Integrand size = 29, antiderivative size = 82 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {35}{9 \sqrt {2+3 x^2}}-\frac {47 x}{18 \sqrt {2+3 x^2}}+\frac {28}{9} \sqrt {2+3 x^2}+\frac {8}{9} x \sqrt {2+3 x^2}+\frac {4 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{3 \sqrt {3}} \] Output:
35/9/(3*x^2+2)^(1/2)-47/18*x/(3*x^2+2)^(1/2)+28/9*(3*x^2+2)^(1/2)+8/9*x*(3 *x^2+2)^(1/2)+4/9*arcsinh(1/2*x*6^(1/2))*3^(1/2)
Time = 0.42 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.74 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {182-15 x+168 x^2+48 x^3}{18 \sqrt {2+3 x^2}}-\frac {4 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{3 \sqrt {3}} \] Input:
Integrate[((1 + 2*x)^2*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]
Output:
(182 - 15*x + 168*x^2 + 48*x^3)/(18*Sqrt[2 + 3*x^2]) - (4*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/(3*Sqrt[3])
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2345, 27, 2346, 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.
\(\displaystyle \int \frac {(2 x+1)^2 \left (4 x^2+3 x+1\right )}{\left (3 x^2+2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {70-47 x}{18 \sqrt {3 x^2+2}}-\frac {1}{2} \int -\frac {8 \left (12 x^2+21 x+7\right )}{9 \sqrt {3 x^2+2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{9} \int \frac {12 x^2+21 x+7}{\sqrt {3 x^2+2}}dx+\frac {70-47 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {4}{9} \left (\frac {1}{6} \int \frac {18 (7 x+1)}{\sqrt {3 x^2+2}}dx+2 \sqrt {3 x^2+2} x\right )+\frac {70-47 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{9} \left (3 \int \frac {7 x+1}{\sqrt {3 x^2+2}}dx+2 \sqrt {3 x^2+2} x\right )+\frac {70-47 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {4}{9} \left (3 \left (\int \frac {1}{\sqrt {3 x^2+2}}dx+\frac {7}{3} \sqrt {3 x^2+2}\right )+2 \sqrt {3 x^2+2} x\right )+\frac {70-47 x}{18 \sqrt {3 x^2+2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {4}{9} \left (3 \left (\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {3}}+\frac {7}{3} \sqrt {3 x^2+2}\right )+2 \sqrt {3 x^2+2} x\right )+\frac {70-47 x}{18 \sqrt {3 x^2+2}}\) |
Input:
Int[((1 + 2*x)^2*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(3/2),x]
Output:
(70 - 47*x)/(18*Sqrt[2 + 3*x^2]) + (4*(2*x*Sqrt[2 + 3*x^2] + 3*((7*Sqrt[2 + 3*x^2])/3 + ArcSinh[Sqrt[3/2]*x]/Sqrt[3])))/9
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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 0.90 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.49
method | result | size |
risch | \(\frac {48 x^{3}+168 x^{2}-15 x +182}{18 \sqrt {3 x^{2}+2}}+\frac {4 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}\) | \(40\) |
trager | \(\frac {48 x^{3}+168 x^{2}-15 x +182}{18 \sqrt {3 x^{2}+2}}+\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{9}\) | \(57\) |
default | \(-\frac {5 x}{6 \sqrt {3 x^{2}+2}}+\frac {4 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{9}+\frac {28 x^{2}}{3 \sqrt {3 x^{2}+2}}+\frac {91}{9 \sqrt {3 x^{2}+2}}+\frac {8 x^{3}}{3 \sqrt {3 x^{2}+2}}\) | \(65\) |
meijerg | \(\frac {\sqrt {2}\, x}{4 \sqrt {\frac {3 x^{2}}{2}+1}}+\frac {20 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}}{2 \sqrt {\frac {3 x^{2}}{2}+1}}+\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )\right )}{9 \sqrt {\pi }}+\frac {7 \sqrt {2}\, \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {\frac {3 x^{2}}{2}+1}}\right )}{6 \sqrt {\pi }}+\frac {28 \sqrt {2}\, \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (6 x^{2}+8\right )}{4 \sqrt {\frac {3 x^{2}}{2}+1}}\right )}{9 \sqrt {\pi }}+\frac {32 \sqrt {3}\, \left (\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \left (\frac {15 x^{2}}{2}+15\right )}{20 \sqrt {\frac {3 x^{2}}{2}+1}}-\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )}{2}\right )}{27 \sqrt {\pi }}\) | \(174\) |
