Integrand size = 24, antiderivative size = 106 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=\frac {1477}{270} (3+2 x)^2 \sqrt {2+3 x^2}+\frac {19}{30} (3+2 x)^3 \sqrt {2+3 x^2}-\frac {1}{15} (3+2 x)^4 \sqrt {2+3 x^2}+\frac {49}{81} (383+99 x) \sqrt {2+3 x^2}+\frac {343 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \]
343/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)+1477/270*(3+2*x)^2*(3*x^2+2)^(1/2)+19 /30*(3+2*x)^3*(3*x^2+2)^(1/2)-1/15*(3+2*x)^4*(3*x^2+2)^(1/2)+49/81*(383+99 *x)*(3*x^2+2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{405} \sqrt {2+3 x^2} \left (-118513-58860 x-12264 x^2+540 x^3+432 x^4\right )-\frac {343 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \]
-1/405*(Sqrt[2 + 3*x^2]*(-118513 - 58860*x - 12264*x^2 + 540*x^3 + 432*x^4 )) - (343*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {687, 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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^4}{\sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \int \frac {(2 x+3)^3 (114 x+241)}{\sqrt {3 x^2+2}}dx-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{12} \int \frac {42 (2 x+3)^2 (211 x+174)}{\sqrt {3 x^2+2}}dx+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \left (\frac {7}{2} \int \frac {(2 x+3)^2 (211 x+174)}{\sqrt {3 x^2+2}}dx+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \left (\frac {7}{2} \left (\frac {1}{9} \int \frac {70 (2 x+3) (99 x+43)}{\sqrt {3 x^2+2}}dx+\frac {211}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \left (\frac {7}{2} \left (\frac {70}{9} \int \frac {(2 x+3) (99 x+43)}{\sqrt {3 x^2+2}}dx+\frac {211}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {1}{15} \left (\frac {7}{2} \left (\frac {70}{9} \left (63 \int \frac {1}{\sqrt {3 x^2+2}}dx+33 \sqrt {3 x^2+2} x+\frac {383}{3} \sqrt {3 x^2+2}\right )+\frac {211}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{15} \left (\frac {7}{2} \left (\frac {70}{9} \left (21 \sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )+33 \sqrt {3 x^2+2} x+\frac {383}{3} \sqrt {3 x^2+2}\right )+\frac {211}{9} \sqrt {3 x^2+2} (2 x+3)^2\right )+\frac {19}{2} \sqrt {3 x^2+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+2}\) |
-1/15*((3 + 2*x)^4*Sqrt[2 + 3*x^2]) + ((19*(3 + 2*x)^3*Sqrt[2 + 3*x^2])/2 + (7*((211*(3 + 2*x)^2*Sqrt[2 + 3*x^2])/9 + (70*((383*Sqrt[2 + 3*x^2])/3 + 33*x*Sqrt[2 + 3*x^2] + 21*Sqrt[3]*ArcSinh[Sqrt[3/2]*x]))/9))/2)/15
3.14.97.3.1 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 = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42
method | result | size |
risch | \(-\frac {\left (432 x^{4}+540 x^{3}-12264 x^{2}-58860 x -118513\right ) \sqrt {3 x^{2}+2}}{405}+\frac {343 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}\) | \(45\) |
trager | \(\left (-\frac {16}{15} x^{4}-\frac {4}{3} x^{3}+\frac {4088}{135} x^{2}+\frac {436}{3} x +\frac {118513}{405}\right ) \sqrt {3 x^{2}+2}+\frac {343 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(61\) |
default | \(\frac {343 \,\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{9}+\frac {118513 \sqrt {3 x^{2}+2}}{405}-\frac {16 x^{4} \sqrt {3 x^{2}+2}}{15}+\frac {4088 x^{2} \sqrt {3 x^{2}+2}}{135}-\frac {4 x^{3} \sqrt {3 x^{2}+2}}{3}+\frac {436 x \sqrt {3 x^{2}+2}}{3}\) | \(79\) |
meijerg | \(135 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )+\frac {333 \sqrt {2}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {\frac {3 x^{2}}{2}+1}\right )}{2 \sqrt {\pi }}+\frac {96 \sqrt {3}\, \left (\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \sqrt {\frac {3 x^{2}}{2}+1}}{2}-\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )\right )}{\sqrt {\pi }}+\frac {88 \sqrt {2}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-6 x^{2}+8\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{6}\right )}{3 \sqrt {\pi }}-\frac {32 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \left (-15 x^{2}+15\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{40}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )}{4}\right )}{27 \sqrt {\pi }}-\frac {32 \sqrt {2}\, \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (\frac {27}{2} x^{4}-12 x^{2}+16\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{15}\right )}{27 \sqrt {\pi }}\) | \(217\) |
-1/405*(432*x^4+540*x^3-12264*x^2-58860*x-118513)*(3*x^2+2)^(1/2)+343/9*ar csinh(1/2*x*6^(1/2))*3^(1/2)
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.57 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{405} \, {\left (432 \, x^{4} + 540 \, x^{3} - 12264 \, x^{2} - 58860 \, x - 118513\right )} \sqrt {3 \, x^{2} + 2} + \frac {343}{18} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \]
-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)
Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=- \frac {16 x^{4} \sqrt {3 x^{2} + 2}}{15} - \frac {4 x^{3} \sqrt {3 x^{2} + 2}}{3} + \frac {4088 x^{2} \sqrt {3 x^{2} + 2}}{135} + \frac {436 x \sqrt {3 x^{2} + 2}}{3} + \frac {118513 \sqrt {3 x^{2} + 2}}{405} + \frac {343 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{9} \]
-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
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.74 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=-\frac {16}{15} \, \sqrt {3 \, x^{2} + 2} x^{4} - \frac {4}{3} \, \sqrt {3 \, x^{2} + 2} x^{3} + \frac {4088}{135} \, \sqrt {3 \, x^{2} + 2} x^{2} + \frac {436}{3} \, \sqrt {3 \, x^{2} + 2} x + \frac {343}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {118513}{405} \, \sqrt {3 \, x^{2} + 2} \]
-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)
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.50 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=-\frac {1}{405} \, {\left (12 \, {\left ({\left (9 \, {\left (4 \, x + 5\right )} x - 1022\right )} x - 4905\right )} x - 118513\right )} \sqrt {3 \, x^{2} + 2} - \frac {343}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \]
-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))
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+3 x^2}} \, dx=\frac {343\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (-\frac {16\,x^4}{5}-4\,x^3+\frac {4088\,x^2}{45}+436\,x+\frac {118513}{135}\right )}{3} \]