Integrand size = 33, antiderivative size = 82 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {89+219 x}{92 \sqrt {3-x+2 x^2}}+\frac {27}{32} \sqrt {3-x+2 x^2}+\frac {5}{8} x \sqrt {3-x+2 x^2}+\frac {213 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {2}} \] Output:
1/92*(89+219*x)/(2*x^2-x+3)^(1/2)+27/32*(2*x^2-x+3)^(1/2)+5/8*x*(2*x^2-x+3 )^(1/2)+213/128*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)
Time = 0.69 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {2575+2511 x+782 x^2+920 x^3}{736 \sqrt {3-x+2 x^2}}+\frac {213 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{64 \sqrt {2}} \] Input:
Integrate[(2 + x + 3*x^2 - x^3 + 5*x^4)/(3 - x + 2*x^2)^(3/2),x]
Output:
(2575 + 2511*x + 782*x^2 + 920*x^3)/(736*Sqrt[3 - x + 2*x^2]) + (213*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/(64*Sqrt[2])
Time = 0.46 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2191, 27, 2192, 27, 1160, 1090, 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.
\(\displaystyle \int \frac {5 x^4-x^3+3 x^2+x+2}{\left (2 x^2-x+3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{23} \int -\frac {23 \left (-20 x^2-6 x+15\right )}{16 \sqrt {2 x^2-x+3}}dx+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {219 x+89}{92 \sqrt {2 x^2-x+3}}-\frac {1}{8} \int \frac {-20 x^2-6 x+15}{\sqrt {2 x^2-x+3}}dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{8} \left (5 x \sqrt {2 x^2-x+3}-\frac {1}{4} \int \frac {6 (20-9 x)}{\sqrt {2 x^2-x+3}}dx\right )+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (5 x \sqrt {2 x^2-x+3}-\frac {3}{2} \int \frac {20-9 x}{\sqrt {2 x^2-x+3}}dx\right )+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{8} \left (5 x \sqrt {2 x^2-x+3}-\frac {3}{2} \left (\frac {71}{4} \int \frac {1}{\sqrt {2 x^2-x+3}}dx-\frac {9}{2} \sqrt {2 x^2-x+3}\right )\right )+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{8} \left (5 x \sqrt {2 x^2-x+3}-\frac {3}{2} \left (\frac {71 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{4 \sqrt {46}}-\frac {9}{2} \sqrt {2 x^2-x+3}\right )\right )+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{8} \left (5 x \sqrt {2 x^2-x+3}-\frac {3}{2} \left (\frac {71 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{4 \sqrt {2}}-\frac {9}{2} \sqrt {2 x^2-x+3}\right )\right )+\frac {219 x+89}{92 \sqrt {2 x^2-x+3}}\) |
Input:
Int[(2 + x + 3*x^2 - x^3 + 5*x^4)/(3 - x + 2*x^2)^(3/2),x]
Output:
(89 + 219*x)/(92*Sqrt[3 - x + 2*x^2]) + (5*x*Sqrt[3 - x + 2*x^2] - (3*((-9 *Sqrt[3 - x + 2*x^2])/2 + (71*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(4*Sqrt[2])))/ 2)/8
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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ (p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int [(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* (2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 2 - 4*a*c, 0] && LtQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(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 + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {920 x^{3}+782 x^{2}+2511 x +2575}{736 \sqrt {2 x^{2}-x +3}}-\frac {213 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}\) | \(45\) |
trager | \(\frac {920 x^{3}+782 x^{2}+2511 x +2575}{736 \sqrt {2 x^{2}-x +3}}+\frac {213 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+4 \sqrt {2 x^{2}-x +3}\right )}{128}\) | \(70\) |
default | \(\frac {901}{256 \sqrt {2 x^{2}-x +3}}+\frac {\frac {123 x}{1472}-\frac {123}{5888}}{\sqrt {2 x^{2}-x +3}}+\frac {213 x}{64 \sqrt {2 x^{2}-x +3}}-\frac {213 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}+\frac {17 x^{2}}{16 \sqrt {2 x^{2}-x +3}}+\frac {5 x^{3}}{4 \sqrt {2 x^{2}-x +3}}\) | \(98\) |
Input:
int((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/736*(920*x^3+782*x^2+2511*x+2575)/(2*x^2-x+3)^(1/2)-213/128*2^(1/2)*arcs inh(4/23*23^(1/2)*(x-1/4))
Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {4899 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (920 \, x^{3} + 782 \, x^{2} + 2511 \, x + 2575\right )} \sqrt {2 \, x^{2} - x + 3}}{5888 \, {\left (2 \, x^{2} - x + 3\right )}} \] Input:
integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x, algorithm="fricas")
Output:
1/5888*(4899*sqrt(2)*(2*x^2 - x + 3)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4* x - 1) - 32*x^2 + 16*x - 25) + 8*(920*x^3 + 782*x^2 + 2511*x + 2575)*sqrt( 2*x^2 - x + 3))/(2*x^2 - x + 3)
\[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((5*x**4-x**3+3*x**2+x+2)/(2*x**2-x+3)**(3/2),x)
Output:
Integral((5*x**4 - x**3 + 3*x**2 + x + 2)/(2*x**2 - x + 3)**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5 \, x^{3}}{4 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {17 \, x^{2}}{16 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {213}{128} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {2511 \, x}{736 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {2575}{736 \, \sqrt {2 \, x^{2} - x + 3}} \] Input:
integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x, algorithm="maxima")
Output:
5/4*x^3/sqrt(2*x^2 - x + 3) + 17/16*x^2/sqrt(2*x^2 - x + 3) - 213/128*sqrt (2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 2511/736*x/sqrt(2*x^2 - x + 3) + 25 75/736/sqrt(2*x^2 - x + 3)
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.76 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {213}{128} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (46 \, {\left (20 \, x + 17\right )} x + 2511\right )} x + 2575}{736 \, \sqrt {2 \, x^{2} - x + 3}} \] Input:
integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x, algorithm="giac")
Output:
213/128*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/ 736*((46*(20*x + 17)*x + 2511)*x + 2575)/sqrt(2*x^2 - x + 3)
Timed out. \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5\,x^4-x^3+3\,x^2+x+2}{{\left (2\,x^2-x+3\right )}^{3/2}} \,d x \] Input:
int((x + 3*x^2 - x^3 + 5*x^4 + 2)/(2*x^2 - x + 3)^(3/2),x)
Output:
int((x + 3*x^2 - x^3 + 5*x^4 + 2)/(2*x^2 - x + 3)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.23 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {3680 \sqrt {2 x^{2}-x +3}\, x^{3}+3128 \sqrt {2 x^{2}-x +3}\, x^{2}+10044 \sqrt {2 x^{2}-x +3}\, x +10300 \sqrt {2 x^{2}-x +3}-9798 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right ) x^{2}+4899 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right ) x -14697 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right )+7008 \sqrt {2}\, x^{2}-3504 \sqrt {2}\, x +10512 \sqrt {2}}{5888 x^{2}-2944 x +8832} \] Input:
int((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x)
Output:
(3680*sqrt(2*x**2 - x + 3)*x**3 + 3128*sqrt(2*x**2 - x + 3)*x**2 + 10044*s qrt(2*x**2 - x + 3)*x + 10300*sqrt(2*x**2 - x + 3) - 9798*sqrt(2)*log((2*s qrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23))*x**2 + 4899*sqrt(2)*log(( 2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23))*x - 14697*sqrt(2)*log( (2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23)) + 7008*sqrt(2)*x**2 - 3504*sqrt(2)*x + 10512*sqrt(2))/(2944*(2*x**2 - x + 3))