Integrand size = 26, antiderivative size = 88 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}}{80 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2}}{72 c^3 d^3}+\frac {(b d+2 c d x)^{13/2}}{208 c^3 d^5} \]
1/80*(-4*a*c+b^2)^2*(2*c*d*x+b*d)^(5/2)/c^3/d-1/72*(-4*a*c+b^2)*(2*c*d*x+b *d)^(9/2)/c^3/d^3+1/208*(2*c*d*x+b*d)^(13/2)/c^3/d^5
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\frac {(d (b+2 c x))^{5/2} \left (117 b^4-936 a b^2 c+1872 a^2 c^2-130 b^2 (b+2 c x)^2+520 a c (b+2 c x)^2+45 (b+2 c x)^4\right )}{9360 c^3 d} \]
((d*(b + 2*c*x))^(5/2)*(117*b^4 - 936*a*b^2*c + 1872*a^2*c^2 - 130*b^2*(b + 2*c*x)^2 + 520*a*c*(b + 2*c*x)^2 + 45*(b + 2*c*x)^4))/(9360*c^3*d)
Time = 0.25 (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.
\(\displaystyle \int \left (a+b x+c x^2\right )^2 (b d+2 c d x)^{3/2} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {\left (4 a c-b^2\right ) (b d+2 c d x)^{7/2}}{8 c^2 d^2}+\frac {\left (4 a c-b^2\right )^2 (b d+2 c d x)^{3/2}}{16 c^2}+\frac {(b d+2 c d x)^{11/2}}{16 c^2 d^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2}}{72 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}}{80 c^3 d}+\frac {(b d+2 c d x)^{13/2}}{208 c^3 d^5}\) |
((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(5/2))/(80*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(9/2))/(72*c^3*d^3) + (b*d + 2*c*d*x)^(13/2)/(208*c^3*d^5)
3.13.69.3.1 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 = 2.57 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {13}{2}}}{13}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}}{16 d^{5} c^{3}}\) | \(83\) |
default | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {13}{2}}}{13}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}}{16 d^{5} c^{3}}\) | \(83\) |
gosper | \(\frac {\left (2 c x +b \right ) \left (45 c^{4} x^{4}+90 b \,c^{3} x^{3}+130 x^{2} c^{3} a +35 b^{2} c^{2} x^{2}+130 a b \,c^{2} x -10 b^{3} c x +117 a^{2} c^{2}-26 a \,b^{2} c +2 b^{4}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{585 c^{3}}\) | \(96\) |
pseudoelliptic | \(\frac {d \left (2 c x +b \right )^{2} \sqrt {d \left (2 c x +b \right )}\, \left (45 c^{4} x^{4}+90 b \,c^{3} x^{3}+130 x^{2} c^{3} a +35 b^{2} c^{2} x^{2}+130 a b \,c^{2} x -10 b^{3} c x +117 a^{2} c^{2}-26 a \,b^{2} c +2 b^{4}\right )}{585 c^{3}}\) | \(98\) |
trager | \(\frac {d \left (180 c^{6} x^{6}+540 b \,c^{5} x^{5}+520 a \,c^{5} x^{4}+545 b^{2} c^{4} x^{4}+1040 a b \,c^{4} x^{3}+190 x^{3} b^{3} c^{3}+468 a^{2} c^{4} x^{2}+546 a \,b^{2} c^{3} x^{2}+3 x^{2} b^{4} c^{2}+468 a^{2} b \,c^{3} x +26 x a \,b^{3} c^{2}-2 x \,b^{5} c +117 a^{2} b^{2} c^{2}-26 a \,b^{4} c +2 b^{6}\right ) \sqrt {2 c d x +b d}}{585 c^{3}}\) | \(161\) |
1/16/d^5/c^3*(1/13*(2*c*d*x+b*d)^(13/2)+1/9*(8*a*c*d^2-2*b^2*d^2)*(2*c*d*x +b*d)^(9/2)+1/5*(4*a*c*d^2-b^2*d^2)^2*(2*c*d*x+b*d)^(5/2))
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (76) = 152\).
