3.1397 \(\int \frac{(5-x) (3+2 x)^4}{\sqrt{2+3 x^2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{1}{15} \sqrt{3 x^2+2} (2 x+3)^4+\frac{19}{30} \sqrt{3 x^2+2} (2 x+3)^3+\frac{1477}{270} \sqrt{3 x^2+2} (2 x+3)^2+\frac{49}{81} (99 x+383) \sqrt{3 x^2+2}+\frac{343 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

[Out]

(1477*(3 + 2*x)^2*Sqrt[2 + 3*x^2])/270 + (19*(3 + 2*x)^3*Sqrt[2 + 3*x^2])/30 - ((3 + 2*x)^4*Sqrt[2 + 3*x^2])/1
5 + (49*(383 + 99*x)*Sqrt[2 + 3*x^2])/81 + (343*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

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Rubi [A]  time = 0.0576403, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {833, 780, 215} \[ -\frac{1}{15} \sqrt{3 x^2+2} (2 x+3)^4+\frac{19}{30} \sqrt{3 x^2+2} (2 x+3)^3+\frac{1477}{270} \sqrt{3 x^2+2} (2 x+3)^2+\frac{49}{81} (99 x+383) \sqrt{3 x^2+2}+\frac{343 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1477*(3 + 2*x)^2*Sqrt[2 + 3*x^2])/270 + (19*(3 + 2*x)^3*Sqrt[2 + 3*x^2])/30 - ((3 + 2*x)^4*Sqrt[2 + 3*x^2])/1
5 + (49*(383 + 99*x)*Sqrt[2 + 3*x^2])/81 + (343*ArcSinh[Sqrt[3/2]*x])/(3*Sqrt[3])

Rule 833

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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}{\sqrt{2+3 x^2}} \, dx &=-\frac{1}{15} (3+2 x)^4 \sqrt{2+3 x^2}+\frac{1}{15} \int \frac{(3+2 x)^3 (241+114 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{19}{30} (3+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{15} (3+2 x)^4 \sqrt{2+3 x^2}+\frac{1}{180} \int \frac{(3+2 x)^2 (7308+8862 x)}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{1477}{270} (3+2 x)^2 \sqrt{2+3 x^2}+\frac{19}{30} (3+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{15} (3+2 x)^4 \sqrt{2+3 x^2}+\frac{\int \frac{(3+2 x) (126420+291060 x)}{\sqrt{2+3 x^2}} \, dx}{1620}\\ &=\frac{1477}{270} (3+2 x)^2 \sqrt{2+3 x^2}+\frac{19}{30} (3+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{15} (3+2 x)^4 \sqrt{2+3 x^2}+\frac{49}{81} (383+99 x) \sqrt{2+3 x^2}+\frac{343}{3} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=\frac{1477}{270} (3+2 x)^2 \sqrt{2+3 x^2}+\frac{19}{30} (3+2 x)^3 \sqrt{2+3 x^2}-\frac{1}{15} (3+2 x)^4 \sqrt{2+3 x^2}+\frac{49}{81} (383+99 x) \sqrt{2+3 x^2}+\frac{343 \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0608506, size = 55, normalized size = 0.52 \[ \frac{1}{405} \left (15435 \sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )-\sqrt{3 x^2+2} \left (432 x^4+540 x^3-12264 x^2-58860 x-118513\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[2 + 3*x^2]*(-118513 - 58860*x - 12264*x^2 + 540*x^3 + 432*x^4)) + 15435*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])/
405

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Maple [A]  time = 0.01, size = 79, normalized size = 0.8 \begin{align*} -{\frac{16\,{x}^{4}}{15}\sqrt{3\,{x}^{2}+2}}+{\frac{4088\,{x}^{2}}{135}\sqrt{3\,{x}^{2}+2}}+{\frac{118513}{405}\sqrt{3\,{x}^{2}+2}}-{\frac{4\,{x}^{3}}{3}\sqrt{3\,{x}^{2}+2}}+{\frac{436\,x}{3}\sqrt{3\,{x}^{2}+2}}+{\frac{343\,\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((5-x)*(3+2*x)^4/(3*x^2+2)^(1/2),x)

[Out]

-16/15*x^4*(3*x^2+2)^(1/2)+4088/135*x^2*(3*x^2+2)^(1/2)+118513/405*(3*x^2+2)^(1/2)-4/3*x^3*(3*x^2+2)^(1/2)+436
/3*x*(3*x^2+2)^(1/2)+343/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)

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Maxima [A]  time = 1.48804, size = 105, normalized size = 0.99 \begin{align*} -\frac{16}{15} \, \sqrt{3 \, x^{2} + 2} x^{4} - \frac{4}{3} \, \sqrt{3 \, x^{2} + 2} x^{3} + \frac{4088}{135} \, \sqrt{3 \, x^{2} + 2} x^{2} + \frac{436}{3} \, \sqrt{3 \, x^{2} + 2} x + \frac{343}{9} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{118513}{405} \, \sqrt{3 \, x^{2} + 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)^(1/2),x, algorithm="maxima")

[Out]

-16/15*sqrt(3*x^2 + 2)*x^4 - 4/3*sqrt(3*x^2 + 2)*x^3 + 4088/135*sqrt(3*x^2 + 2)*x^2 + 436/3*sqrt(3*x^2 + 2)*x
+ 343/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 118513/405*sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.84486, size = 184, normalized size = 1.74 \begin{align*} -\frac{1}{405} \,{\left (432 \, x^{4} + 540 \, x^{3} - 12264 \, x^{2} - 58860 \, x - 118513\right )} \sqrt{3 \, x^{2} + 2} + \frac{343}{18} \, \sqrt{3} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\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)^(1/2),x, algorithm="fricas")

[Out]

-1/405*(432*x^4 + 540*x^3 - 12264*x^2 - 58860*x - 118513)*sqrt(3*x^2 + 2) + 343/18*sqrt(3)*log(-sqrt(3)*sqrt(3
*x^2 + 2)*x - 3*x^2 - 1)

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Sympy [A]  time = 2.14295, size = 97, normalized size = 0.92 \begin{align*} - \frac{16 x^{4} \sqrt{3 x^{2} + 2}}{15} - \frac{4 x^{3} \sqrt{3 x^{2} + 2}}{3} + \frac{4088 x^{2} \sqrt{3 x^{2} + 2}}{135} + \frac{436 x \sqrt{3 x^{2} + 2}}{3} + \frac{118513 \sqrt{3 x^{2} + 2}}{405} + \frac{343 \sqrt{3} \operatorname{asinh}{\left (\frac{\sqrt{6} x}{2} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*x**4*sqrt(3*x**2 + 2)/15 - 4*x**3*sqrt(3*x**2 + 2)/3 + 4088*x**2*sqrt(3*x**2 + 2)/135 + 436*x*sqrt(3*x**2
+ 2)/3 + 118513*sqrt(3*x**2 + 2)/405 + 343*sqrt(3)*asinh(sqrt(6)*x/2)/9

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Giac [A]  time = 1.1416, size = 72, normalized size = 0.68 \begin{align*} -\frac{1}{405} \,{\left (12 \,{\left ({\left (9 \,{\left (4 \, x + 5\right )} x - 1022\right )} x - 4905\right )} x - 118513\right )} \sqrt{3 \, x^{2} + 2} - \frac{343}{9} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\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)^(1/2),x, algorithm="giac")

[Out]

-1/405*(12*((9*(4*x + 5)*x - 1022)*x - 4905)*x - 118513)*sqrt(3*x^2 + 2) - 343/9*sqrt(3)*log(-sqrt(3)*x + sqrt
(3*x^2 + 2))