Integrand size = 26, antiderivative size = 121 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {\left (b^2-4 a c\right )^3}{576 c^4 d (b d+2 c d x)^{9/2}}-\frac {3 \left (b^2-4 a c\right )^2}{320 c^4 d^3 (b d+2 c d x)^{5/2}}+\frac {3 \left (b^2-4 a c\right )}{64 c^4 d^5 \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{192 c^4 d^7} \] Output:
1/576*(-4*a*c+b^2)^3/c^4/d/(2*c*d*x+b*d)^(9/2)-3/320*(-4*a*c+b^2)^2/c^4/d^ 3/(2*c*d*x+b*d)^(5/2)+3/64*(-4*a*c+b^2)/c^4/d^5/(2*c*d*x+b*d)^(1/2)+1/192* (2*c*d*x+b*d)^(3/2)/c^4/d^7
Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {5 b^6-60 a b^4 c+240 a^2 b^2 c^2-320 a^3 c^3-27 b^4 (b+2 c x)^2+216 a b^2 c (b+2 c x)^2-432 a^2 c^2 (b+2 c x)^2+135 b^2 (b+2 c x)^4-540 a c (b+2 c x)^4+15 (b+2 c x)^6}{2880 c^4 d (d (b+2 c x))^{9/2}} \] Input:
Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x]
Output:
(5*b^6 - 60*a*b^4*c + 240*a^2*b^2*c^2 - 320*a^3*c^3 - 27*b^4*(b + 2*c*x)^2 + 216*a*b^2*c*(b + 2*c*x)^2 - 432*a^2*c^2*(b + 2*c*x)^2 + 135*b^2*(b + 2* c*x)^4 - 540*a*c*(b + 2*c*x)^4 + 15*(b + 2*c*x)^6)/(2880*c^4*d*(d*(b + 2*c *x))^(9/2))
Time = 0.25 (sec) , antiderivative size = 121, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {3 \left (4 a c-b^2\right )}{64 c^3 d^4 (b d+2 c d x)^{3/2}}+\frac {3 \left (4 a c-b^2\right )^2}{64 c^3 d^2 (b d+2 c d x)^{7/2}}+\frac {\left (4 a c-b^2\right )^3}{64 c^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {b d+2 c d x}}{64 c^3 d^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \left (b^2-4 a c\right )}{64 c^4 d^5 \sqrt {b d+2 c d x}}-\frac {3 \left (b^2-4 a c\right )^2}{320 c^4 d^3 (b d+2 c d x)^{5/2}}+\frac {\left (b^2-4 a c\right )^3}{576 c^4 d (b d+2 c d x)^{9/2}}+\frac {(b d+2 c d x)^{3/2}}{192 c^4 d^7}\) |
Input:
Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x]
Output:
(b^2 - 4*a*c)^3/(576*c^4*d*(b*d + 2*c*d*x)^(9/2)) - (3*(b^2 - 4*a*c)^2)/(3 20*c^4*d^3*(b*d + 2*c*d*x)^(5/2)) + (3*(b^2 - 4*a*c))/(64*c^4*d^5*Sqrt[b*d + 2*c*d*x]) + (b*d + 2*c*d*x)^(3/2)/(192*c^4*d^7)
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 = 0.99 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {3 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{6} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{64 c^{4} d^{7}}\) | \(132\) |
default | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {3 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{6} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{64 c^{4} d^{7}}\) | \(132\) |
pseudoelliptic | \(\frac {15 x^{6} c^{6}+\left (45 b \,x^{5}-135 a \,x^{4}\right ) c^{5}+\left (90 b^{2} x^{4}-270 a b \,x^{3}-27 a^{2} x^{2}\right ) c^{4}+\left (105 b^{3} x^{3}-189 a \,b^{2} x^{2}-27 a^{2} b x -5 a^{3}\right ) c^{3}+\left (63 b^{4} x^{2}-54 a \,b^{3} x -3 a^{2} b^{2}\right ) c^{2}-6 b^{4} \left (-3 b x +a \right ) c +2 b^{6}}{45 \sqrt {d \left (2 c x +b \right )}\, d^{5} \left (2 c x +b \right )^{4} c^{4}}\) | \(160\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-15 x^{6} c^{6}-45 x^{5} b \,c^{5}+135 a \,c^{5} x^{4}-90 x^{4} b^{2} c^{4}+270 a b \,c^{4} x^{3}-105 b^{3} c^{3} x^{3}+27 a^{2} c^{4} x^{2}+189 a \,b^{2} c^{3} x^{2}-63 c^{2} x^{2} b^{4}+27 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-18 x c \,b^{5}+5 a^{3} c^{3}+3 a^{2} b^{2} c^{2}+6 a \,b^{4} c -2 b^{6}\right )}{45 c^{4} \left (2 c d x +b d \right )^{\frac {11}{2}}}\) | \(174\) |
orering | \(-\frac {\left (2 c x +b \right ) \left (-15 x^{6} c^{6}-45 x^{5} b \,c^{5}+135 a \,c^{5} x^{4}-90 x^{4} b^{2} c^{4}+270 a b \,c^{4} x^{3}-105 b^{3} c^{3} x^{3}+27 a^{2} c^{4} x^{2}+189 a \,b^{2} c^{3} x^{2}-63 c^{2} x^{2} b^{4}+27 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-18 x c \,b^{5}+5 a^{3} c^{3}+3 a^{2} b^{2} c^{2}+6 a \,b^{4} c -2 b^{6}\right )}{45 c^{4} \left (2 c d x +b d \right )^{\frac {11}{2}}}\) | \(174\) |
