Integrand size = 33, antiderivative size = 82 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=-\frac {2 (2321+2449 x)}{875 \sqrt {3+2 x+5 x^2}}-\frac {261}{250} \sqrt {3+2 x+5 x^2}-\frac {7}{50} x \sqrt {3+2 x+5 x^2}+\frac {149 \text {arcsinh}\left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {5}} \]
149/125*arcsinh(1/14*(1+5*x)*14^(1/2))*5^(1/2)-2/875*(2321+2449*x)/(5*x^2+ 2*x+3)^(1/2)-261/250*(5*x^2+2*x+3)^(1/2)-7/50*x*(5*x^2+2*x+3)^(1/2)
Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=\frac {-2953-2837 x-1925 x^2-245 x^3}{350 \sqrt {3+2 x+5 x^2}}-\frac {149 \log \left (-1-5 x+\sqrt {5} \sqrt {3+2 x+5 x^2}\right )}{25 \sqrt {5}} \]
(-2953 - 2837*x - 1925*x^2 - 245*x^3)/(350*Sqrt[3 + 2*x + 5*x^2]) - (149*L og[-1 - 5*x + Sqrt[5]*Sqrt[3 + 2*x + 5*x^2]])/(25*Sqrt[5])
Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11, 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.
| \(\displaystyle \int \frac {\left (-7 x^2+4 x+1\right ) \left (x^2+5 x+2\right )}{\left (5 x^2+2 x+3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2191 |
| \(\displaystyle \frac {1}{28} \int \frac {28 \left (-175 x^2-705 x+562\right )}{125 \sqrt {5 x^2+2 x+3}}dx-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{125} \int \frac {-175 x^2-705 x+562}{\sqrt {5 x^2+2 x+3}}dx-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 2192 |
| \(\displaystyle \frac {1}{125} \left (\frac {1}{10} \int \frac {5 (1229-1305 x)}{\sqrt {5 x^2+2 x+3}}dx-\frac {35}{2} x \sqrt {5 x^2+2 x+3}\right )-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{125} \left (\frac {1}{2} \int \frac {1229-1305 x}{\sqrt {5 x^2+2 x+3}}dx-\frac {35}{2} x \sqrt {5 x^2+2 x+3}\right )-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {1}{125} \left (\frac {1}{2} \left (1490 \int \frac {1}{\sqrt {5 x^2+2 x+3}}dx-261 \sqrt {5 x^2+2 x+3}\right )-\frac {35}{2} x \sqrt {5 x^2+2 x+3}\right )-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {1}{125} \left (\frac {1}{2} \left (149 \sqrt {\frac {5}{14}} \int \frac {1}{\sqrt {\frac {1}{56} (10 x+2)^2+1}}d(10 x+2)-261 \sqrt {5 x^2+2 x+3}\right )-\frac {35}{2} x \sqrt {5 x^2+2 x+3}\right )-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{125} \left (\frac {1}{2} \left (298 \sqrt {5} \text {arcsinh}\left (\frac {10 x+2}{2 \sqrt {14}}\right )-261 \sqrt {5 x^2+2 x+3}\right )-\frac {35}{2} x \sqrt {5 x^2+2 x+3}\right )-\frac {2 (2449 x+2321)}{875 \sqrt {5 x^2+2 x+3}}\) |
(-2*(2321 + 2449*x))/(875*Sqrt[3 + 2*x + 5*x^2]) + ((-35*x*Sqrt[3 + 2*x + 5*x^2])/2 + (-261*Sqrt[3 + 2*x + 5*x^2] + 298*Sqrt[5]*ArcSinh[(2 + 10*x)/( 2*Sqrt[14])])/2)/125
3.4.94.3.1 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 = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.55
| method | result | size |
| risch | \(-\frac {245 x^{3}+1925 x^{2}+2837 x +2953}{350 \sqrt {5 x^{2}+2 x +3}}+\frac {149 \sqrt {5}\, \operatorname {arcsinh}\left (\frac {5 \sqrt {14}\, \left (x +\frac {1}{5}\right )}{14}\right )}{125}\) | \(45\) |
| trager | \(-\frac {245 x^{3}+1925 x^{2}+2837 x +2953}{350 \sqrt {5 x^{2}+2 x +3}}-\frac {149 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +5 \sqrt {5 x^{2}+2 x +3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )\right )}{125}\) | \(72\) |
