Integrand size = 33, antiderivative size = 101 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=\frac {463}{125} \sqrt {3+2 x+5 x^2}+\frac {59}{30} x \sqrt {3+2 x+5 x^2}-\frac {571}{300} x^2 \sqrt {3+2 x+5 x^2}-\frac {7}{20} x^3 \sqrt {3+2 x+5 x^2}-\frac {1901 \text {arcsinh}\left (\frac {1+5 x}{\sqrt {14}}\right )}{250 \sqrt {5}} \]
-1901/1250*arcsinh(1/14*(1+5*x)*14^(1/2))*5^(1/2)+463/125*(5*x^2+2*x+3)^(1 /2)+59/30*x*(5*x^2+2*x+3)^(1/2)-571/300*x^2*(5*x^2+2*x+3)^(1/2)-7/20*x^3*( 5*x^2+2*x+3)^(1/2)
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=\frac {\sqrt {3+2 x+5 x^2} \left (5556+2950 x-2855 x^2-525 x^3\right )}{1500}+\frac {1901 \log \left (-1-5 x+\sqrt {5} \sqrt {3+2 x+5 x^2}\right )}{250 \sqrt {5}} \]
(Sqrt[3 + 2*x + 5*x^2]*(5556 + 2950*x - 2855*x^2 - 525*x^3))/1500 + (1901* Log[-1 - 5*x + Sqrt[5]*Sqrt[3 + 2*x + 5*x^2]])/(250*Sqrt[5])
Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2192, 2192, 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.
\(\displaystyle \int \frac {\left (-7 x^2+4 x+1\right ) \left (x^2+5 x+2\right )}{\sqrt {5 x^2+2 x+3}} \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{20} \int \frac {-571 x^3+203 x^2+260 x+40}{\sqrt {5 x^2+2 x+3}}dx-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{20} \left (\frac {1}{15} \int \frac {2 \left (2950 x^2+3663 x+300\right )}{\sqrt {5 x^2+2 x+3}}dx-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \int \frac {2950 x^2+3663 x+300}{\sqrt {5 x^2+2 x+3}}dx-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \left (\frac {1}{10} \int -\frac {30 (195-926 x)}{\sqrt {5 x^2+2 x+3}}dx+295 \sqrt {5 x^2+2 x+3} x\right )-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \left (295 x \sqrt {5 x^2+2 x+3}-3 \int \frac {195-926 x}{\sqrt {5 x^2+2 x+3}}dx\right )-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \left (295 x \sqrt {5 x^2+2 x+3}-3 \left (\frac {1901}{5} \int \frac {1}{\sqrt {5 x^2+2 x+3}}dx-\frac {926}{5} \sqrt {5 x^2+2 x+3}\right )\right )-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \left (295 x \sqrt {5 x^2+2 x+3}-3 \left (\frac {1901 \int \frac {1}{\sqrt {\frac {1}{56} (10 x+2)^2+1}}d(10 x+2)}{10 \sqrt {70}}-\frac {926}{5} \sqrt {5 x^2+2 x+3}\right )\right )-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{20} \left (\frac {2}{15} \left (295 x \sqrt {5 x^2+2 x+3}-3 \left (\frac {1901 \text {arcsinh}\left (\frac {10 x+2}{2 \sqrt {14}}\right )}{5 \sqrt {5}}-\frac {926}{5} \sqrt {5 x^2+2 x+3}\right )\right )-\frac {571}{15} x^2 \sqrt {5 x^2+2 x+3}\right )-\frac {7}{20} x^3 \sqrt {5 x^2+2 x+3}\) |
(-7*x^3*Sqrt[3 + 2*x + 5*x^2])/20 + ((-571*x^2*Sqrt[3 + 2*x + 5*x^2])/15 + (2*(295*x*Sqrt[3 + 2*x + 5*x^2] - 3*((-926*Sqrt[3 + 2*x + 5*x^2])/5 + (19 01*ArcSinh[(2 + 10*x)/(2*Sqrt[14])])/(5*Sqrt[5]))))/15)/20
3.4.88.3.1 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 = 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 = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.45
method | result | size |
