Integrand size = 27, antiderivative size = 82 \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {415}{32} \sqrt {3-x+2 x^2}+\frac {25}{8} x \sqrt {3-x+2 x^2}-\frac {223 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {2}} \]
-223/128*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+121/92*(19-7*x)/(2*x^2-x+3 )^(1/2)+415/32*(2*x^2-x+3)^(1/2)+25/8*x*(2*x^2-x+3)^(1/2)
Time = 0.47 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {47027-9421 x+16790 x^2+4600 x^3}{736 \sqrt {3-x+2 x^2}}-\frac {223 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{64 \sqrt {2}} \]
(47027 - 9421*x + 16790*x^2 + 4600*x^3)/(736*Sqrt[3 - x + 2*x^2]) - (223*L og[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/(64*Sqrt[2])
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, 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.
\(\displaystyle \int \frac {\left (5 x^2+3 x+2\right )^2}{\left (2 x^2-x+3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{23} \int \frac {23 \left (100 x^2+170 x+51\right )}{16 \sqrt {2 x^2-x+3}}dx+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \frac {100 x^2+170 x+51}{\sqrt {2 x^2-x+3}}dx+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \int -\frac {2 (48-415 x)}{\sqrt {2 x^2-x+3}}dx+25 \sqrt {2 x^2-x+3} x\right )+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (25 x \sqrt {2 x^2-x+3}-\frac {1}{2} \int \frac {48-415 x}{\sqrt {2 x^2-x+3}}dx\right )+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {223}{4} \int \frac {1}{\sqrt {2 x^2-x+3}}dx+\frac {415}{2} \sqrt {2 x^2-x+3}\right )+25 \sqrt {2 x^2-x+3} x\right )+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {223 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{4 \sqrt {46}}+\frac {415}{2} \sqrt {2 x^2-x+3}\right )+25 \sqrt {2 x^2-x+3} x\right )+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {223 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{4 \sqrt {2}}+\frac {415}{2} \sqrt {2 x^2-x+3}\right )+25 \sqrt {2 x^2-x+3} x\right )+\frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}\) |
(121*(19 - 7*x))/(92*Sqrt[3 - x + 2*x^2]) + (25*x*Sqrt[3 - x + 2*x^2] + (( 415*Sqrt[3 - x + 2*x^2])/2 + (223*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(4*Sqrt[2] ))/2)/8
3.1.88.3.1 Defintions of rubi rules used
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 = 0.79 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {4600 x^{3}+16790 x^{2}-9421 x +47027}{736 \sqrt {2 x^{2}-x +3}}+\frac {223 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}\) | \(45\) |
trager | \(\frac {4600 x^{3}+16790 x^{2}-9421 x +47027}{736 \sqrt {2 x^{2}-x +3}}+\frac {223 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {2 x^{2}-x +3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{128}\) | \(72\) |
default | \(-\frac {13713 \left (-1+4 x \right )}{5888 \sqrt {2 x^{2}-x +3}}+\frac {15761}{256 \sqrt {2 x^{2}-x +3}}+\frac {25 x^{3}}{4 \sqrt {2 x^{2}-x +3}}+\frac {365 x^{2}}{16 \sqrt {2 x^{2}-x +3}}-\frac {223 x}{64 \sqrt {2 x^{2}-x +3}}+\frac {223 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}\) | \(98\) |
1/736*(4600*x^3+16790*x^2-9421*x+47027)/(2*x^2-x+3)^(1/2)+223/128*2^(1/2)* arcsinh(4/23*23^(1/2)*(x-1/4))
Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5129 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (4600 \, x^{3} + 16790 \, x^{2} - 9421 \, x + 47027\right )} \sqrt {2 \, x^{2} - x + 3}}{5888 \, {\left (2 \, x^{2} - x + 3\right )}} \]
1/5888*(5129*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*(4600*x^3 + 16790*x^2 - 9421*x + 47027)* sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)
\[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {\left (5 x^{2} + 3 x + 2\right )^{2}}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {25 \, x^{3}}{4 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {365 \, x^{2}}{16 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {223}{128} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {9421 \, x}{736 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {47027}{736 \, \sqrt {2 \, x^{2} - x + 3}} \]
25/4*x^3/sqrt(2*x^2 - x + 3) + 365/16*x^2/sqrt(2*x^2 - x + 3) + 223/128*sq rt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 9421/736*x/sqrt(2*x^2 - x + 3) + 47027/736/sqrt(2*x^2 - x + 3)
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.76 \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {223}{128} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (230 \, {\left (20 \, x + 73\right )} x - 9421\right )} x + 47027}{736 \, \sqrt {2 \, x^{2} - x + 3}} \]
-223/128*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1 /736*((230*(20*x + 73)*x - 9421)*x + 47027)/sqrt(2*x^2 - x + 3)
Timed out. \[ \int \frac {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (5\,x^2+3\,x+2\right )}^2}{{\left (2\,x^2-x+3\right )}^{3/2}} \,d x \]