Integrand size = 26, antiderivative size = 88 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac {(b d+2 c d x)^{11/2}}{176 c^3 d^5} \]
1/48*(-4*a*c+b^2)^2*(2*c*d*x+b*d)^(3/2)/c^3/d-1/56*(-4*a*c+b^2)*(2*c*d*x+b *d)^(7/2)/c^3/d^3+1/176*(2*c*d*x+b*d)^(11/2)/c^3/d^5
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\frac {(d (b+2 c x))^{3/2} \left (77 b^4-616 a b^2 c+1232 a^2 c^2-66 b^2 (b+2 c x)^2+264 a c (b+2 c x)^2+21 (b+2 c x)^4\right )}{3696 c^3 d} \]
((d*(b + 2*c*x))^(3/2)*(77*b^4 - 616*a*b^2*c + 1232*a^2*c^2 - 66*b^2*(b + 2*c*x)^2 + 264*a*c*(b + 2*c*x)^2 + 21*(b + 2*c*x)^4))/(3696*c^3*d)
Time = 0.24 (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.
\(\displaystyle \int \left (a+b x+c x^2\right )^2 \sqrt {b d+2 c d x} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {\left (4 a c-b^2\right ) (b d+2 c d x)^{5/2}}{8 c^2 d^2}+\frac {\left (4 a c-b^2\right )^2 \sqrt {b d+2 c d x}}{16 c^2}+\frac {(b d+2 c d x)^{9/2}}{16 c^2 d^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}+\frac {(b d+2 c d x)^{11/2}}{176 c^3 d^5}\) |
((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2))/(48*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(56*c^3*d^3) + (b*d + 2*c*d*x)^(11/2)/(176*c^3*d^5)
3.13.70.3.1 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 = 2.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}}{16 d^{5} c^{3}}\) | \(83\) |
default | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}}{16 d^{5} c^{3}}\) | \(83\) |
pseudoelliptic | \(\frac {\left (2 c x +b \right ) \sqrt {d \left (2 c x +b \right )}\, \left (21 c^{4} x^{4}+42 b \,c^{3} x^{3}+66 x^{2} c^{3} a +15 b^{2} c^{2} x^{2}+66 a b \,c^{2} x -6 b^{3} c x +77 a^{2} c^{2}-22 a \,b^{2} c +2 b^{4}\right )}{231 c^{3}}\) | \(95\) |
gosper | \(\frac {\left (2 c x +b \right ) \left (21 c^{4} x^{4}+42 b \,c^{3} x^{3}+66 x^{2} c^{3} a +15 b^{2} c^{2} x^{2}+66 a b \,c^{2} x -6 b^{3} c x +77 a^{2} c^{2}-22 a \,b^{2} c +2 b^{4}\right ) \sqrt {2 c d x +b d}}{231 c^{3}}\) | \(96\) |
trager | \(\frac {\left (42 c^{5} x^{5}+105 b \,x^{4} c^{4}+132 a \,c^{4} x^{3}+72 b^{2} c^{3} x^{3}+198 a b \,c^{3} x^{2}+3 x^{2} b^{3} c^{2}+154 a^{2} c^{3} x +22 a \,c^{2} b^{2} x -2 c x \,b^{4}+77 a^{2} b \,c^{2}-22 a \,b^{3} c +2 b^{5}\right ) \sqrt {2 c d x +b d}}{231 c^{3}}\) | \(123\) |
1/16/d^5/c^3*(1/11*(2*c*d*x+b*d)^(11/2)+1/7*(8*a*c*d^2-2*b^2*d^2)*(2*c*d*x +b*d)^(7/2)+1/3*(4*a*c*d^2-b^2*d^2)^2*(2*c*d*x+b*d)^(3/2))
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.38 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\frac {{\left (42 \, c^{5} x^{5} + 105 \, b c^{4} x^{4} + 2 \, b^{5} - 22 \, a b^{3} c + 77 \, a^{2} b c^{2} + 12 \, {\left (6 \, b^{2} c^{3} + 11 \, a c^{4}\right )} x^{3} + 3 \, {\left (b^{3} c^{2} + 66 \, a b c^{3}\right )} x^{2} - 2 \, {\left (b^{4} c - 11 \, a b^{2} c^{2} - 77 \, a^{2} c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{231 \, c^{3}} \]
1/231*(42*c^5*x^5 + 105*b*c^4*x^4 + 2*b^5 - 22*a*b^3*c + 77*a^2*b*c^2 + 12 *(6*b^2*c^3 + 11*a*c^4)*x^3 + 3*(b^3*c^2 + 66*a*b*c^3)*x^2 - 2*(b^4*c - 11 *a*b^2*c^2 - 77*a^2*c^3)*x)*sqrt(2*c*d*x + b*d)/c^3
Time = 0.88 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\begin {cases} \frac {\frac {\left (b d + 2 c d x\right )^{\frac {3}{2}} \cdot \left (16 a^{2} c^{2} - 8 a b^{2} c + b^{4}\right )}{48 c^{2}} + \frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {7}{2}}}{56 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {11}{2}}}{176 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\sqrt {b d} \left (a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((((b*d + 2*c*d*x)**(3/2)*(16*a**2*c**2 - 8*a*b**2*c + b**4)/(48* c**2) + (4*a*c - b**2)*(b*d + 2*c*d*x)**(7/2)/(56*c**2*d**2) + (b*d + 2*c* d*x)**(11/2)/(176*c**2*d**4))/(c*d), Ne(c*d, 0)), (sqrt(b*d)*(a**2*x + a*b *x**2 + b*c*x**4/2 + c**2*x**5/5 + x**3*(2*a*c + b**2)/3), True))