| default | \(-\frac {751 \left (10 x +2\right )}{3500 \sqrt {5 x^{2}+2 x +3}}-\frac {1001}{125 \sqrt {5 x^{2}+2 x +3}}-\frac {7 x^{3}}{10 \sqrt {5 x^{2}+2 x +3}}-\frac {11 x^{2}}{2 \sqrt {5 x^{2}+2 x +3}}-\frac {149 x}{25 \sqrt {5 x^{2}+2 x +3}}+\frac {149 \sqrt {5}\, \operatorname {arcsinh}\left (\frac {5 \sqrt {14}\, \left (x +\frac {1}{5}\right )}{14}\right )}{125}\) | \(98\) |
-1/350*(245*x^3+1925*x^2+2837*x+2953)/(5*x^2+2*x+3)^(1/2)+149/125*5^(1/2)* arcsinh(5/14*14^(1/2)*(x+1/5))
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=\frac {1043 \, \sqrt {5} {\left (5 \, x^{2} + 2 \, x + 3\right )} \log \left (-\sqrt {5} \sqrt {5 \, x^{2} + 2 \, x + 3} {\left (5 \, x + 1\right )} - 25 \, x^{2} - 10 \, x - 8\right ) - 5 \, {\left (245 \, x^{3} + 1925 \, x^{2} + 2837 \, x + 2953\right )} \sqrt {5 \, x^{2} + 2 \, x + 3}}{1750 \, {\left (5 \, x^{2} + 2 \, x + 3\right )}} \]
1/1750*(1043*sqrt(5)*(5*x^2 + 2*x + 3)*log(-sqrt(5)*sqrt(5*x^2 + 2*x + 3)* (5*x + 1) - 25*x^2 - 10*x - 8) - 5*(245*x^3 + 1925*x^2 + 2837*x + 2953)*sq rt(5*x^2 + 2*x + 3))/(5*x^2 + 2*x + 3)
\[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {13 x}{5 x^{2} \sqrt {5 x^{2} + 2 x + 3} + 2 x \sqrt {5 x^{2} + 2 x + 3} + 3 \sqrt {5 x^{2} + 2 x + 3}}\right )\, dx - \int \left (- \frac {7 x^{2}}{5 x^{2} \sqrt {5 x^{2} + 2 x + 3} + 2 x \sqrt {5 x^{2} + 2 x + 3} + 3 \sqrt {5 x^{2} + 2 x + 3}}\right )\, dx - \int \frac {31 x^{3}}{5 x^{2} \sqrt {5 x^{2} + 2 x + 3} + 2 x \sqrt {5 x^{2} + 2 x + 3} + 3 \sqrt {5 x^{2} + 2 x + 3}}\, dx - \int \frac {7 x^{4}}{5 x^{2} \sqrt {5 x^{2} + 2 x + 3} + 2 x \sqrt {5 x^{2} + 2 x + 3} + 3 \sqrt {5 x^{2} + 2 x + 3}}\, dx - \int \left (- \frac {2}{5 x^{2} \sqrt {5 x^{2} + 2 x + 3} + 2 x \sqrt {5 x^{2} + 2 x + 3} + 3 \sqrt {5 x^{2} + 2 x + 3}}\right )\, dx \]
-Integral(-13*x/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3 ) + 3*sqrt(5*x**2 + 2*x + 3)), x) - Integral(-7*x**2/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x) - I ntegral(31*x**3/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3 ) + 3*sqrt(5*x**2 + 2*x + 3)), x) - Integral(7*x**4/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x) - In tegral(-2/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3* sqrt(5*x**2 + 2*x + 3)), x)
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=-\frac {7 \, x^{3}}{10 \, \sqrt {5 \, x^{2} + 2 \, x + 3}} - \frac {11 \, x^{2}}{2 \, \sqrt {5 \, x^{2} + 2 \, x + 3}} + \frac {149}{125} \, \sqrt {5} \operatorname {arsinh}\left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) - \frac {2837 \, x}{350 \, \sqrt {5 \, x^{2} + 2 \, x + 3}} - \frac {2953}{350 \, \sqrt {5 \, x^{2} + 2 \, x + 3}} \]
-7/10*x^3/sqrt(5*x^2 + 2*x + 3) - 11/2*x^2/sqrt(5*x^2 + 2*x + 3) + 149/125 *sqrt(5)*arcsinh(1/14*sqrt(14)*(5*x + 1)) - 2837/350*x/sqrt(5*x^2 + 2*x + 3) - 2953/350/sqrt(5*x^2 + 2*x + 3)
Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.76 \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=-\frac {149}{125} \, \sqrt {5} \log \left (-\sqrt {5} {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x + 3}\right )} - 1\right ) - \frac {{\left (35 \, {\left (7 \, x + 55\right )} x + 2837\right )} x + 2953}{350 \, \sqrt {5 \, x^{2} + 2 \, x + 3}} \]
-149/125*sqrt(5)*log(-sqrt(5)*(sqrt(5)*x - sqrt(5*x^2 + 2*x + 3)) - 1) - 1 /350*((35*(7*x + 55)*x + 2837)*x + 2953)/sqrt(5*x^2 + 2*x + 3)
Timed out. \[ \int \frac {\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx=\int \frac {\left (x^2+5\,x+2\right )\,\left (-7\,x^2+4\,x+1\right )}{{\left (5\,x^2+2\,x+3\right )}^{3/2}} \,d x \]