Integrand size = 40, antiderivative size = 105 \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {247}{16} \sqrt {3-x+2 x^2}+\frac {5}{4} x \sqrt {3-x+2 x^2}-\frac {1471 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]
-4/69*(346-533*x)/(2*x^2-x+3)^(3/2)-1471/64*arcsinh(1/23*(1-4*x)*23^(1/2)) *2^(1/2)+4/1587*(18982-20383*x)/(2*x^2-x+3)^(1/2)+247/16*(2*x^2-x+3)^(1/2) +5/4*x*(2*x^2-x+3)^(1/2)
Time = 0.54 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {6663133-6410082 x+8639625 x^2-3764360 x^3+1440996 x^4+126960 x^5}{25392 \left (3-x+2 x^2\right )^{3/2}}-\frac {1471 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{32 \sqrt {2}} \]
(6663133 - 6410082*x + 8639625*x^2 - 3764360*x^3 + 1440996*x^4 + 126960*x^ 5)/(25392*(3 - x + 2*x^2)^(3/2)) - (1471*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4* x^2]])/(32*Sqrt[2])
Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {2191, 27, 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.
\(\displaystyle \int \frac {(2 x+5)^2 \left (5 x^4-x^3+3 x^2+x+2\right )}{\left (2 x^2-x+3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{69} \int -\frac {-690 x^4-3657 x^3-4830 x^2+1725 x+290}{2 \left (2 x^2-x+3\right )^{3/2}}dx-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{69} \int \frac {-690 x^4-3657 x^3-4830 x^2+1725 x+290}{\left (2 x^2-x+3\right )^{3/2}}dx-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{69} \left (\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}-\frac {2}{23} \int -\frac {1587 \left (5 x^2+29 x+42\right )}{2 \sqrt {2 x^2-x+3}}dx\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{69} \left (69 \int \frac {5 x^2+29 x+42}{\sqrt {2 x^2-x+3}}dx+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{69} \left (69 \left (\frac {1}{4} \int \frac {247 x+306}{2 \sqrt {2 x^2-x+3}}dx+\frac {5}{4} \sqrt {2 x^2-x+3} x\right )+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{69} \left (69 \left (\frac {1}{8} \int \frac {247 x+306}{\sqrt {2 x^2-x+3}}dx+\frac {5}{4} \sqrt {2 x^2-x+3} x\right )+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{69} \left (69 \left (\frac {1}{8} \left (\frac {1471}{4} \int \frac {1}{\sqrt {2 x^2-x+3}}dx+\frac {247}{2} \sqrt {2 x^2-x+3}\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\right )+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{69} \left (69 \left (\frac {1}{8} \left (\frac {1471 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{4 \sqrt {46}}+\frac {247}{2} \sqrt {2 x^2-x+3}\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\right )+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{69} \left (69 \left (\frac {1}{8} \left (\frac {1471 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{4 \sqrt {2}}+\frac {247}{2} \sqrt {2 x^2-x+3}\right )+\frac {5}{4} \sqrt {2 x^2-x+3} x\right )+\frac {4 (18982-20383 x)}{23 \sqrt {2 x^2-x+3}}\right )-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}\) |
(-4*(346 - 533*x))/(69*(3 - x + 2*x^2)^(3/2)) + ((4*(18982 - 20383*x))/(23 *Sqrt[3 - x + 2*x^2]) + 69*((5*x*Sqrt[3 - x + 2*x^2])/4 + ((247*Sqrt[3 - x + 2*x^2])/2 + (1471*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(4*Sqrt[2]))/8))/69
3.4.58.3.1 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]
Timed out.
hanged
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.16 \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {2334477 \, \sqrt {2} {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (126960 \, x^{5} + 1440996 \, x^{4} - 3764360 \, x^{3} + 8639625 \, x^{2} - 6410082 \, x + 6663133\right )} \sqrt {2 \, x^{2} - x + 3}}{203136 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]
1/203136*(2334477*sqrt(2)*(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)*log(-4*sqrt(2 )*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 8*(126960*x^5 + 14 40996*x^4 - 3764360*x^3 + 8639625*x^2 - 6410082*x + 6663133)*sqrt(2*x^2 - x + 3))/(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)
\[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {\left (2 x + 5\right )^{2} \cdot \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{\left (2 x^{2} - x + 3\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (84) = 168\).
Time = 0.29 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.09 \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {5 \, x^{5}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {227 \, x^{4}}{4 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {1471}{50784} \, x {\left (\frac {284 \, x}{\sqrt {2 \, x^{2} - x + 3}} - \frac {3174 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {71}{\sqrt {2 \, x^{2} - x + 3}} + \frac {805 \, x}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {3243}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} + \frac {1471}{64} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {104441}{25392} \, \sqrt {2 \, x^{2} - x + 3} - \frac {383581 \, x}{12696 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {321 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {15965}{4232 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {4147 \, x}{46 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {42883}{138 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
5*x^5/(2*x^2 - x + 3)^(3/2) + 227/4*x^4/(2*x^2 - x + 3)^(3/2) + 1471/50784 *x*(284*x/sqrt(2*x^2 - x + 3) - 3174*x^2/(2*x^2 - x + 3)^(3/2) - 71/sqrt(2 *x^2 - x + 3) + 805*x/(2*x^2 - x + 3)^(3/2) - 3243/(2*x^2 - x + 3)^(3/2)) + 1471/64*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 104441/25392*sqrt(2*x ^2 - x + 3) - 383581/12696*x/sqrt(2*x^2 - x + 3) + 321*x^2/(2*x^2 - x + 3) ^(3/2) - 15965/4232/sqrt(2*x^2 - x + 3) - 4147/46*x/(2*x^2 - x + 3)^(3/2) + 42883/138/(2*x^2 - x + 3)^(3/2)
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68 \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=-\frac {1471}{64} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left ({\left (4 \, {\left (1587 \, {\left (20 \, x + 227\right )} x - 941090\right )} x + 8639625\right )} x - 6410082\right )} x + 6663133}{25392 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
-1471/64*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1 /25392*(((4*(1587*(20*x + 227)*x - 941090)*x + 8639625)*x - 6410082)*x + 6 663133)/(2*x^2 - x + 3)^(3/2)
Timed out. \[ \int \frac {(5+2 x)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (2\,x+5\right )}^2\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{{\left (2\,x^2-x+3\right )}^{5/2}} \,d x \]