Integrand size = 29, antiderivative size = 95 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {199}{81 \left (2+3 x^2\right )^{3/2}}+\frac {31 x}{18 \left (2+3 x^2\right )^{3/2}}-\frac {76}{27 \sqrt {2+3 x^2}}-\frac {155 x}{18 \sqrt {2+3 x^2}}+\frac {32}{27} \sqrt {2+3 x^2}+\frac {8 \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {3}} \] Output:
199/81/(3*x^2+2)^(3/2)+31/18*x/(3*x^2+2)^(3/2)-76/27/(3*x^2+2)^(1/2)-155/1 8*x/(3*x^2+2)^(1/2)+32/27*(3*x^2+2)^(1/2)+8/3*arcsinh(1/2*x*6^(1/2))*3^(1/ 2)
Time = 0.51 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {254-2511 x+936 x^2-4185 x^3+1728 x^4}{162 \left (2+3 x^2\right )^{3/2}}-\frac {8 \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right )}{\sqrt {3}} \] Input:
Integrate[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(5/2),x]
Output:
(254 - 2511*x + 936*x^2 - 4185*x^3 + 1728*x^4)/(162*(2 + 3*x^2)^(3/2)) - ( 8*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]])/Sqrt[3]
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2345, 27, 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.
| \(\displaystyle \int \frac {(2 x+1)^3 \left (4 x^2+3 x+1\right )}{\left (3 x^2+2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}-\frac {1}{6} \int \frac {2 \left (-96 x^3-216 x^2-140 x+11\right )}{3 \left (3 x^2+2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}-\frac {1}{9} \int \frac {-96 x^3-216 x^2-140 x+11}{\left (3 x^2+2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{2} \int \frac {16 (4 x+9)}{\sqrt {3 x^2+2}}dx-\frac {465 x+152}{6 \sqrt {3 x^2+2}}\right )+\frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (8 \int \frac {4 x+9}{\sqrt {3 x^2+2}}dx-\frac {465 x+152}{6 \sqrt {3 x^2+2}}\right )+\frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {1}{9} \left (8 \left (9 \int \frac {1}{\sqrt {3 x^2+2}}dx+\frac {4}{3} \sqrt {3 x^2+2}\right )-\frac {465 x+152}{6 \sqrt {3 x^2+2}}\right )+\frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{9} \left (8 \left (3 \sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )+\frac {4}{3} \sqrt {3 x^2+2}\right )-\frac {465 x+152}{6 \sqrt {3 x^2+2}}\right )+\frac {279 x+398}{162 \left (3 x^2+2\right )^{3/2}}\) |
Input:
Int[((1 + 2*x)^3*(1 + 3*x + 4*x^2))/(2 + 3*x^2)^(5/2),x]
Output:
(398 + 279*x)/(162*(2 + 3*x^2)^(3/2)) + (-1/6*(152 + 465*x)/Sqrt[2 + 3*x^2 ] + 8*((4*Sqrt[2 + 3*x^2])/3 + 3*Sqrt[3]*ArcSinh[Sqrt[3/2]*x]))/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]
Time = 0.91 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.47
| method | result | size |
| risch | \(\frac {1728 x^{4}-4185 x^{3}+936 x^{2}-2511 x +254}{162 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {8 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{3}\) | \(45\) |
| trager | \(\frac {1728 x^{4}-4185 x^{3}+936 x^{2}-2511 x +254}{162 \left (3 x^{2}+2\right )^{\frac {3}{2}}}-\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{3}\) | \(63\) |
| default | \(-\frac {65 x}{18 \left (3 x^{2}+2\right )^{\frac {3}{2}}}-\frac {107 x}{18 \sqrt {3 x^{2}+2}}+\frac {127}{81 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {52 x^{2}}{9 \left (3 x^{2}+2\right )^{\frac {3}{2}}}-\frac {8 x^{3}}{\left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {8 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{3}+\frac {32 x^{4}}{3 \left (3 x^{2}+2\right )^{\frac {3}{2}}}\) | \(91\) |
| meijerg | \(\frac {\sqrt {2}\, x \left (3 x^{2}+3\right )}{24 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}+\frac {17 \sqrt {2}\, x^{3}}{12 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \left (\frac {\sqrt {\pi }}{2}-\frac {\sqrt {\pi }}{2 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}\right )}{2 \sqrt {\pi }}+\frac {68 \sqrt {2}\, \left (\sqrt {\pi }-\frac {\sqrt {\pi }\, \left (18 x^{2}+8\right )}{8 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}\right )}{27 \sqrt {\pi }}+\frac {16 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \left (30 x^{2}+15\right )}{20 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )}{2}\right )}{9 \sqrt {\pi }}+\frac {64 \sqrt {2}\, \left (-4 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (\frac {27}{2} x^{4}+36 x^{2}+16\right )}{4 \left (\frac {3 x^{2}}{2}+1\right )^{\frac {3}{2}}}\right )}{81 \sqrt {\pi }}\) | \(194\) |
Input:
int((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/162*(1728*x^4-4185*x^3+936*x^2-2511*x+254)/(3*x^2+2)^(3/2)+8/3*arcsinh(1 /2*6^(1/2)*x)*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {216 \, \sqrt {3} {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + {\left (1728 \, x^{4} - 4185 \, x^{3} + 936 \, x^{2} - 2511 \, x + 254\right )} \sqrt {3 \, x^{2} + 2}}{162 \, {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )}} \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(5/2),x, algorithm="fricas")
Output:
1/162*(216*sqrt(3)*(9*x^4 + 12*x^2 + 4)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3 *x^2 - 1) + (1728*x^4 - 4185*x^3 + 936*x^2 - 2511*x + 254)*sqrt(3*x^2 + 2) )/(9*x^4 + 12*x^2 + 4)
