Integrand size = 25, antiderivative size = 45 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}-\frac {5 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{2 \sqrt {2}} \] Output:
1/23*(-55-33*x)/(2*x^2-x+3)^(1/2)-5/4*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/ 2)
Time = 0.48 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.22 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {11 (5+3 x)}{23 \sqrt {3-x+2 x^2}}-\frac {5 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{2 \sqrt {2}} \] Input:
Integrate[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^(3/2),x]
Output:
(-11*(5 + 3*x))/(23*Sqrt[3 - x + 2*x^2]) - (5*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/(2*Sqrt[2])
Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2191, 27, 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.
\(\displaystyle \int \frac {5 x^2+3 x+2}{\left (2 x^2-x+3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2}{23} \int \frac {115}{4 \sqrt {2 x^2-x+3}}dx-\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{2} \int \frac {1}{\sqrt {2 x^2-x+3}}dx-\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {5 \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)}{2 \sqrt {46}}-\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {5 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{2 \sqrt {2}}-\frac {11 (3 x+5)}{23 \sqrt {2 x^2-x+3}}\) |
Input:
Int[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^(3/2),x]
Output:
(-11*(5 + 3*x))/(23*Sqrt[3 - x + 2*x^2]) + (5*ArcSinh[(-1 + 4*x)/Sqrt[23]] )/(2*Sqrt[2])
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[(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]
Time = 1.98 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {11 \left (3 x +5\right )}{23 \sqrt {2 x^{2}-x +3}}+\frac {5 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{4}\) | \(35\) |
trager | \(-\frac {11 \left (3 x +5\right )}{23 \sqrt {2 x^{2}-x +3}}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+4 \sqrt {2 x^{2}-x +3}\right )}{4}\) | \(60\) |
default | \(\frac {\frac {49 x}{46}-\frac {49}{184}}{\sqrt {2 x^{2}-x +3}}-\frac {17}{8 \sqrt {2 x^{2}-x +3}}-\frac {5 x}{2 \sqrt {2 x^{2}-x +3}}+\frac {5 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{4}\) | \(64\) |
Input:
int((5*x^2+3*x+2)/(2*x^2-x+3)^(3/2),x,method=_RETURNVERBOSE)
Output:
-11/23*(3*x+5)/(2*x^2-x+3)^(1/2)+5/4*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4) )
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (36) = 72\).
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.82 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {115 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) - 88 \, \sqrt {2 \, x^{2} - x + 3} {\left (3 \, x + 5\right )}}{184 \, {\left (2 \, x^{2} - x + 3\right )}} \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(3/2),x, algorithm="fricas")
Output:
1/184*(115*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) - 88*sqrt(2*x^2 - x + 3)*(3*x + 5))/(2*x^2 - x + 3)
\[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5 x^{2} + 3 x + 2}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((5*x**2+3*x+2)/(2*x**2-x+3)**(3/2),x)
Output:
Integral((5*x**2 + 3*x + 2)/(2*x**2 - x + 3)**(3/2), x)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5}{4} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {33 \, x}{23 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {55}{23 \, \sqrt {2 \, x^{2} - x + 3}} \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(3/2),x, algorithm="maxima")
Output:
5/4*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 33/23*x/sqrt(2*x^2 - x + 3) - 55/23/sqrt(2*x^2 - x + 3)
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {5}{4} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) - \frac {11 \, {\left (3 \, x + 5\right )}}{23 \, \sqrt {2 \, x^{2} - x + 3}} \] Input:
integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(3/2),x, algorithm="giac")
Output:
-5/4*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) - 11/23 *(3*x + 5)/sqrt(2*x^2 - x + 3)
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.93 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{4}+\frac {3\,\left (2\,x-12\right )}{23\,\sqrt {2\,x^2-x+3}}-\frac {10\,\left (\frac {11\,x}{2}+\frac {3}{2}\right )}{23\,\sqrt {2\,x^2-x+3}}+\frac {16\,x-4}{23\,\sqrt {2\,x^2-x+3}} \] Input:
int((3*x + 5*x^2 + 2)/(2*x^2 - x + 3)^(3/2),x)
Output:
(5*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/4 + (3*(2 *x - 12))/(23*(2*x^2 - x + 3)^(1/2)) - (10*((11*x)/2 + 3/2))/(23*(2*x^2 - x + 3)^(1/2)) + (16*x - 4)/(23*(2*x^2 - x + 3)^(1/2))
Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.36 \[ \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {-132 \sqrt {2 x^{2}-x +3}\, x -220 \sqrt {2 x^{2}-x +3}+230 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right ) x^{2}-115 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right ) x +345 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right )-132 \sqrt {2}\, x^{2}+66 \sqrt {2}\, x -198 \sqrt {2}}{184 x^{2}-92 x +276} \] Input:
int((5*x^2+3*x+2)/(2*x^2-x+3)^(3/2),x)
Output:
( - 132*sqrt(2*x**2 - x + 3)*x - 220*sqrt(2*x**2 - x + 3) + 230*sqrt(2)*lo g((2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23))*x**2 - 115*sqrt(2)* log((2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23))*x + 345*sqrt(2)*l og((2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23)) - 132*sqrt(2)*x**2 + 66*sqrt(2)*x - 198*sqrt(2))/(92*(2*x**2 - x + 3))