Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{144 c^3 d (b d+2 c d x)^{9/2}}+\frac {b^2-4 a c}{40 c^3 d^3 (b d+2 c d x)^{5/2}}-\frac {1}{16 c^3 d^5 \sqrt {b d+2 c d x}} \] Output:
-1/144*(-4*a*c+b^2)^2/c^3/d/(2*c*d*x+b*d)^(9/2)+1/40*(-4*a*c+b^2)/c^3/d^3/ (2*c*d*x+b*d)^(5/2)-1/16/c^3/d^5/(2*c*d*x+b*d)^(1/2)
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=\frac {-5 b^4+40 a b^2 c-80 a^2 c^2+18 b^2 (b+2 c x)^2-72 a c (b+2 c x)^2-45 (b+2 c x)^4}{720 c^3 d (d (b+2 c x))^{9/2}} \] Input:
Integrate[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(11/2),x]
Output:
(-5*b^4 + 40*a*b^2*c - 80*a^2*c^2 + 18*b^2*(b + 2*c*x)^2 - 72*a*c*(b + 2*c *x)^2 - 45*(b + 2*c*x)^4)/(720*c^3*d*(d*(b + 2*c*x))^(9/2))
Time = 0.22 (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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {4 a c-b^2}{8 c^2 d^2 (b d+2 c d x)^{7/2}}+\frac {\left (4 a c-b^2\right )^2}{16 c^2 (b d+2 c d x)^{11/2}}+\frac {1}{16 c^2 d^4 (b d+2 c d x)^{3/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2-4 a c}{40 c^3 d^3 (b d+2 c d x)^{5/2}}-\frac {\left (b^2-4 a c\right )^2}{144 c^3 d (b d+2 c d x)^{9/2}}-\frac {1}{16 c^3 d^5 \sqrt {b d+2 c d x}}\) |
Input:
Int[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(11/2),x]
Output:
-1/144*(b^2 - 4*a*c)^2/(c^3*d*(b*d + 2*c*d*x)^(9/2)) + (b^2 - 4*a*c)/(40*c ^3*d^3*(b*d + 2*c*d*x)^(5/2)) - 1/(16*c^3*d^5*Sqrt[b*d + 2*c*d*x])
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 = 0.92 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {1}{\sqrt {2 c d x +b d}}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{16 c^{3} d^{5}}\) | \(84\) |
default | \(\frac {-\frac {1}{\sqrt {2 c d x +b d}}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{16 c^{3} d^{5}}\) | \(84\) |
pseudoelliptic | \(-\frac {9 c^{4} x^{4}+\frac {18 x^{2} \left (5 b x +a \right ) c^{3}}{5}+\left (\frac {63}{5} b^{2} x^{2}+\frac {18}{5} a b x +a^{2}\right ) c^{2}+\frac {2 b^{2} \left (9 b x +a \right ) c}{5}+\frac {2 b^{4}}{5}}{9 \sqrt {d \left (2 c x +b \right )}\, d^{5} \left (2 c x +b \right )^{4} c^{3}}\) | \(88\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (45 c^{4} x^{4}+90 b \,c^{3} x^{3}+18 a \,c^{3} x^{2}+63 b^{2} c^{2} x^{2}+18 a b \,c^{2} x +18 b^{3} c x +5 a^{2} c^{2}+2 c a \,b^{2}+2 b^{4}\right )}{45 c^{3} \left (2 c d x +b d \right )^{\frac {11}{2}}}\) | \(96\) |
orering | \(-\frac {\left (2 c x +b \right ) \left (45 c^{4} x^{4}+90 b \,c^{3} x^{3}+18 a \,c^{3} x^{2}+63 b^{2} c^{2} x^{2}+18 a b \,c^{2} x +18 b^{3} c x +5 a^{2} c^{2}+2 c a \,b^{2}+2 b^{4}\right )}{45 c^{3} \left (2 c d x +b d \right )^{\frac {11}{2}}}\) | \(96\) |
trager | \(-\frac {\left (45 c^{4} x^{4}+90 b \,c^{3} x^{3}+18 a \,c^{3} x^{2}+63 b^{2} c^{2} x^{2}+18 a b \,c^{2} x +18 b^{3} c x +5 a^{2} c^{2}+2 c a \,b^{2}+2 b^{4}\right ) \sqrt {2 c d x +b d}}{45 d^{6} \left (2 c x +b \right )^{5} c^{3}}\) | \(101\) |
Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(11/2),x,method=_RETURNVERBOSE)
Output:
1/16/c^3/d^5*(-1/(2*c*d*x+b*d)^(1/2)-2/5*d^2*(4*a*c-b^2)/(2*c*d*x+b*d)^(5/ 2)-1/9*d^4*(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)^(9/2))
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (76) = 152\).
Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=-\frac {{\left (45 \, c^{4} x^{4} + 90 \, b c^{3} x^{3} + 2 \, b^{4} + 2 \, a b^{2} c + 5 \, a^{2} c^{2} + 9 \, {\left (7 \, b^{2} c^{2} + 2 \, a c^{3}\right )} x^{2} + 18 \, {\left (b^{3} c + a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, {\left (32 \, c^{8} d^{6} x^{5} + 80 \, b c^{7} d^{6} x^{4} + 80 \, b^{2} c^{6} d^{6} x^{3} + 40 \, b^{3} c^{5} d^{6} x^{2} + 10 \, b^{4} c^{4} d^{6} x + b^{5} c^{3} d^{6}\right )}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(11/2),x, algorithm="fricas")
Output:
-1/45*(45*c^4*x^4 + 90*b*c^3*x^3 + 2*b^4 + 2*a*b^2*c + 5*a^2*c^2 + 9*(7*b^ 2*c^2 + 2*a*c^3)*x^2 + 18*(b^3*c + a*b*c^2)*x)*sqrt(2*c*d*x + b*d)/(32*c^8 *d^6*x^5 + 80*b*c^7*d^6*x^4 + 80*b^2*c^6*d^6*x^3 + 40*b^3*c^5*d^6*x^2 + 10 *b^4*c^4*d^6*x + b^5*c^3*d^6)
Leaf count of result is larger than twice the leaf count of optimal. 966 vs. \(2 (83) = 166\).
Time = 0.94 (sec) , antiderivative size = 966, normalized size of antiderivative = 10.98 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx =\text {Too large to display} \] Input:
integrate((c*x**2+b*x+a)**2/(2*c*d*x+b*d)**(11/2),x)
Output:
Piecewise((-5*a**2*c**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3*d**6 + 450*b**4* c**4*d**6*x + 1800*b**3*c**5*d**6*x**2 + 3600*b**2*c**6*d**6*x**3 + 3600*b *c**7*d**6*x**4 + 1440*c**8*d**6*x**5) - 2*a*b**2*c*sqrt(b*d + 2*c*d*x)/(4 5*b**5*c**3*d**6 + 450*b**4*c**4*d**6*x + 1800*b**3*c**5*d**6*x**2 + 3600* b**2*c**6*d**6*x**3 + 3600*b*c**7*d**6*x**4 + 1440*c**8*d**6*x**5) - 18*a* b*c**2*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3*d**6 + 450*b**4*c**4*d**6*x + 1 800*b**3*c**5*d**6*x**2 + 3600*b**2*c**6*d**6*x**3 + 3600*b*c**7*d**6*x**4 + 1440*c**8*d**6*x**5) - 18*a*c**3*x**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3 *d**6 + 450*b**4*c**4*d**6*x + 1800*b**3*c**5*d**6*x**2 + 3600*b**2*c**6*d **6*x**3 + 3600*b*c**7*d**6*x**4 + 1440*c**8*d**6*x**5) - 2*b**4*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3*d**6 + 450*b**4*c**4*d**6*x + 1800*b**3*c**5*d**6 *x**2 + 3600*b**2*c**6*d**6*x**3 + 3600*b*c**7*d**6*x**4 + 1440*c**8*d**6* x**5) - 18*b**3*c*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3*d**6 + 450*b**4*c**4 *d**6*x + 1800*b**3*c**5*d**6*x**2 + 3600*b**2*c**6*d**6*x**3 + 3600*b*c** 7*d**6*x**4 + 1440*c**8*d**6*x**5) - 63*b**2*c**2*x**2*sqrt(b*d + 2*c*d*x) /(45*b**5*c**3*d**6 + 450*b**4*c**4*d**6*x + 1800*b**3*c**5*d**6*x**2 + 36 00*b**2*c**6*d**6*x**3 + 3600*b*c**7*d**6*x**4 + 1440*c**8*d**6*x**5) - 90 *b*c**3*x**3*sqrt(b*d + 2*c*d*x)/(45*b**5*c**3*d**6 + 450*b**4*c**4*d**6*x + 1800*b**3*c**5*d**6*x**2 + 3600*b**2*c**6*d**6*x**3 + 3600*b*c**7*d**6* x**4 + 1440*c**8*d**6*x**5) - 45*c**4*x**4*sqrt(b*d + 2*c*d*x)/(45*b**5...
Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=\frac {18 \, {\left (2 \, c d x + b d\right )}^{2} {\left (b^{2} - 4 \, a c\right )} d^{2} - 5 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{4} - 45 \, {\left (2 \, c d x + b d\right )}^{4}}{720 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{3} d^{5}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(11/2),x, algorithm="maxima")
Output:
1/720*(18*(2*c*d*x + b*d)^2*(b^2 - 4*a*c)*d^2 - 5*(b^4 - 8*a*b^2*c + 16*a^ 2*c^2)*d^4 - 45*(2*c*d*x + b*d)^4)/((2*c*d*x + b*d)^(9/2)*c^3*d^5)
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=-\frac {5 \, b^{4} d^{4} - 40 \, a b^{2} c d^{4} + 80 \, a^{2} c^{2} d^{4} - 18 \, {\left (2 \, c d x + b d\right )}^{2} b^{2} d^{2} + 72 \, {\left (2 \, c d x + b d\right )}^{2} a c d^{2} + 45 \, {\left (2 \, c d x + b d\right )}^{4}}{720 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{3} d^{5}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(11/2),x, algorithm="giac")
Output:
-1/720*(5*b^4*d^4 - 40*a*b^2*c*d^4 + 80*a^2*c^2*d^4 - 18*(2*c*d*x + b*d)^2 *b^2*d^2 + 72*(2*c*d*x + b*d)^2*a*c*d^2 + 45*(2*c*d*x + b*d)^4)/((2*c*d*x + b*d)^(9/2)*c^3*d^5)
Time = 5.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=-\frac {{\left (b+2\,c\,x\right )}^4+\left (\frac {8\,a\,c}{5}-\frac {2\,b^2}{5}\right )\,{\left (b+2\,c\,x\right )}^2+\frac {b^4}{9}+\frac {16\,a^2\,c^2}{9}-\frac {8\,a\,b^2\,c}{9}}{16\,c^3\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{9/2}} \] Input:
int((a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(11/2),x)
Output:
-((b + 2*c*x)^4 + ((8*a*c)/5 - (2*b^2)/5)*(b + 2*c*x)^2 + b^4/9 + (16*a^2* c^2)/9 - (8*a*b^2*c)/9)/(16*c^3*d*(b*d + 2*c*d*x)^(9/2))
Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{11/2}} \, dx=\frac {\sqrt {d}\, \left (-45 c^{4} x^{4}-90 b \,c^{3} x^{3}-18 a \,c^{3} x^{2}-63 b^{2} c^{2} x^{2}-18 a b \,c^{2} x -18 b^{3} c x -5 a^{2} c^{2}-2 a \,b^{2} c -2 b^{4}\right )}{45 \sqrt {2 c x +b}\, c^{3} d^{6} \left (16 c^{4} x^{4}+32 b \,c^{3} x^{3}+24 b^{2} c^{2} x^{2}+8 b^{3} c x +b^{4}\right )} \] Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(11/2),x)
Output:
(sqrt(d)*( - 5*a**2*c**2 - 2*a*b**2*c - 18*a*b*c**2*x - 18*a*c**3*x**2 - 2 *b**4 - 18*b**3*c*x - 63*b**2*c**2*x**2 - 90*b*c**3*x**3 - 45*c**4*x**4))/ (45*sqrt(b + 2*c*x)*c**3*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))