Integrand size = 29, antiderivative size = 97 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=-\frac {199}{27} \sqrt {2+3 x^2}-\frac {1}{3} x \sqrt {2+3 x^2}+6 x^3 \sqrt {2+3 x^2}+\frac {76}{81} \left (2+3 x^2\right )^{3/2}+\frac {32}{135} \left (2+3 x^2\right )^{5/2}+\frac {5 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \] Output:
-199/27*(3*x^2+2)^(1/2)-1/3*x*(3*x^2+2)^(1/2)+6*x^3*(3*x^2+2)^(1/2)+76/81* (3*x^2+2)^(3/2)+32/135*(3*x^2+2)^(5/2)+5/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)
Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{405} \sqrt {2+3 x^2} \left (-1841-135 x+2292 x^2+2430 x^3+864 x^4\right )-\frac {5 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \] Input:
Integrate[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/Sqrt[2 + 3*x^2],x]
Output:
(Sqrt[2 + 3*x^2]*(-1841 - 135*x + 2292*x^2 + 2430*x^3 + 864*x^4))/405 - (5 *Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2185, 27, 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.
\(\displaystyle \int \frac {(2 x+1)^3 \left (4 x^2+3 x+1\right )}{\sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 2185 |
\(\displaystyle \frac {1}{60} \int -\frac {4 (17-39 x) (2 x+1)^3}{\sqrt {3 x^2+2}}dx+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{15} (2 x+1)^4 \sqrt {3 x^2+2}-\frac {1}{15} \int \frac {(17-39 x) (2 x+1)^3}{\sqrt {3 x^2+2}}dx\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \left (\frac {13}{4} (2 x+1)^3 \sqrt {3 x^2+2}-\frac {1}{12} \int \frac {3 (2 x+1)^2 (19 x+224)}{\sqrt {3 x^2+2}}dx\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \left (\frac {13}{4} (2 x+1)^3 \sqrt {3 x^2+2}-\frac {1}{4} \int \frac {(2 x+1)^2 (19 x+224)}{\sqrt {3 x^2+2}}dx\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 687 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{4} \left (-\frac {1}{9} \int \frac {2 (2 x+1) (2073 x+932)}{\sqrt {3 x^2+2}}dx-\frac {19}{9} \sqrt {3 x^2+2} (2 x+1)^2\right )+\frac {13}{4} \sqrt {3 x^2+2} (2 x+1)^3\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{4} \left (-\frac {2}{9} \int \frac {(2 x+1) (2073 x+932)}{\sqrt {3 x^2+2}}dx-\frac {19}{9} \sqrt {3 x^2+2} (2 x+1)^2\right )+\frac {13}{4} \sqrt {3 x^2+2} (2 x+1)^3\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{4} \left (-\frac {2}{9} \left (-450 \int \frac {1}{\sqrt {3 x^2+2}}dx+691 \sqrt {3 x^2+2} x+\frac {3937}{3} \sqrt {3 x^2+2}\right )-\frac {19}{9} \sqrt {3 x^2+2} (2 x+1)^2\right )+\frac {13}{4} \sqrt {3 x^2+2} (2 x+1)^3\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{4} \left (-\frac {2}{9} \left (-150 \sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )+691 \sqrt {3 x^2+2} x+\frac {3937}{3} \sqrt {3 x^2+2}\right )-\frac {19}{9} \sqrt {3 x^2+2} (2 x+1)^2\right )+\frac {13}{4} \sqrt {3 x^2+2} (2 x+1)^3\right )+\frac {2}{15} \sqrt {3 x^2+2} (2 x+1)^4\) |
Input:
Int[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/Sqrt[2 + 3*x^2],x]
Output:
(2*(1 + 2*x)^4*Sqrt[2 + 3*x^2])/15 + ((13*(1 + 2*x)^3*Sqrt[2 + 3*x^2])/4 + ((-19*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/9 - (2*((3937*Sqrt[2 + 3*x^2])/3 + 691 *x*Sqrt[2 + 3*x^2] - 150*Sqrt[3]*ArcSinh[Sqrt[3/2]*x]))/9)/4)/15
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])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x )^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p )*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d , e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && !(EqQ[d, 0] && True) && !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.89 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.46
method | result | size |
risch | \(\frac {\left (864 x^{4}+2430 x^{3}+2292 x^{2}-135 x -1841\right ) \sqrt {3 x^{2}+2}}{405}+\frac {5 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}\) | \(45\) |
trager | \(\left (\frac {32}{15} x^{4}+6 x^{3}+\frac {764}{135} x^{2}-\frac {1}{3} x -\frac {1841}{405}\right ) \sqrt {3 x^{2}+2}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(62\) |
default | \(\frac {5 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}-\frac {1841 \sqrt {3 x^{2}+2}}{405}-\frac {x \sqrt {3 x^{2}+2}}{3}+\frac {764 x^{2} \sqrt {3 x^{2}+2}}{135}+6 x^{3} \sqrt {3 x^{2}+2}+\frac {32 x^{4} \sqrt {3 x^{2}+2}}{15}\) | \(79\) |
