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]

-38/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)+1/54*(398+279*x)/(3*x^2+2)^(1/2)+292/81*(3*x^2+2)^(1/2)+4*x*(3*x^2+2)^(1/
2)+32/27*x^2*(3*x^2+2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.85, size = 76, normalized size = 0.87 \[ \frac {342 \, \sqrt {3} {\left (3 \, x^{2} + 2\right )} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + {\left (576 \, x^{4} + 1944 \, x^{3} + 2136 \, x^{2} + 2133 \, x + 2362\right )} \sqrt {3 \, x^{2} + 2}}{162 \, {\left (3 \, x^{2} + 2\right )}} \]

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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giac [A]  time = 0.21, size = 54, normalized size = 0.62 \[ \frac {38}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {3 \, {\left (8 \, {\left (3 \, {\left (8 \, x + 27\right )} x + 89\right )} x + 711\right )} x + 2362}{162 \, \sqrt {3 \, x^{2} + 2}} \]

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)

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maple [A]  time = 0.01, size = 79, normalized size = 0.91 \[ \frac {32 x^{4}}{9 \sqrt {3 x^{2}+2}}+\frac {12 x^{3}}{\sqrt {3 x^{2}+2}}+\frac {356 x^{2}}{27 \sqrt {3 x^{2}+2}}+\frac {79 x}{6 \sqrt {3 x^{2}+2}}-\frac {38 \sqrt {3}\, \arcsinh \left (\frac {\sqrt {6}\, x}{2}\right )}{9}+\frac {1181}{81 \sqrt {3 x^{2}+2}} \]

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*6^(1/2)*x)*3^(1/2)

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maxima [A]  time = 0.96, size = 78, normalized size = 0.90 \[ \frac {32 \, x^{4}}{9 \, \sqrt {3 \, x^{2} + 2}} + \frac {12 \, x^{3}}{\sqrt {3 \, x^{2} + 2}} + \frac {356 \, x^{2}}{27 \, \sqrt {3 \, x^{2} + 2}} - \frac {38}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {79 \, x}{6 \, \sqrt {3 \, x^{2} + 2}} + \frac {1181}{81 \, \sqrt {3 \, x^{2} + 2}} \]

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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mupad [B]  time = 0.06, size = 110, normalized size = 1.26 \[ \frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {32\,x^2}{9}+12\,x+\frac {292}{27}\right )}{3}-\frac {38\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{9}-\frac {\sqrt {3}\,\sqrt {6}\,\left (-1194+\sqrt {6}\,279{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{1944\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}-\frac {\sqrt {3}\,\sqrt {6}\,\left (1194+\sqrt {6}\,279{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{1944\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3^(1/2)*(x^2 + 2/3)^(1/2)*(12*x + (32*x^2)/9 + 292/27))/3 - (38*3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2))/9 - (3^
(1/2)*6^(1/2)*(6^(1/2)*279i - 1194)*(x^2 + 2/3)^(1/2)*1i)/(1944*(x + (6^(1/2)*1i)/3)) - (3^(1/2)*6^(1/2)*(6^(1
/2)*279i + 1194)*(x^2 + 2/3)^(1/2)*1i)/(1944*(x - (6^(1/2)*1i)/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 x + 1\right )^{3} \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} + 2\right )^{\frac {3}{2}}}\, dx \]

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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