3.1420 \(\int \frac{(5-x) (3+2 x)^4}{(2+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=87 \[ -\frac{7 (2-7 x) (2 x+3)^3}{18 \left (3 x^2+2\right )^{3/2}}-\frac{(318-1783 x) (2 x+3)}{54 \sqrt{3 x^2+2}}-\frac{2027}{81} \sqrt{3 x^2+2}-\frac{16 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{9 \sqrt{3}} \]

[Out]

(-7*(2 - 7*x)*(3 + 2*x)^3)/(18*(2 + 3*x^2)^(3/2)) - ((318 - 1783*x)*(3 + 2*x))/(54*Sqrt[2 + 3*x^2]) - (2027*Sq
rt[2 + 3*x^2])/81 - (16*ArcSinh[Sqrt[3/2]*x])/(9*Sqrt[3])

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Rubi [A]  time = 0.0386323, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {819, 641, 215} \[ -\frac{7 (2-7 x) (2 x+3)^3}{18 \left (3 x^2+2\right )^{3/2}}-\frac{(318-1783 x) (2 x+3)}{54 \sqrt{3 x^2+2}}-\frac{2027}{81} \sqrt{3 x^2+2}-\frac{16 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{9 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*(2 - 7*x)*(3 + 2*x)^3)/(18*(2 + 3*x^2)^(3/2)) - ((318 - 1783*x)*(3 + 2*x))/(54*Sqrt[2 + 3*x^2]) - (2027*Sq
rt[2 + 3*x^2])/81 - (16*ArcSinh[Sqrt[3/2]*x])/(9*Sqrt[3])

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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{(5-x) (3+2 x)^4}{\left (2+3 x^2\right )^{5/2}} \, dx &=-\frac{7 (2-7 x) (3+2 x)^3}{18 \left (2+3 x^2\right )^{3/2}}+\frac{1}{18} \int \frac{(342-122 x) (3+2 x)^2}{\left (2+3 x^2\right )^{3/2}} \, dx\\ &=-\frac{7 (2-7 x) (3+2 x)^3}{18 \left (2+3 x^2\right )^{3/2}}-\frac{(318-1783 x) (3+2 x)}{54 \sqrt{2+3 x^2}}+\frac{1}{108} \int \frac{-192-8108 x}{\sqrt{2+3 x^2}} \, dx\\ &=-\frac{7 (2-7 x) (3+2 x)^3}{18 \left (2+3 x^2\right )^{3/2}}-\frac{(318-1783 x) (3+2 x)}{54 \sqrt{2+3 x^2}}-\frac{2027}{81} \sqrt{2+3 x^2}-\frac{16}{9} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=-\frac{7 (2-7 x) (3+2 x)^3}{18 \left (2+3 x^2\right )^{3/2}}-\frac{(318-1783 x) (3+2 x)}{54 \sqrt{2+3 x^2}}-\frac{2027}{81} \sqrt{2+3 x^2}-\frac{16 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{9 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.067431, size = 63, normalized size = 0.72 \[ -\frac{864 x^4-57285 x^3+16560 x^2+96 \sqrt{3} \left (3 x^2+2\right )^{3/2} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )-33381 x+25342}{162 \left (3 x^2+2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(25342 - 33381*x + 16560*x^2 - 57285*x^3 + 864*x^4 + 96*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 = 0.007, size = 91, normalized size = 1.1 \begin{align*} -{\frac{16\,{x}^{4}}{3} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{920\,{x}^{2}}{9} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}-{\frac{12671}{81} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,{x}^{3}}{9} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{2111\,x}{18}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}-{\frac{16\,\sqrt{3}}{27}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }-{\frac{57\,x}{2} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/3*x^4/(3*x^2+2)^(3/2)-920/9*x^2/(3*x^2+2)^(3/2)-12671/81/(3*x^2+2)^(3/2)+16/9*x^3/(3*x^2+2)^(3/2)+2111/18*
x/(3*x^2+2)^(1/2)-16/27*arcsinh(1/2*x*6^(1/2))*3^(1/2)-57/2*x/(3*x^2+2)^(3/2)

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Maxima [A]  time = 1.48677, size = 142, normalized size = 1.63 \begin{align*} -\frac{16 \, x^{4}}{3 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} + \frac{16}{27} \, 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{16}{27} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{6269 \, x}{54 \, \sqrt{3 \, x^{2} + 2}} - \frac{920 \, x^{2}}{9 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{57 \, x}{2 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{12671}{81 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/3*x^4/(3*x^2 + 2)^(3/2) + 16/27*x*(9*x^2/(3*x^2 + 2)^(3/2) + 4/(3*x^2 + 2)^(3/2)) - 16/27*sqrt(3)*arcsinh(
1/2*sqrt(6)*x) + 6269/54*x/sqrt(3*x^2 + 2) - 920/9*x^2/(3*x^2 + 2)^(3/2) - 57/2*x/(3*x^2 + 2)^(3/2) - 12671/81
/(3*x^2 + 2)^(3/2)

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Fricas [A]  time = 1.57319, size = 236, normalized size = 2.71 \begin{align*} \frac{48 \, \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 (864 \, x^{4} - 57285 \, x^{3} + 16560 \, x^{2} - 33381 \, x + 25342\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((5-x)*(3+2*x)^4/(3*x^2+2)^(5/2),x, algorithm="fricas")

[Out]

1/162*(48*sqrt(3)*(9*x^4 + 12*x^2 + 4)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) - (864*x^4 - 57285*x^3 + 165
60*x^2 - 33381*x + 25342)*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{999 x}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx - \int - \frac{864 x^{2}}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx - \int - \frac{264 x^{3}}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx - \int \frac{16 x^{4}}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx - \int \frac{16 x^{5}}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx - \int - \frac{405}{9 x^{4} \sqrt{3 x^{2} + 2} + 12 x^{2} \sqrt{3 x^{2} + 2} + 4 \sqrt{3 x^{2} + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-999*x/(9*x**4*sqrt(3*x**2 + 2) + 12*x**2*sqrt(3*x**2 + 2) + 4*sqrt(3*x**2 + 2)), x) - Integral(-864
*x**2/(9*x**4*sqrt(3*x**2 + 2) + 12*x**2*sqrt(3*x**2 + 2) + 4*sqrt(3*x**2 + 2)), x) - Integral(-264*x**3/(9*x*
*4*sqrt(3*x**2 + 2) + 12*x**2*sqrt(3*x**2 + 2) + 4*sqrt(3*x**2 + 2)), x) - Integral(16*x**4/(9*x**4*sqrt(3*x**
2 + 2) + 12*x**2*sqrt(3*x**2 + 2) + 4*sqrt(3*x**2 + 2)), x) - Integral(16*x**5/(9*x**4*sqrt(3*x**2 + 2) + 12*x
**2*sqrt(3*x**2 + 2) + 4*sqrt(3*x**2 + 2)), x) - Integral(-405/(9*x**4*sqrt(3*x**2 + 2) + 12*x**2*sqrt(3*x**2
+ 2) + 4*sqrt(3*x**2 + 2)), x)

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Giac [A]  time = 1.18714, size = 70, normalized size = 0.8 \begin{align*} \frac{16}{27} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) - \frac{9 \,{\left ({\left ({\left (96 \, x - 6365\right )} x + 1840\right )} x - 3709\right )} x + 25342}{162 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/27*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) - 1/162*(9*(((96*x - 6365)*x + 1840)*x - 3709)*x + 25342)/(3*x
^2 + 2)^(3/2)