meijerg | \(\frac {\sqrt {3}\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )}{3}+\frac {34 \sqrt {3}\, \left (\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \sqrt {\frac {3 x^{2}}{2}+1}}{2}-\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )\right )}{9 \sqrt {\pi }}+\frac {3 \sqrt {2}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {\frac {3 x^{2}}{2}+1}\right )}{2 \sqrt {\pi }}+\frac {68 \sqrt {2}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-6 x^{2}+8\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{6}\right )}{9 \sqrt {\pi }}+\frac {16 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \left (-15 x^{2}+15\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{40}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )}{4}\right )}{3 \sqrt {\pi }}+\frac {64 \sqrt {2}\, \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (\frac {27}{2} x^{4}-12 x^{2}+16\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{15}\right )}{27 \sqrt {\pi }}\) | \(217\) |
Input:
int((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/405*(864*x^4+2430*x^3+2292*x^2-135*x-1841)*(3*x^2+2)^(1/2)+5/9*arcsinh(1 /2*6^(1/2)*x)*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{405} \, {\left (864 \, x^{4} + 2430 \, x^{3} + 2292 \, x^{2} - 135 \, x - 1841\right )} \sqrt {3 \, x^{2} + 2} + \frac {5}{18} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="fricas")
Output:
1/405*(864*x^4 + 2430*x^3 + 2292*x^2 - 135*x - 1841)*sqrt(3*x^2 + 2) + 5/1 8*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)
Time = 0.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {32 x^{4} \sqrt {3 x^{2} + 2}}{15} + 6 x^{3} \sqrt {3 x^{2} + 2} + \frac {764 x^{2} \sqrt {3 x^{2} + 2}}{135} - \frac {x \sqrt {3 x^{2} + 2}}{3} - \frac {1841 \sqrt {3 x^{2} + 2}}{405} + \frac {5 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{9} \] Input:
integrate((1+2*x)**3*(4*x**2+3*x+1)/(3*x**2+2)**(1/2),x)
Output:
32*x**4*sqrt(3*x**2 + 2)/15 + 6*x**3*sqrt(3*x**2 + 2) + 764*x**2*sqrt(3*x* *2 + 2)/135 - x*sqrt(3*x**2 + 2)/3 - 1841*sqrt(3*x**2 + 2)/405 + 5*sqrt(3) *asinh(sqrt(6)*x/2)/9
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {32}{15} \, \sqrt {3 \, x^{2} + 2} x^{4} + 6 \, \sqrt {3 \, x^{2} + 2} x^{3} + \frac {764}{135} \, \sqrt {3 \, x^{2} + 2} x^{2} - \frac {1}{3} \, \sqrt {3 \, x^{2} + 2} x + \frac {5}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) - \frac {1841}{405} \, \sqrt {3 \, x^{2} + 2} \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="maxima")
Output:
32/15*sqrt(3*x^2 + 2)*x^4 + 6*sqrt(3*x^2 + 2)*x^3 + 764/135*sqrt(3*x^2 + 2 )*x^2 - 1/3*sqrt(3*x^2 + 2)*x + 5/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 1841/ 405*sqrt(3*x^2 + 2)
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.56 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{405} \, {\left (3 \, {\left (2 \, {\left (9 \, {\left (16 \, x + 45\right )} x + 382\right )} x - 45\right )} x - 1841\right )} \sqrt {3 \, x^{2} + 2} - \frac {5}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x, algorithm="giac")
Output:
1/405*(3*(2*(9*(16*x + 45)*x + 382)*x - 45)*x - 1841)*sqrt(3*x^2 + 2) - 5/ 9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.46 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {5\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {32\,x^4}{5}+18\,x^3+\frac {764\,x^2}{45}-x-\frac {1841}{135}\right )}{3} \] Input:
int(((2*x + 1)^3*(3*x + 4*x^2 + 1))/(3*x^2 + 2)^(1/2),x)
Output:
(5*3^(1/2)*asinh((6^(1/2)*x)/2))/9 + (3^(1/2)*(x^2 + 2/3)^(1/2)*((764*x^2) /45 - x + 18*x^3 + (32*x^4)/5 - 1841/135))/3
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.87 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {32 \sqrt {3 x^{2}+2}\, x^{4}}{15}+6 \sqrt {3 x^{2}+2}\, x^{3}+\frac {764 \sqrt {3 x^{2}+2}\, x^{2}}{135}-\frac {\sqrt {3 x^{2}+2}\, x}{3}-\frac {1841 \sqrt {3 x^{2}+2}}{405}+\frac {5 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )}{9} \] Input:
int((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(1/2),x)
Output:
(864*sqrt(3*x**2 + 2)*x**4 + 2430*sqrt(3*x**2 + 2)*x**3 + 2292*sqrt(3*x**2 + 2)*x**2 - 135*sqrt(3*x**2 + 2)*x - 1841*sqrt(3*x**2 + 2) + 225*sqrt(3)* log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2)))/405