Integrand size = 24, antiderivative size = 78 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=\frac {131}{6} x \sqrt {2+3 x^2}-\frac {1}{15} (3+2 x)^2 \left (2+3 x^2\right )^{3/2}+\frac {2}{135} (431+99 x) \left (2+3 x^2\right )^{3/2}+\frac {131 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \]
-1/15*(3+2*x)^2*(3*x^2+2)^(3/2)+2/135*(431+99*x)*(3*x^2+2)^(3/2)+131/9*arc sinh(1/2*x*6^(1/2))*3^(1/2)+131/6*x*(3*x^2+2)^(1/2)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=-\frac {1}{270} \sqrt {2+3 x^2} \left (-3124-6255 x-4542 x^2-540 x^3+216 x^4\right )-\frac {131 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \]
-1/270*(Sqrt[2 + 3*x^2]*(-3124 - 6255*x - 4542*x^2 - 540*x^3 + 216*x^4)) - (131*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {687, 676, 211, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (5-x) (2 x+3)^2 \sqrt {3 x^2+2} \, dx\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \int (2 x+3) (132 x+233) \sqrt {3 x^2+2}dx-\frac {1}{15} (2 x+3)^2 \left (3 x^2+2\right )^{3/2}\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {1}{15} \left (655 \int \sqrt {3 x^2+2}dx+22 x \left (3 x^2+2\right )^{3/2}+\frac {862}{9} \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+2\right )^{3/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{15} \left (655 \left (\int \frac {1}{\sqrt {3 x^2+2}}dx+\frac {1}{2} \sqrt {3 x^2+2} x\right )+22 x \left (3 x^2+2\right )^{3/2}+\frac {862}{9} \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+2\right )^{3/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{15} \left (655 \left (\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {3}}+\frac {1}{2} \sqrt {3 x^2+2} x\right )+22 x \left (3 x^2+2\right )^{3/2}+\frac {862}{9} \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+2\right )^{3/2}\) |
-1/15*((3 + 2*x)^2*(2 + 3*x^2)^(3/2)) + ((862*(2 + 3*x^2)^(3/2))/9 + 22*x* (2 + 3*x^2)^(3/2) + 655*((x*Sqrt[2 + 3*x^2])/2 + ArcSinh[Sqrt[3/2]*x]/Sqrt [3]))/15
3.14.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(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]
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] + Simp[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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && Eq Q[f, 0])
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.58
method | result | size |
risch | \(-\frac {\left (216 x^{4}-540 x^{3}-4542 x^{2}-6255 x -3124\right ) \sqrt {3 x^{2}+2}}{270}+\frac {131 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}\) | \(45\) |
trager | \(\left (-\frac {4}{5} x^{4}+2 x^{3}+\frac {757}{45} x^{2}+\frac {139}{6} x +\frac {1562}{135}\right ) \sqrt {3 x^{2}+2}+\frac {131 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(61\) |
default | \(\frac {131 x \sqrt {3 x^{2}+2}}{6}+\frac {131 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}+\frac {781 \left (3 x^{2}+2\right )^{\frac {3}{2}}}{135}-\frac {4 x^{2} \left (3 x^{2}+2\right )^{\frac {3}{2}}}{15}+\frac {2 x \left (3 x^{2}+2\right )^{\frac {3}{2}}}{3}\) | \(63\) |
meijerg | \(-\frac {15 \sqrt {3}\, \left (-\sqrt {6}\, \sqrt {\pi }\, x \sqrt {\frac {3 x^{2}}{2}+1}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )\right )}{2 \sqrt {\pi }}-\frac {17 \sqrt {2}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (3 x^{2}+2\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{3}\right )}{2 \sqrt {\pi }}-\frac {8 \sqrt {3}\, \left (-\frac {\sqrt {6}\, \sqrt {\pi }\, x \left (9 x^{2}+3\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{12}+\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )}{2}\right )}{9 \sqrt {\pi }}+\frac {4 \sqrt {2}\, \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}} \left (-\frac {9 x^{2}}{2}+2\right )}{15}\right )}{9 \sqrt {\pi }}\) | \(163\) |
-1/270*(216*x^4-540*x^3-4542*x^2-6255*x-3124)*(3*x^2+2)^(1/2)+131/9*arcsin h(1/2*x*6^(1/2))*3^(1/2)
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=-\frac {1}{270} \, {\left (216 \, x^{4} - 540 \, x^{3} - 4542 \, x^{2} - 6255 \, x - 3124\right )} \sqrt {3 \, x^{2} + 2} + \frac {131}{18} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \]
-1/270*(216*x^4 - 540*x^3 - 4542*x^2 - 6255*x - 3124)*sqrt(3*x^2 + 2) + 13 1/18*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.22 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=- \frac {4 x^{4} \sqrt {3 x^{2} + 2}}{5} + 2 x^{3} \sqrt {3 x^{2} + 2} + \frac {757 x^{2} \sqrt {3 x^{2} + 2}}{45} + \frac {139 x \sqrt {3 x^{2} + 2}}{6} + \frac {1562 \sqrt {3 x^{2} + 2}}{135} + \frac {131 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{9} \]
-4*x**4*sqrt(3*x**2 + 2)/5 + 2*x**3*sqrt(3*x**2 + 2) + 757*x**2*sqrt(3*x** 2 + 2)/45 + 139*x*sqrt(3*x**2 + 2)/6 + 1562*sqrt(3*x**2 + 2)/135 + 131*sqr t(3)*asinh(sqrt(6)*x/2)/9
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.79 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=-\frac {4}{15} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} x^{2} + \frac {2}{3} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} x + \frac {781}{135} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} + \frac {131}{6} \, \sqrt {3 \, x^{2} + 2} x + \frac {131}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) \]
-4/15*(3*x^2 + 2)^(3/2)*x^2 + 2/3*(3*x^2 + 2)^(3/2)*x + 781/135*(3*x^2 + 2 )^(3/2) + 131/6*sqrt(3*x^2 + 2)*x + 131/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x)
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.69 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=-\frac {1}{270} \, {\left (3 \, {\left (2 \, {\left (18 \, {\left (2 \, x - 5\right )} x - 757\right )} x - 2085\right )} x - 3124\right )} \sqrt {3 \, x^{2} + 2} - \frac {131}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \]
-1/270*(3*(2*(18*(2*x - 5)*x - 757)*x - 2085)*x - 3124)*sqrt(3*x^2 + 2) - 131/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))
Time = 10.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.58 \[ \int (5-x) (3+2 x)^2 \sqrt {2+3 x^2} \, dx=\frac {131\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (-\frac {12\,x^4}{5}+6\,x^3+\frac {757\,x^2}{15}+\frac {139\,x}{2}+\frac {1562}{45}\right )}{3} \]