Time = 0.24 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.86 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\frac {{\left (180 \, c^{6} d x^{6} + 540 \, b c^{5} d x^{5} + 5 \, {\left (109 \, b^{2} c^{4} + 104 \, a c^{5}\right )} d x^{4} + 10 \, {\left (19 \, b^{3} c^{3} + 104 \, a b c^{4}\right )} d x^{3} + 3 \, {\left (b^{4} c^{2} + 182 \, a b^{2} c^{3} + 156 \, a^{2} c^{4}\right )} d x^{2} - 2 \, {\left (b^{5} c - 13 \, a b^{3} c^{2} - 234 \, a^{2} b c^{3}\right )} d x + {\left (2 \, b^{6} - 26 \, a b^{4} c + 117 \, a^{2} b^{2} c^{2}\right )} d\right )} \sqrt {2 \, c d x + b d}}{585 \, c^{3}} \]
1/585*(180*c^6*d*x^6 + 540*b*c^5*d*x^5 + 5*(109*b^2*c^4 + 104*a*c^5)*d*x^4 + 10*(19*b^3*c^3 + 104*a*b*c^4)*d*x^3 + 3*(b^4*c^2 + 182*a*b^2*c^3 + 156* a^2*c^4)*d*x^2 - 2*(b^5*c - 13*a*b^3*c^2 - 234*a^2*b*c^3)*d*x + (2*b^6 - 2 6*a*b^4*c + 117*a^2*b^2*c^2)*d)*sqrt(2*c*d*x + b*d)/c^3
Time = 0.99 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\begin {cases} \frac {\frac {\left (b d + 2 c d x\right )^{\frac {5}{2}} \cdot \left (16 a^{2} c^{2} - 8 a b^{2} c + b^{4}\right )}{80 c^{2}} + \frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {9}{2}}}{72 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {13}{2}}}{208 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\left (b d\right )^{\frac {3}{2}} \left (a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((((b*d + 2*c*d*x)**(5/2)*(16*a**2*c**2 - 8*a*b**2*c + b**4)/(80* c**2) + (4*a*c - b**2)*(b*d + 2*c*d*x)**(9/2)/(72*c**2*d**2) + (b*d + 2*c* d*x)**(13/2)/(208*c**2*d**4))/(c*d), Ne(c*d, 0)), ((b*d)**(3/2)*(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 = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=-\frac {130 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 117 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} d^{4} - 45 \, {\left (2 \, c d x + b d\right )}^{\frac {13}{2}}}{9360 \, c^{3} d^{5}} \]
-1/9360*(130*(2*c*d*x + b*d)^(9/2)*(b^2 - 4*a*c)*d^2 - 117*(b^4 - 8*a*b^2* c + 16*a^2*c^2)*(2*c*d*x + b*d)^(5/2)*d^4 - 45*(2*c*d*x + b*d)^(13/2))/(c^ 3*d^5)
Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (76) = 152\).
Time = 0.28 (sec) , antiderivative size = 867, normalized size of antiderivative = 9.85 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\frac {720720 \, \sqrt {2 \, c d x + b d} a^{2} b^{2} d - 480480 \, {\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} b - \frac {240240 \, {\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^{3}}{c} + \frac {48048 \, {\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^{2}}{d} + \frac {12012 \, {\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^{4}}{c^{2} d} + \frac {120120 \, {\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^{2}}{c d} - \frac {15444 \, {\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^{3}}{c^{2} d^{2}} - \frac {41184 \, {\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 b}{c d^{2}} + \frac {1859 \, {\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^{2}}{c^{2} d^{3}} + \frac {1144 \, {\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 )} a}{c d^{3}} - \frac {390 \, {\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 )} b}{c^{2} d^{4}} + \frac {15 \, {\left (3003 \, \sqrt {2 \, c d x + b d} b^{6} d^{6} - 6006 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{5} d^{5} + 9009 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{4} d^{4} - 8580 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{3} d^{3} + 5005 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} b^{2} d^{2} - 1638 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}} b d + 231 \, {\left (2 \, c d x + b d\right )}^{\frac {13}{2}}\right )}}{c^{2} d^{5}}}{720720 \, c} \]
1/720720*(720720*sqrt(2*c*d*x + b*d)*a^2*b^2*d - 480480*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a^2*b - 240240*(3*sqrt(2*c*d*x + b*d)*b* d - (2*c*d*x + b*d)^(3/2))*a*b^3/c + 48048*(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^2/d + 12012*( 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^4/(c^2*d) + 120120*(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^2/(c*d) - 15444*(3 5*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^3/(c^2*d^2) - 41184*(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*b/(c*d^2) + 1859*(315*sq rt(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^2/(c^2*d^3) + 1144*(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))*a/(c*d^3) - 390*(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))*b/(c^2*d^4) + 15*(3003*sqrt( 2*c*d*x + b*d)*b^6*d^6 - 6006*(2*c*d*x + b*d)^(3/2)*b^5*d^5 + 9009*(2*c...
Time = 9.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (45\,{\left (b\,d+2\,c\,d\,x\right )}^4+117\,b^4\,d^4-130\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+1872\,a^2\,c^2\,d^4+520\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-936\,a\,b^2\,c\,d^4\right )}{9360\,c^3\,d^5} \]