Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{112 c^3 d (b d+2 c d x)^{7/2}}+\frac {b^2-4 a c}{24 c^3 d^3 (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{16 c^3 d^5} \]
-1/112*(-4*a*c+b^2)^2/c^3/d/(2*c*d*x+b*d)^(7/2)+1/24*(-4*a*c+b^2)/c^3/d^3/ (2*c*d*x+b*d)^(3/2)+1/16*(2*c*d*x+b*d)^(1/2)/c^3/d^5
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\frac {-3 b^4+24 a b^2 c-48 a^2 c^2+14 b^2 (b+2 c x)^2-56 a c (b+2 c x)^2+21 (b+2 c x)^4}{336 c^3 d (d (b+2 c x))^{7/2}} \]
(-3*b^4 + 24*a*b^2*c - 48*a^2*c^2 + 14*b^2*(b + 2*c*x)^2 - 56*a*c*(b + 2*c *x)^2 + 21*(b + 2*c*x)^4)/(336*c^3*d*(d*(b + 2*c*x))^(7/2))
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1107, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {4 a c-b^2}{8 c^2 d^2 (b d+2 c d x)^{5/2}}+\frac {\left (4 a c-b^2\right )^2}{16 c^2 (b d+2 c d x)^{9/2}}+\frac {1}{16 c^2 d^4 \sqrt {b d+2 c d x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2-4 a c}{24 c^3 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right )^2}{112 c^3 d (b d+2 c d x)^{7/2}}+\frac {\sqrt {b d+2 c d x}}{16 c^3 d^5}\) |
-1/112*(b^2 - 4*a*c)^2/(c^3*d*(b*d + 2*c*d*x)^(7/2)) + (b^2 - 4*a*c)/(24*c ^3*d^3*(b*d + 2*c*d*x)^(3/2)) + Sqrt[b*d + 2*c*d*x]/(16*c^3*d^5)
3.13.75.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b*e, 0] && IGtQ[p, 0] && !(EqQ[ m, 3] && NeQ[p, 1])
Time = 2.56 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {\sqrt {2 c d x +b d}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{7 \left (2 c d x +b d \right )^{\frac {7}{2}}}}{16 c^{3} d^{5}}\) | \(82\) |
default | \(\frac {\sqrt {2 c d x +b d}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{7 \left (2 c d x +b d \right )^{\frac {7}{2}}}}{16 c^{3} d^{5}}\) | \(82\) |
pseudoelliptic | \(-\frac {-7 c^{4} x^{4}+\frac {14 x^{2} \left (-3 b x +a \right ) c^{3}}{3}+\left (-\frac {35}{3} b^{2} x^{2}+\frac {14}{3} a b x +a^{2}\right ) c^{2}+\frac {2 b^{2} \left (-7 b x +a \right ) c}{3}-\frac {2 b^{4}}{3}}{7 \sqrt {d \left (2 c x +b \right )}\, d^{4} \left (2 c x +b \right )^{3} c^{3}}\) | \(88\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-21 c^{4} x^{4}-42 b \,c^{3} x^{3}+14 x^{2} c^{3} a -35 b^{2} c^{2} x^{2}+14 a b \,c^{2} x -14 b^{3} c x +3 a^{2} c^{2}+2 a \,b^{2} c -2 b^{4}\right )}{21 c^{3} \left (2 c d x +b d \right )^{\frac {9}{2}}}\) | \(96\) |
trager | \(-\frac {\left (-21 c^{4} x^{4}-42 b \,c^{3} x^{3}+14 x^{2} c^{3} a -35 b^{2} c^{2} x^{2}+14 a b \,c^{2} x -14 b^{3} c x +3 a^{2} c^{2}+2 a \,b^{2} c -2 b^{4}\right ) \sqrt {2 c d x +b d}}{21 d^{5} \left (2 c x +b \right )^{4} c^{3}}\) | \(101\) |
1/16/c^3/d^5*((2*c*d*x+b*d)^(1/2)-2/3*d^2*(4*a*c-b^2)/(2*c*d*x+b*d)^(3/2)- 1/7*d^4*(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)^(7/2))
Time = 0.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\frac {{\left (21 \, c^{4} x^{4} + 42 \, b c^{3} x^{3} + 2 \, b^{4} - 2 \, a b^{2} c - 3 \, a^{2} c^{2} + 7 \, {\left (5 \, b^{2} c^{2} - 2 \, a c^{3}\right )} x^{2} + 14 \, {\left (b^{3} c - a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{21 \, {\left (16 \, c^{7} d^{5} x^{4} + 32 \, b c^{6} d^{5} x^{3} + 24 \, b^{2} c^{5} d^{5} x^{2} + 8 \, b^{3} c^{4} d^{5} x + b^{4} c^{3} d^{5}\right )}} \]
1/21*(21*c^4*x^4 + 42*b*c^3*x^3 + 2*b^4 - 2*a*b^2*c - 3*a^2*c^2 + 7*(5*b^2 *c^2 - 2*a*c^3)*x^2 + 14*(b^3*c - a*b*c^2)*x)*sqrt(2*c*d*x + b*d)/(16*c^7* d^5*x^4 + 32*b*c^6*d^5*x^3 + 24*b^2*c^5*d^5*x^2 + 8*b^3*c^4*d^5*x + b^4*c^ 3*d^5)
Leaf count of result is larger than twice the leaf count of optimal. 826 vs. \(2 (82) = 164\).