\[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\int \frac {\left (2 x + 1\right )^{3} \cdot \left (4 x^{2} + 3 x + 1\right )}{\left (3 x^{2} + 2\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((1+2*x)**3*(4*x**2+3*x+1)/(3*x**2+2)**(5/2),x)
Output:
Integral((2*x + 1)**3*(4*x**2 + 3*x + 1)/(3*x**2 + 2)**(5/2), x)
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {32 \, x^{4}}{3 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} - \frac {8}{3} \, x {\left (\frac {9 \, x^{2}}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} + \frac {4}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}}\right )} + \frac {8}{3} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) - \frac {11 \, x}{18 \, \sqrt {3 \, x^{2} + 2}} + \frac {52 \, x^{2}}{9 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} - \frac {65 \, x}{18 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} + \frac {127}{81 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(5/2),x, algorithm="maxima")
Output:
32/3*x^4/(3*x^2 + 2)^(3/2) - 8/3*x*(9*x^2/(3*x^2 + 2)^(3/2) + 4/(3*x^2 + 2 )^(3/2)) + 8/3*sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 11/18*x/sqrt(3*x^2 + 2) + 52/9*x^2/(3*x^2 + 2)^(3/2) - 65/18*x/(3*x^2 + 2)^(3/2) + 127/81/(3*x^2 + 2 )^(3/2)
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.56 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=-\frac {8}{3} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {9 \, {\left ({\left (3 \, {\left (64 \, x - 155\right )} x + 104\right )} x - 279\right )} x + 254}{162 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \] Input:
integrate((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(5/2),x, algorithm="giac")
Output:
-8/3*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/162*(9*((3*(64*x - 155) *x + 104)*x - 279)*x + 254)/(3*x^2 + 2)^(3/2)
Time = 0.06 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.23 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {32\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{27}+\frac {8\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{3}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {-\frac {31}{16}+\frac {\sqrt {6}\,199{}\mathrm {i}}{144}}{x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}-\frac {\sqrt {6}\,\left (-\frac {31}{24}+\frac {\sqrt {6}\,199{}\mathrm {i}}{216}\right )\,1{}\mathrm {i}}{2\,{\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {\frac {31}{16}+\frac {\sqrt {6}\,199{}\mathrm {i}}{144}}{x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}+\frac {\sqrt {6}\,\left (\frac {31}{24}+\frac {\sqrt {6}\,199{}\mathrm {i}}{216}\right )\,1{}\mathrm {i}}{2\,{\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}+\frac {\sqrt {3}\,\sqrt {6}\,\left (-1824+\sqrt {6}\,1953{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{7776\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}+\frac {\sqrt {3}\,\sqrt {6}\,\left (1824+\sqrt {6}\,1953{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{7776\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \] Input:
int(((2*x + 1)^3*(3*x + 4*x^2 + 1))/(3*x^2 + 2)^(5/2),x)
Output:
(32*3^(1/2)*(x^2 + 2/3)^(1/2))/27 + (8*3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2 ))/3 - (3^(1/2)*(x^2 + 2/3)^(1/2)*(((6^(1/2)*199i)/144 - 31/16)/(x - (6^(1 /2)*1i)/3) - (6^(1/2)*((6^(1/2)*199i)/216 - 31/24)*1i)/(2*(x - (6^(1/2)*1i )/3)^2)))/27 + (3^(1/2)*(x^2 + 2/3)^(1/2)*(((6^(1/2)*199i)/144 + 31/16)/(x + (6^(1/2)*1i)/3) + (6^(1/2)*((6^(1/2)*199i)/216 + 31/24)*1i)/(2*(x + (6^ (1/2)*1i)/3)^2)))/27 + (3^(1/2)*6^(1/2)*(6^(1/2)*1953i - 1824)*(x^2 + 2/3) ^(1/2)*1i)/(7776*(x + (6^(1/2)*1i)/3)) + (3^(1/2)*6^(1/2)*(6^(1/2)*1953i + 1824)*(x^2 + 2/3)^(1/2)*1i)/(7776*(x - (6^(1/2)*1i)/3))
Time = 0.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.78 \[ \int \frac {(1+2 x)^3 \left (1+3 x+4 x^2\right )}{\left (2+3 x^2\right )^{5/2}} \, dx=\frac {1728 \sqrt {3 x^{2}+2}\, x^{4}-4185 \sqrt {3 x^{2}+2}\, x^{3}+936 \sqrt {3 x^{2}+2}\, x^{2}-2511 \sqrt {3 x^{2}+2}\, x +254 \sqrt {3 x^{2}+2}+3888 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right ) x^{4}+5184 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right ) x^{2}+1728 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )+837 \sqrt {3}\, x^{4}+1116 \sqrt {3}\, x^{2}+372 \sqrt {3}}{1458 x^{4}+1944 x^{2}+648} \] Input:
int((1+2*x)^3*(4*x^2+3*x+1)/(3*x^2+2)^(5/2),x)
Output:
(1728*sqrt(3*x**2 + 2)*x**4 - 4185*sqrt(3*x**2 + 2)*x**3 + 936*sqrt(3*x**2 + 2)*x**2 - 2511*sqrt(3*x**2 + 2)*x + 254*sqrt(3*x**2 + 2) + 3888*sqrt(3) *log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2))*x**4 + 5184*sqrt(3)*log((sqrt (3*x**2 + 2) + sqrt(3)*x)/sqrt(2))*x**2 + 1728*sqrt(3)*log((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2)) + 837*sqrt(3)*x**4 + 1116*sqrt(3)*x**2 + 372*sqrt (3))/(162*(9*x**4 + 12*x**2 + 4))