Integrand size = 24, antiderivative size = 84 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=\frac {31}{36} (3+2 x)^2 \sqrt {2+3 x^2}-\frac {1}{12} (3+2 x)^3 \sqrt {2+3 x^2}+\frac {5}{54} (809+171 x) \sqrt {2+3 x^2}+\frac {275 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \]
275/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)+31/36*(3+2*x)^2*(3*x^2+2)^(1/2)-1/12* (3+2*x)^3*(3*x^2+2)^(1/2)+5/54*(809+171*x)*(3*x^2+2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{27} \sqrt {2+3 x^2} \left (-2171-585 x-12 x^2+18 x^3\right )-\frac {275 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \]
-1/27*(Sqrt[2 + 3*x^2]*(-2171 - 585*x - 12*x^2 + 18*x^3)) - (275*Log[-(Sqr t[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {687, 27, 687, 27, 676, 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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^3}{\sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{12} \int \frac {3 (2 x+3)^2 (31 x+64)}{\sqrt {3 x^2+2}}dx-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {(2 x+3)^2 (31 x+64)}{\sqrt {3 x^2+2}}dx-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{9} \int \frac {10 (2 x+3) (171 x+148)}{\sqrt {3 x^2+2}}dx+\frac {31}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {10}{9} \int \frac {(2 x+3) (171 x+148)}{\sqrt {3 x^2+2}}dx+\frac {31}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {1}{4} \left (\frac {10}{9} \left (330 \int \frac {1}{\sqrt {3 x^2+2}}dx+57 \sqrt {3 x^2+2} x+\frac {809}{3} \sqrt {3 x^2+2}\right )+\frac {31}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{4} \left (\frac {10}{9} \left (110 \sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )+57 \sqrt {3 x^2+2} x+\frac {809}{3} \sqrt {3 x^2+2}\right )+\frac {31}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+2}\) |
-1/12*((3 + 2*x)^3*Sqrt[2 + 3*x^2]) + ((31*(3 + 2*x)^2*Sqrt[2 + 3*x^2])/9 + (10*((809*Sqrt[2 + 3*x^2])/3 + 57*x*Sqrt[2 + 3*x^2] + 110*Sqrt[3]*ArcSin h[Sqrt[3/2]*x]))/9)/4
3.14.98.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.48
method | result | size |
risch | \(-\frac {\left (18 x^{3}-12 x^{2}-585 x -2171\right ) \sqrt {3 x^{2}+2}}{27}+\frac {275 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}\) | \(40\) |
trager | \(\left (-\frac {2}{3} x^{3}+\frac {4}{9} x^{2}+\frac {65}{3} x +\frac {2171}{27}\right ) \sqrt {3 x^{2}+2}-\frac {275 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(57\) |
default | \(\frac {275 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}+\frac {2171 \sqrt {3 x^{2}+2}}{27}-\frac {2 x^{3} \sqrt {3 x^{2}+2}}{3}+\frac {65 x \sqrt {3 x^{2}+2}}{3}+\frac {4 x^{2} \sqrt {3 x^{2}+2}}{9}\) | \(65\) |
meijerg | \(45 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )+\frac {81 \sqrt {2}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {\frac {3 x^{2}}{2}+1}\right )}{2 \sqrt {\pi }}+\frac {14 \sqrt {3}\, \left (\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \sqrt {\frac {3 x^{2}}{2}+1}}{2}-\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )\right )}{\sqrt {\pi }}+\frac {4 \sqrt {2}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-6 x^{2}+8\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{6}\right )}{9 \sqrt {\pi }}-\frac {16 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \left (-15 x^{2}+15\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{40}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )}{4}\right )}{27 \sqrt {\pi }}\) | \(177\) |
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{27} \, {\left (18 \, x^{3} - 12 \, x^{2} - 585 \, x - 2171\right )} \sqrt {3 \, x^{2} + 2} + \frac {275}{18} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \]
-1/27*(18*x^3 - 12*x^2 - 585*x - 2171)*sqrt(3*x^2 + 2) + 275/18*sqrt(3)*lo g(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=- \frac {2 x^{3} \sqrt {3 x^{2} + 2}}{3} + \frac {4 x^{2} \sqrt {3 x^{2} + 2}}{9} + \frac {65 x \sqrt {3 x^{2} + 2}}{3} + \frac {2171 \sqrt {3 x^{2} + 2}}{27} + \frac {275 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{9} \]
-2*x**3*sqrt(3*x**2 + 2)/3 + 4*x**2*sqrt(3*x**2 + 2)/9 + 65*x*sqrt(3*x**2 + 2)/3 + 2171*sqrt(3*x**2 + 2)/27 + 275*sqrt(3)*asinh(sqrt(6)*x/2)/9
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=-\frac {2}{3} \, \sqrt {3 \, x^{2} + 2} x^{3} + \frac {4}{9} \, \sqrt {3 \, x^{2} + 2} x^{2} + \frac {65}{3} \, \sqrt {3 \, x^{2} + 2} x + \frac {275}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {2171}{27} \, \sqrt {3 \, x^{2} + 2} \]
-2/3*sqrt(3*x^2 + 2)*x^3 + 4/9*sqrt(3*x^2 + 2)*x^2 + 65/3*sqrt(3*x^2 + 2)* x + 275/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 2171/27*sqrt(3*x^2 + 2)
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.58 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{27} \, {\left (3 \, {\left (2 \, {\left (3 \, x - 2\right )} x - 195\right )} x - 2171\right )} \sqrt {3 \, x^{2} + 2} - \frac {275}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \]
-1/27*(3*(2*(3*x - 2)*x - 195)*x - 2171)*sqrt(3*x^2 + 2) - 275/9*sqrt(3)*l og(-sqrt(3)*x + sqrt(3*x^2 + 2))
Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.48 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+3 x^2}} \, dx=\frac {275\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (-2\,x^3+\frac {4\,x^2}{3}+65\,x+\frac {2171}{9}\right )}{3} \]