Time = 0.73 (sec) , antiderivative size = 826, normalized size of antiderivative = 9.39 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\begin {cases} - \frac {3 a^{2} c^{2} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} - \frac {2 a b^{2} c \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} - \frac {14 a b c^{2} x \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} - \frac {14 a c^{3} x^{2} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} + \frac {2 b^{4} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} + \frac {14 b^{3} c x \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} + \frac {35 b^{2} c^{2} x^{2} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} + \frac {42 b c^{3} x^{3} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} + \frac {21 c^{4} x^{4} \sqrt {b d + 2 c d x}}{21 b^{4} c^{3} d^{5} + 168 b^{3} c^{4} d^{5} x + 504 b^{2} c^{5} d^{5} x^{2} + 672 b c^{6} d^{5} x^{3} + 336 c^{7} d^{5} x^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{\left (b d\right )^{\frac {9}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-3*a**2*c**2*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3* c**4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d* *5*x**4) - 2*a*b**2*c*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3*c* *4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d**5 *x**4) - 14*a*b*c**2*x*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3*c **4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d** 5*x**4) - 14*a*c**3*x**2*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3 *c**4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d **5*x**4) + 2*b**4*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3*c**4* d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d**5*x* *4) + 14*b**3*c*x*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3*c**4*d **5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d**5*x** 4) + 35*b**2*c**2*x**2*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3*c **4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d** 5*x**4) + 42*b*c**3*x**3*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3 *c**4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d **5*x**4) + 21*c**4*x**4*sqrt(b*d + 2*c*d*x)/(21*b**4*c**3*d**5 + 168*b**3 *c**4*d**5*x + 504*b**2*c**5*d**5*x**2 + 672*b*c**6*d**5*x**3 + 336*c**7*d **5*x**4), Ne(c, 0)), ((a**2*x + a*b*x**2 + b**2*x**3/3)/(b*d)**(9/2), Tru e))
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\frac {\frac {21 \, \sqrt {2 \, c d x + b d}}{c^{2} d^{4}} + \frac {14 \, {\left (2 \, c d x + b d\right )}^{2} {\left (b^{2} - 4 \, a c\right )} - 3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{2}}{{\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{2}}}{336 \, c d} \]
1/336*(21*sqrt(2*c*d*x + b*d)/(c^2*d^4) + (14*(2*c*d*x + b*d)^2*(b^2 - 4*a *c) - 3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^2)/((2*c*d*x + b*d)^(7/2)*c^2*d^2 ))/(c*d)
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\frac {\sqrt {2 \, c d x + b d}}{16 \, c^{3} d^{5}} - \frac {3 \, b^{4} d^{2} - 24 \, a b^{2} c d^{2} + 48 \, a^{2} c^{2} d^{2} - 14 \, {\left (2 \, c d x + b d\right )}^{2} b^{2} + 56 \, {\left (2 \, c d x + b d\right )}^{2} a c}{336 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{3} d^{3}} \]
1/16*sqrt(2*c*d*x + b*d)/(c^3*d^5) - 1/336*(3*b^4*d^2 - 24*a*b^2*c*d^2 + 4 8*a^2*c^2*d^2 - 14*(2*c*d*x + b*d)^2*b^2 + 56*(2*c*d*x + b*d)^2*a*c)/((2*c *d*x + b*d)^(7/2)*c^3*d^3)
Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{9/2}} \, dx=\frac {-3\,a^2\,c^2-2\,a\,b^2\,c-14\,a\,b\,c^2\,x-14\,a\,c^3\,x^2+2\,b^4+14\,b^3\,c\,x+35\,b^2\,c^2\,x^2+42\,b\,c^3\,x^3+21\,c^4\,x^4}{21\,c^3\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}} \]