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

Optimal. Leaf size=87 \[ \frac{32}{27} \sqrt{3 x^2+2} x^2+4 \sqrt{3 x^2+2} x+\frac{292}{81} \sqrt{3 x^2+2}+\frac{279 x+398}{54 \sqrt{3 x^2+2}}-\frac{38 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

[Out]

(398 + 279*x)/(54*Sqrt[2 + 3*x^2]) + (292*Sqrt[2 + 3*x^2])/81 + 4*x*Sqrt[2 + 3*x^2] + (32*x^2*Sqrt[2 + 3*x^2])
/27 - (38*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.103597, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1814, 1815, 641, 215} \[ \frac{32}{27} \sqrt{3 x^2+2} x^2+4 \sqrt{3 x^2+2} x+\frac{292}{81} \sqrt{3 x^2+2}+\frac{279 x+398}{54 \sqrt{3 x^2+2}}-\frac{38 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(398 + 279*x)/(54*Sqrt[2 + 3*x^2]) + (292*Sqrt[2 + 3*x^2])/81 + 4*x*Sqrt[2 + 3*x^2] + (32*x^2*Sqrt[2 + 3*x^2])
/27 - (38*ArcSinh[Sqrt[3/2]*x])/(3*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 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[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 )^{3/2}} \, dx &=\frac{398+279 x}{54 \sqrt{2+3 x^2}}-\frac{1}{2} \int \frac{\frac{28}{3}-\frac{280 x}{9}-48 x^2-\frac{64 x^3}{3}}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{54 \sqrt{2+3 x^2}}+\frac{32}{27} x^2 \sqrt{2+3 x^2}-\frac{1}{18} \int \frac{84-\frac{584 x}{3}-432 x^2}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{54 \sqrt{2+3 x^2}}+4 x \sqrt{2+3 x^2}+\frac{32}{27} x^2 \sqrt{2+3 x^2}-\frac{1}{108} \int \frac{1368-1168 x}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{54 \sqrt{2+3 x^2}}+\frac{292}{81} \sqrt{2+3 x^2}+4 x \sqrt{2+3 x^2}+\frac{32}{27} x^2 \sqrt{2+3 x^2}-\frac{38}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{398+279 x}{54 \sqrt{2+3 x^2}}+\frac{292}{81} \sqrt{2+3 x^2}+4 x \sqrt{2+3 x^2}+\frac{32}{27} x^2 \sqrt{2+3 x^2}-\frac{38 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0572413, size = 58, normalized size = 0.67 \[ \frac{576 x^4+1944 x^3+2136 x^2-684 \sqrt{9 x^2+6} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )+2133 x+2362}{162 \sqrt{3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2362 + 2133*x + 2136*x^2 + 1944*x^3 + 576*x^4 - 684*Sqrt[6 + 9*x^2]*ArcSinh[Sqrt[3/2]*x])/(162*Sqrt[2 + 3*x^2
])

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Maple [A]  time = 0.056, size = 79, normalized size = 0.9 \begin{align*}{\frac{32\,{x}^{4}}{9}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{356\,{x}^{2}}{27}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{1181}{81}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+12\,{\frac{{x}^{3}}{\sqrt{3\,{x}^{2}+2}}}+{\frac{79\,x}{6}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}-{\frac{38\,\sqrt{3}}{9}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) } \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)^(3/2),x)

[Out]

32/9*x^4/(3*x^2+2)^(1/2)+356/27*x^2/(3*x^2+2)^(1/2)+1181/81/(3*x^2+2)^(1/2)+12*x^3/(3*x^2+2)^(1/2)+79/6*x/(3*x
^2+2)^(1/2)-38/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)

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Maxima [A]  time = 1.50175, size = 105, normalized size = 1.21 \begin{align*} \frac{32 \, x^{4}}{9 \, \sqrt{3 \, x^{2} + 2}} + \frac{12 \, x^{3}}{\sqrt{3 \, x^{2} + 2}} + \frac{356 \, x^{2}}{27 \, \sqrt{3 \, x^{2} + 2}} - \frac{38}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{79 \, x}{6 \, \sqrt{3 \, x^{2} + 2}} + \frac{1181}{81 \, \sqrt{3 \, x^{2} + 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)^(3/2),x, algorithm="maxima")

[Out]

32/9*x^4/sqrt(3*x^2 + 2) + 12*x^3/sqrt(3*x^2 + 2) + 356/27*x^2/sqrt(3*x^2 + 2) - 38/9*sqrt(3)*arcsinh(1/2*sqrt
(6)*x) + 79/6*x/sqrt(3*x^2 + 2) + 1181/81/sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.60615, size = 208, normalized size = 2.39 \begin{align*} \frac{342 \, \sqrt{3}{\left (3 \, x^{2} + 2\right )} \log \left (\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) +{\left (576 \, x^{4} + 1944 \, x^{3} + 2136 \, x^{2} + 2133 \, x + 2362\right )} \sqrt{3 \, x^{2} + 2}}{162 \,{\left (3 \, x^{2} + 2\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)^(3/2),x, algorithm="fricas")

[Out]

1/162*(342*sqrt(3)*(3*x^2 + 2)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + (576*x^4 + 1944*x^3 + 2136*x^2 + 2
133*x + 2362)*sqrt(3*x^2 + 2))/(3*x^2 + 2)

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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{3}{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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.31974, size = 73, normalized size = 0.84 \begin{align*} \frac{38}{9} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) + \frac{3 \,{\left (8 \,{\left (3 \,{\left (8 \, x + 27\right )} x + 89\right )} x + 711\right )} x + 2362}{162 \, \sqrt{3 \, x^{2} + 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)^(3/2),x, algorithm="giac")

[Out]

38/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/162*(3*(8*(3*(8*x + 27)*x + 89)*x + 711)*x + 2362)/sqrt(3*x
^2 + 2)