + 9
36*x^2 - 2511*x + 254)*sqrt(3*x^2 + 2))/(9*x^4 + 12*x^2 + 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} + 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+2)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.21445, size = 72, normalized size = 0.99 \begin{align*} -\frac{8}{3} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) + \frac{9 \,{\left ({\left (3 \,{\left (64 \, x - 155\right )} x + 104\right )} x - 279\right )} x + 254}{162 \,{\left (3 \, x^{2} + 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+2)^(5/2),x, algorithm="giac")

[Out]

-8/3*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/162*(9*((3*(64*x - 155)*x + 104)*x - 279)*x + 254)/(3*x^2 +
2)^(3/2)