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=-\frac {66 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 77 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d^{4} - 21 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{3696 \, c^{3} d^{5}} \]
-1/3696*(66*(2*c*d*x + b*d)^(7/2)*(b^2 - 4*a*c)*d^2 - 77*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(3/2)*d^4 - 21*(2*c*d*x + b*d)^(11/2))/(c^3* d^5)
Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (76) = 152\).
Time = 0.27 (sec) , antiderivative size = 580, normalized size of antiderivative = 6.59 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\frac {55440 \, \sqrt {2 \, c d x + b d} a^{2} b - \frac {18480 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a^{2}}{d} - \frac {18480 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a b^{2}}{c d} + \frac {924 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} b^{3}}{c^{2} d^{2}} + \frac {5544 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} a b}{c d^{2}} - \frac {792 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} b^{2}}{c^{2} d^{3}} - \frac {792 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} a}{c d^{3}} + \frac {55 \, {\left (315 \, \sqrt {2 \, c d x + b d} b^{4} d^{4} - 420 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{3} d^{3} + 378 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} d^{2} - 180 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b d + 35 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}\right )} b}{c^{2} d^{4}} - \frac {5 \, {\left (693 \, \sqrt {2 \, c d x + b d} b^{5} d^{5} - 1155 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} d^{4} + 1386 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{3} d^{3} - 990 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{2} d^{2} + 385 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} b d - 63 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}\right )}}{c^{2} d^{5}}}{55440 \, c} \]
1/55440*(55440*sqrt(2*c*d*x + b*d)*a^2*b - 18480*(3*sqrt(2*c*d*x + b*d)*b* d - (2*c*d*x + b*d)^(3/2))*a^2/d - 18480*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c *d*x + b*d)^(3/2))*a*b^2/(c*d) + 924*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10* (2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*b^3/(c^2*d^2) + 5544* (15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d* x + b*d)^(5/2))*a*b/(c*d^2) - 792*(35*sqrt(2*c*d*x + b*d)*b^3*d^3 - 35*(2* c*d*x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b *d)^(7/2))*b^2/(c^2*d^3) - 792*(35*sqrt(2*c*d*x + b*d)*b^3*d^3 - 35*(2*c*d *x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b*d) ^(7/2))*a/(c*d^3) + 55*(315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 420*(2*c*d*x + b *d)^(3/2)*b^3*d^3 + 378*(2*c*d*x + b*d)^(5/2)*b^2*d^2 - 180*(2*c*d*x + b*d )^(7/2)*b*d + 35*(2*c*d*x + b*d)^(9/2))*b/(c^2*d^4) - 5*(693*sqrt(2*c*d*x + b*d)*b^5*d^5 - 1155*(2*c*d*x + b*d)^(3/2)*b^4*d^4 + 1386*(2*c*d*x + b*d) ^(5/2)*b^3*d^3 - 990*(2*c*d*x + b*d)^(7/2)*b^2*d^2 + 385*(2*c*d*x + b*d)^( 9/2)*b*d - 63*(2*c*d*x + b*d)^(11/2))/(c^2*d^5))/c
Time = 9.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (21\,{\left (b\,d+2\,c\,d\,x\right )}^4+77\,b^4\,d^4-66\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+1232\,a^2\,c^2\,d^4+264\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-616\,a\,b^2\,c\,d^4\right )}{3696\,c^3\,d^5} \]