risch | \(-\frac {\left (525 x^{3}+2855 x^{2}-2950 x -5556\right ) \sqrt {5 x^{2}+2 x +3}}{1500}-\frac {1901 \sqrt {5}\, \operatorname {arcsinh}\left (\frac {5 \sqrt {14}\, \left (x +\frac {1}{5}\right )}{14}\right )}{1250}\) | \(45\) |
trager | \(\left (-\frac {7}{20} x^{3}-\frac {571}{300} x^{2}+\frac {59}{30} x +\frac {463}{125}\right ) \sqrt {5 x^{2}+2 x +3}-\frac {1901 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )+5 \sqrt {5 x^{2}+2 x +3}\right )}{1250}\) | \(69\) |
default | \(-\frac {1901 \sqrt {5}\, \operatorname {arcsinh}\left (\frac {5 \sqrt {14}\, \left (x +\frac {1}{5}\right )}{14}\right )}{1250}+\frac {463 \sqrt {5 x^{2}+2 x +3}}{125}-\frac {7 x^{3} \sqrt {5 x^{2}+2 x +3}}{20}-\frac {571 x^{2} \sqrt {5 x^{2}+2 x +3}}{300}+\frac {59 x \sqrt {5 x^{2}+2 x +3}}{30}\) | \(79\) |
-1/1500*(525*x^3+2855*x^2-2950*x-5556)*(5*x^2+2*x+3)^(1/2)-1901/1250*5^(1/ 2)*arcsinh(5/14*14^(1/2)*(x+1/5))
Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=-\frac {1}{1500} \, {\left (525 \, x^{3} + 2855 \, x^{2} - 2950 \, x - 5556\right )} \sqrt {5 \, x^{2} + 2 \, x + 3} + \frac {1901}{2500} \, \sqrt {5} \log \left (\sqrt {5} \sqrt {5 \, x^{2} + 2 \, x + 3} {\left (5 \, x + 1\right )} - 25 \, x^{2} - 10 \, x - 8\right ) \]
-1/1500*(525*x^3 + 2855*x^2 - 2950*x - 5556)*sqrt(5*x^2 + 2*x + 3) + 1901/ 2500*sqrt(5)*log(sqrt(5)*sqrt(5*x^2 + 2*x + 3)*(5*x + 1) - 25*x^2 - 10*x - 8)
Time = 0.38 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.57 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=\sqrt {5 x^{2} + 2 x + 3} \left (- \frac {7 x^{3}}{20} - \frac {571 x^{2}}{300} + \frac {59 x}{30} + \frac {463}{125}\right ) - \frac {1901 \sqrt {5} \operatorname {asinh}{\left (\frac {5 \sqrt {14} \left (x + \frac {1}{5}\right )}{14} \right )}}{1250} \]
sqrt(5*x**2 + 2*x + 3)*(-7*x**3/20 - 571*x**2/300 + 59*x/30 + 463/125) - 1 901*sqrt(5)*asinh(5*sqrt(14)*(x + 1/5)/14)/1250
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=-\frac {7}{20} \, \sqrt {5 \, x^{2} + 2 \, x + 3} x^{3} - \frac {571}{300} \, \sqrt {5 \, x^{2} + 2 \, x + 3} x^{2} + \frac {59}{30} \, \sqrt {5 \, x^{2} + 2 \, x + 3} x - \frac {1901}{1250} \, \sqrt {5} \operatorname {arsinh}\left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + \frac {463}{125} \, \sqrt {5 \, x^{2} + 2 \, x + 3} \]
-7/20*sqrt(5*x^2 + 2*x + 3)*x^3 - 571/300*sqrt(5*x^2 + 2*x + 3)*x^2 + 59/3 0*sqrt(5*x^2 + 2*x + 3)*x - 1901/1250*sqrt(5)*arcsinh(1/14*sqrt(14)*(5*x + 1)) + 463/125*sqrt(5*x^2 + 2*x + 3)
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=-\frac {1}{1500} \, {\left (5 \, {\left ({\left (105 \, x + 571\right )} x - 590\right )} x - 5556\right )} \sqrt {5 \, x^{2} + 2 \, x + 3} + \frac {1901}{1250} \, \sqrt {5} \log \left (-\sqrt {5} {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x + 3}\right )} - 1\right ) \]
-1/1500*(5*((105*x + 571)*x - 590)*x - 5556)*sqrt(5*x^2 + 2*x + 3) + 1901/ 1250*sqrt(5)*log(-sqrt(5)*(sqrt(5)*x - sqrt(5*x^2 + 2*x + 3)) - 1)
Timed out. \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\sqrt {3+2 x+5 x^2}} \, dx=\int \frac {\left (x^2+5\,x+2\right )\,\left (-7\,x^2+4\,x+1\right )}{\sqrt {5\,x^2+2\,x+3}} \,d x \]