Integrand size = 27, antiderivative size = 66 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {1}{9 \sqrt {2+3 x^2}}-\frac {17 x}{6 \sqrt {2+3 x^2}}+\frac {8}{9} \sqrt {2+3 x^2}+\frac {10 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \] Output:
1/9/(3*x^2+2)^(1/2)-17/6*x/(3*x^2+2)^(1/2)+8/9*(3*x^2+2)^(1/2)+10/9*arcsin h(1/2*x*6^(1/2))*3^(1/2)
Time = 0.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {34-51 x+48 x^2}{18 \sqrt {2+3 x^2}}-\frac {10 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \] Input:
Integrate[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]
Output:
(34 - 51*x + 48*x^2)/(18*Sqrt[2 + 3*x^2]) - (10*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2345, 27, 455, 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.
\(\displaystyle \int \frac {(2 x+1) \left (4 x^2+3 x+1\right )}{\left (3 x^2+2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {2-51 x}{18 \sqrt {3 x^2+2}}-\frac {1}{2} \int -\frac {4 (4 x+5)}{3 \sqrt {3 x^2+2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int \frac {4 x+5}{\sqrt {3 x^2+2}}dx+\frac {2-51 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {2}{3} \left (5 \int \frac {1}{\sqrt {3 x^2+2}}dx+\frac {4}{3} \sqrt {3 x^2+2}\right )+\frac {2-51 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {2}{3} \left (\frac {5 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {3}}+\frac {4}{3} \sqrt {3 x^2+2}\right )+\frac {2-51 x}{18 \sqrt {3 x^2+2}}\) |
Input:
Int[((1 + 2*x)*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]
Output:
(2 - 51*x)/(18*Sqrt[2 + 3*x^2]) + (2*((4*Sqrt[2 + 3*x^2])/3 + (5*ArcSinh[S qrt[3/2]*x])/Sqrt[3]))/3
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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.77 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.53
method | result | size |
risch | \(\frac {48 x^{2}-51 x +34}{18 \sqrt {3 x^{2}+2}}+\frac {10 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}\) | \(35\) |
default | \(-\frac {17 x}{6 \sqrt {3 x^{2}+2}}+\frac {10 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}+\frac {8 x^{2}}{3 \sqrt {3 x^{2}+2}}+\frac {17}{9 \sqrt {3 x^{2}+2}}\) | \(51\) |
trager | \(\frac {48 x^{2}-51 x +34}{18 \sqrt {3 x^{2}+2}}-\frac {10 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(53\) |
meijerg | \(\frac {\sqrt {2}\, x}{4 \sqrt {\frac {3 x^{2}}{2}+1}}+\frac {10 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}}{2 \sqrt {\frac {3 x^{2}}{2}+1}}+\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )\right )}{9 \sqrt {\pi }}+\frac {5 \sqrt {2}\, \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {\frac {3 x^{2}}{2}+1}}\right )}{6 \sqrt {\pi }}+\frac {8 \sqrt {2}\, \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (6 x^{2}+8\right )}{4 \sqrt {\frac {3 x^{2}}{2}+1}}\right )}{9 \sqrt {\pi }}\) | \(122\) |
Input:
int((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/18*(48*x^2-51*x+34)/(3*x^2+2)^(1/2)+10/9*arcsinh(1/2*6^(1/2)*x)*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {10 \, \sqrt {3} {\left (3 \, x^{2} + 2\right )} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + {\left (48 \, x^{2} - 51 \, x + 34\right )} \sqrt {3 \, x^{2} + 2}}{18 \, {\left (3 \, x^{2} + 2\right )}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="fricas")
Output:
1/18*(10*sqrt(3)*(3*x^2 + 2)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + (48*x^2 - 51*x + 34)*sqrt(3*x^2 + 2))/(3*x^2 + 2)
Time = 5.50 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.73 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {30 \sqrt {3} x^{2} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{27 x^{2} + 18} + \frac {8 x^{2}}{3 \sqrt {3 x^{2} + 2}} - \frac {30 x \sqrt {3 x^{2} + 2}}{27 x^{2} + 18} + \frac {x}{2 \sqrt {3 x^{2} + 2}} + \frac {20 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{27 x^{2} + 18} + \frac {17}{9 \sqrt {3 x^{2} + 2}} \] Input:
integrate((1+2*x)*(4*x**2+3*x+1)/(3*x**2+2)**(3/2),x)
Output:
30*sqrt(3)*x**2*asinh(sqrt(6)*x/2)/(27*x**2 + 18) + 8*x**2/(3*sqrt(3*x**2 + 2)) - 30*x*sqrt(3*x**2 + 2)/(27*x**2 + 18) + x/(2*sqrt(3*x**2 + 2)) + 20 *sqrt(3)*asinh(sqrt(6)*x/2)/(27*x**2 + 18) + 17/(9*sqrt(3*x**2 + 2))
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {8 \, x^{2}}{3 \, \sqrt {3 \, x^{2} + 2}} + \frac {10}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) - \frac {17 \, x}{6 \, \sqrt {3 \, x^{2} + 2}} + \frac {17}{9 \, \sqrt {3 \, x^{2} + 2}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="maxima")
Output:
8/3*x^2/sqrt(3*x^2 + 2) + 10/9*sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 17/6*x/sqr t(3*x^2 + 2) + 17/9/sqrt(3*x^2 + 2)
Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.67 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=-\frac {10}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {3 \, {\left (16 \, x - 17\right )} x + 34}{18 \, \sqrt {3 \, x^{2} + 2}} \] Input:
integrate((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="giac")
Output:
-10/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/18*(3*(16*x - 17)*x + 34)/sqrt(3*x^2 + 2)
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.52 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {8\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{9}+\frac {10\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\sqrt {6}\,\left (-6+\sqrt {6}\,51{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{648\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}+\frac {\sqrt {3}\,\sqrt {6}\,\left (6+\sqrt {6}\,51{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{648\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \] Input:
int(((2*x + 1)*(3*x + 4*x^2 + 1))/(3*x^2 + 2)^(3/2),x)
Output:
(8*3^(1/2)*(x^2 + 2/3)^(1/2))/9 + (10*3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2) )/9 + (3^(1/2)*6^(1/2)*(6^(1/2)*51i - 6)*(x^2 + 2/3)^(1/2)*1i)/(648*(x - ( 6^(1/2)*1i)/3)) + (3^(1/2)*6^(1/2)*(6^(1/2)*51i + 6)*(x^2 + 2/3)^(1/2)*1i) /(648*(x + (6^(1/2)*1i)/3))
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.59 \[ \int \frac {(1+2 x) \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {48 \sqrt {3 x^{2}+2}\, x^{2}-51 \sqrt {3 x^{2}+2}\, x +34 \sqrt {3 x^{2}+2}+60 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right ) x^{2}+40 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )-51 \sqrt {3}\, x^{2}-34 \sqrt {3}}{54 x^{2}+36} \] Input:
int((1+2*x)*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x)
Output:
(48*sqrt(3*x**2 + 2)*x**2 - 51*sqrt(3*x**2 + 2)*x + 34*sqrt(3*x**2 + 2) + 60*sqrt(3)*log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2))*x**2 + 40*sqrt(3)*l og((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2)) - 51*sqrt(3)*x**2 - 34*sqrt(3)) /(18*(3*x**2 + 2))