Input:
int((1+2*x)^2*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/18*(48*x^3+168*x^2-15*x+182)/(3*x^2+2)^(1/2)+4/9*arcsinh(1/2*6^(1/2)*x)* 3^(1/2)
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {4 \, \sqrt {3} {\left (3 \, x^{2} + 2\right )} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + {\left (48 \, x^{3} + 168 \, x^{2} - 15 \, x + 182\right )} \sqrt {3 \, x^{2} + 2}}{18 \, {\left (3 \, x^{2} + 2\right )}} \] Input:
integrate((1+2*x)^2*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="fricas")
Output:
1/18*(4*sqrt(3)*(3*x^2 + 2)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + (48*x^3 + 168*x^2 - 15*x + 182)*sqrt(3*x^2 + 2))/(3*x^2 + 2)
\[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\int \frac {\left (2 x + 1\right )^{2} \cdot \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} + 2\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((1+2*x)**2*(4*x**2+3*x+1)/(3*x**2+2)**(3/2),x)
Output:
Integral((2*x + 1)**2*(4*x**2 + 3*x + 1)/(3*x**2 + 2)**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {8 \, x^{3}}{3 \, \sqrt {3 \, x^{2} + 2}} + \frac {28 \, x^{2}}{3 \, \sqrt {3 \, x^{2} + 2}} + \frac {4}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) - \frac {5 \, x}{6 \, \sqrt {3 \, x^{2} + 2}} + \frac {91}{9 \, \sqrt {3 \, x^{2} + 2}} \] Input:
integrate((1+2*x)^2*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="maxima")
Output:
8/3*x^3/sqrt(3*x^2 + 2) + 28/3*x^2/sqrt(3*x^2 + 2) + 4/9*sqrt(3)*arcsinh(1 /2*sqrt(6)*x) - 5/6*x/sqrt(3*x^2 + 2) + 91/9/sqrt(3*x^2 + 2)
Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.60 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=-\frac {4}{9} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {3 \, {\left (8 \, {\left (2 \, x + 7\right )} x - 5\right )} x + 182}{18 \, \sqrt {3 \, x^{2} + 2}} \] Input:
integrate((1+2*x)^2*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x, algorithm="giac")
Output:
-4/9*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/18*(3*(8*(2*x + 7)*x - 5)*x + 182)/sqrt(3*x^2 + 2)
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.28 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {4\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{9}+\frac {\sqrt {3}\,\left (\frac {8\,x}{3}+\frac {28}{3}\right )\,\sqrt {x^2+\frac {2}{3}}}{3}+\frac {\sqrt {3}\,\sqrt {6}\,\left (-630+\sqrt {6}\,141{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{1944\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}+\frac {\sqrt {3}\,\sqrt {6}\,\left (630+\sqrt {6}\,141{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{1944\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \] Input:
int(((2*x + 1)^2*(3*x + 4*x^2 + 1))/(3*x^2 + 2)^(3/2),x)
Output:
(4*3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2))/9 + (3^(1/2)*((8*x)/3 + 28/3)*(x^ 2 + 2/3)^(1/2))/3 + (3^(1/2)*6^(1/2)*(6^(1/2)*141i - 630)*(x^2 + 2/3)^(1/2 )*1i)/(1944*(x - (6^(1/2)*1i)/3)) + (3^(1/2)*6^(1/2)*(6^(1/2)*141i + 630)* (x^2 + 2/3)^(1/2)*1i)/(1944*(x + (6^(1/2)*1i)/3))
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{3/2}} \, dx=\frac {48 \sqrt {3 x^{2}+2}\, x^{3}+168 \sqrt {3 x^{2}+2}\, x^{2}-15 \sqrt {3 x^{2}+2}\, x +182 \sqrt {3 x^{2}+2}+24 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right ) x^{2}+16 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )-39 \sqrt {3}\, x^{2}-26 \sqrt {3}}{54 x^{2}+36} \] Input:
int((1+2*x)^2*(4*x^2+3*x+1)/(3*x^2+2)^(3/2),x)
Output:
(48*sqrt(3*x**2 + 2)*x**3 + 168*sqrt(3*x**2 + 2)*x**2 - 15*sqrt(3*x**2 + 2 )*x + 182*sqrt(3*x**2 + 2) + 24*sqrt(3)*log((sqrt(3*x**2 + 2) + sqrt(3)*x) /sqrt(2))*x**2 + 16*sqrt(3)*log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2)) - 39*sqrt(3)*x**2 - 26*sqrt(3))/(18*(3*x**2 + 2))