trager | \(-\frac {\left (-15 x^{6} c^{6}-45 x^{5} b \,c^{5}+135 a \,c^{5} x^{4}-90 x^{4} b^{2} c^{4}+270 a b \,c^{4} x^{3}-105 b^{3} c^{3} x^{3}+27 a^{2} c^{4} x^{2}+189 a \,b^{2} c^{3} x^{2}-63 c^{2} x^{2} b^{4}+27 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-18 x c \,b^{5}+5 a^{3} c^{3}+3 a^{2} b^{2} c^{2}+6 a \,b^{4} c -2 b^{6}\right ) \sqrt {2 c d x +b d}}{45 d^{6} \left (2 c x +b \right )^{5} c^{4}}\) | \(179\) |
Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x,method=_RETURNVERBOSE)
Output:
1/64/c^4/d^7*(1/3*(2*c*d*x+b*d)^(3/2)-3*d^2*(4*a*c-b^2)/(2*c*d*x+b*d)^(1/2 )-3/5*d^4*(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)^(5/2)-1/9*d^6*(64*a^3*c ^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(2*c*d*x+b*d)^(9/2))
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (105) = 210\).
Time = 0.09 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (15 \, c^{6} x^{6} + 45 \, b c^{5} x^{5} + 2 \, b^{6} - 6 \, a b^{4} c - 3 \, a^{2} b^{2} c^{2} - 5 \, a^{3} c^{3} + 45 \, {\left (2 \, b^{2} c^{4} - 3 \, a c^{5}\right )} x^{4} + 15 \, {\left (7 \, b^{3} c^{3} - 18 \, a b c^{4}\right )} x^{3} + 9 \, {\left (7 \, b^{4} c^{2} - 21 \, a b^{2} c^{3} - 3 \, a^{2} c^{4}\right )} x^{2} + 9 \, {\left (2 \, b^{5} c - 6 \, a b^{3} c^{2} - 3 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, {\left (32 \, c^{9} d^{6} x^{5} + 80 \, b c^{8} d^{6} x^{4} + 80 \, b^{2} c^{7} d^{6} x^{3} + 40 \, b^{3} c^{6} d^{6} x^{2} + 10 \, b^{4} c^{5} d^{6} x + b^{5} c^{4} d^{6}\right )}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="fricas")
Output:
1/45*(15*c^6*x^6 + 45*b*c^5*x^5 + 2*b^6 - 6*a*b^4*c - 3*a^2*b^2*c^2 - 5*a^ 3*c^3 + 45*(2*b^2*c^4 - 3*a*c^5)*x^4 + 15*(7*b^3*c^3 - 18*a*b*c^4)*x^3 + 9 *(7*b^4*c^2 - 21*a*b^2*c^3 - 3*a^2*c^4)*x^2 + 9*(2*b^5*c - 6*a*b^3*c^2 - 3 *a^2*b*c^3)*x)*sqrt(2*c*d*x + b*d)/(32*c^9*d^6*x^5 + 80*b*c^8*d^6*x^4 + 80 *b^2*c^7*d^6*x^3 + 40*b^3*c^6*d^6*x^2 + 10*b^4*c^5*d^6*x + b^5*c^4*d^6)
Leaf count of result is larger than twice the leaf count of optimal. 1731 vs. \(2 (119) = 238\).
Time = 1.01 (sec) , antiderivative size = 1731, normalized size of antiderivative = 14.31 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\text {Too large to display} \] Input:
integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**(11/2),x)
Output:
Piecewise((-5*a**3*c**3*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4* c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b *c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 3*a**2*b**2*c**2*sqrt(b*d + 2*c*d *x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 27*a**2*b*c**3*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d **6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8* d**6*x**4 + 1440*c**9*d**6*x**5) - 27*a**2*c**4*x**2*sqrt(b*d + 2*c*d*x)/( 45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600 *b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 6*a* b**4*c*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 180 0*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 54*a*b**3*c**2*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4 *d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d **6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 189*a*b**2*c**3* x**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800* b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1 440*c**9*d**6*x**5) - 270*a*b*c**4*x**3*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4* d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d* *6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 135*a*c**5*x**...
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {\frac {15 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{c^{3} d^{6}} + \frac {135 \, {\left (2 \, c d x + b d\right )}^{4} {\left (b^{2} - 4 \, a c\right )} - 27 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{2} d^{2} + 5 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{4}}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{3} d^{4}}}{2880 \, c d} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="maxima")
Output:
1/2880*(15*(2*c*d*x + b*d)^(3/2)/(c^3*d^6) + (135*(2*c*d*x + b*d)^4*(b^2 - 4*a*c) - 27*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^2*d^2 + 5*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^4)/((2*c*d*x + b*d)^(9/2)*c ^3*d^4))/(c*d)
Time = 0.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{192 \, c^{4} d^{7}} + \frac {5 \, b^{6} d^{4} - 60 \, a b^{4} c d^{4} + 240 \, a^{2} b^{2} c^{2} d^{4} - 320 \, a^{3} c^{3} d^{4} - 27 \, {\left (2 \, c d x + b d\right )}^{2} b^{4} d^{2} + 216 \, {\left (2 \, c d x + b d\right )}^{2} a b^{2} c d^{2} - 432 \, {\left (2 \, c d x + b d\right )}^{2} a^{2} c^{2} d^{2} + 135 \, {\left (2 \, c d x + b d\right )}^{4} b^{2} - 540 \, {\left (2 \, c d x + b d\right )}^{4} a c}{2880 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{4} d^{5}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="giac")
Output:
1/192*(2*c*d*x + b*d)^(3/2)/(c^4*d^7) + 1/2880*(5*b^6*d^4 - 60*a*b^4*c*d^4 + 240*a^2*b^2*c^2*d^4 - 320*a^3*c^3*d^4 - 27*(2*c*d*x + b*d)^2*b^4*d^2 + 216*(2*c*d*x + b*d)^2*a*b^2*c*d^2 - 432*(2*c*d*x + b*d)^2*a^2*c^2*d^2 + 13 5*(2*c*d*x + b*d)^4*b^2 - 540*(2*c*d*x + b*d)^4*a*c)/((2*c*d*x + b*d)^(9/2 )*c^4*d^5)
Time = 0.05 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{192\,c^4\,d^7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (\frac {48\,a^2\,c^2\,d^2}{5}-\frac {24\,a\,b^2\,c\,d^2}{5}+\frac {3\,b^4\,d^2}{5}\right )+{\left (b\,d+2\,c\,d\,x\right )}^4\,\left (12\,a\,c-3\,b^2\right )-\frac {b^6\,d^4}{9}+\frac {64\,a^3\,c^3\,d^4}{9}-\frac {16\,a^2\,b^2\,c^2\,d^4}{3}+\frac {4\,a\,b^4\,c\,d^4}{3}}{64\,c^4\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{9/2}} \] Input:
int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x)
Output:
(b*d + 2*c*d*x)^(3/2)/(192*c^4*d^7) - ((b*d + 2*c*d*x)^2*((3*b^4*d^2)/5 + (48*a^2*c^2*d^2)/5 - (24*a*b^2*c*d^2)/5) + (b*d + 2*c*d*x)^4*(12*a*c - 3*b ^2) - (b^6*d^4)/9 + (64*a^3*c^3*d^4)/9 - (16*a^2*b^2*c^2*d^4)/3 + (4*a*b^4 *c*d^4)/3)/(64*c^4*d^5*(b*d + 2*c*d*x)^(9/2))
Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {\sqrt {d}\, \left (15 c^{6} x^{6}+45 b \,c^{5} x^{5}-135 a \,c^{5} x^{4}+90 b^{2} c^{4} x^{4}-270 a b \,c^{4} x^{3}+105 b^{3} c^{3} x^{3}-27 a^{2} c^{4} x^{2}-189 a \,b^{2} c^{3} x^{2}+63 b^{4} c^{2} x^{2}-27 a^{2} b \,c^{3} x -54 a \,b^{3} c^{2} x +18 b^{5} c x -5 a^{3} c^{3}-3 a^{2} b^{2} c^{2}-6 a \,b^{4} c +2 b^{6}\right )}{45 \sqrt {2 c x +b}\, c^{4} d^{6} \left (16 c^{4} x^{4}+32 b \,c^{3} x^{3}+24 b^{2} c^{2} x^{2}+8 b^{3} c x +b^{4}\right )} \] Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x)
Output:
(sqrt(d)*( - 5*a**3*c**3 - 3*a**2*b**2*c**2 - 27*a**2*b*c**3*x - 27*a**2*c **4*x**2 - 6*a*b**4*c - 54*a*b**3*c**2*x - 189*a*b**2*c**3*x**2 - 270*a*b* c**4*x**3 - 135*a*c**5*x**4 + 2*b**6 + 18*b**5*c*x + 63*b**4*c**2*x**2 + 1 05*b**3*c**3*x**3 + 90*b**2*c**4*x**4 + 45*b*c**5*x**5 + 15*c**6*x**6))/(4 5*sqrt(b + 2*c*x)*c**4*d**6*(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b* c**3*x**3 + 16*c**4*x**4))