Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{16 c^3 d \sqrt {b d+2 c d x}}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{24 c^3 d^3}+\frac {(b d+2 c d x)^{7/2}}{112 c^3 d^5} \] Output:
-1/16*(-4*a*c+b^2)^2/c^3/d/(2*c*d*x+b*d)^(1/2)-1/24*(-4*a*c+b^2)*(2*c*d*x+ b*d)^(3/2)/c^3/d^3+1/112*(2*c*d*x+b*d)^(7/2)/c^3/d^5
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {-21 b^4+168 a b^2 c-336 a^2 c^2-14 b^2 (b+2 c x)^2+56 a c (b+2 c x)^2+3 (b+2 c x)^4}{336 c^3 d \sqrt {d (b+2 c x)}} \] Input:
Integrate[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(3/2),x]
Output:
(-21*b^4 + 168*a*b^2*c - 336*a^2*c^2 - 14*b^2*(b + 2*c*x)^2 + 56*a*c*(b + 2*c*x)^2 + 3*(b + 2*c*x)^4)/(336*c^3*d*Sqrt[d*(b + 2*c*x)])
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {\left (4 a c-b^2\right ) \sqrt {b d+2 c d x}}{8 c^2 d^2}+\frac {\left (4 a c-b^2\right )^2}{16 c^2 (b d+2 c d x)^{3/2}}+\frac {(b d+2 c d x)^{5/2}}{16 c^2 d^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{24 c^3 d^3}-\frac {\left (b^2-4 a c\right )^2}{16 c^3 d \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{7/2}}{112 c^3 d^5}\) |
Input:
Int[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(3/2),x]
Output:
-1/16*(b^2 - 4*a*c)^2/(c^3*d*Sqrt[b*d + 2*c*d*x]) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2))/(24*c^3*d^3) + (b*d + 2*c*d*x)^(7/2)/(112*c^3*d^5)
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 = 0.88 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(\frac {3 c^{4} x^{4}+6 b \,c^{3} x^{3}+14 a \,c^{3} x^{2}+b^{2} c^{2} x^{2}+14 a b \,c^{2} x -2 b^{3} c x -21 a^{2} c^{2}+14 c a \,b^{2}-2 b^{4}}{21 d \sqrt {d \left (2 c x +b \right )}\, c^{3}}\) | \(91\) |
derivativedivides | \(\frac {\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{\sqrt {2 c d x +b d}}}{16 c^{3} d^{5}}\) | \(95\) |
default | \(\frac {\frac {8 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{\sqrt {2 c d x +b d}}}{16 c^{3} d^{5}}\) | \(95\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-14 a \,c^{3} x^{2}-b^{2} c^{2} x^{2}-14 a b \,c^{2} x +2 b^{3} c x +21 a^{2} c^{2}-14 c a \,b^{2}+2 b^{4}\right )}{21 c^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}\) | \(96\) |
orering | \(-\frac {\left (2 c x +b \right ) \left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-14 a \,c^{3} x^{2}-b^{2} c^{2} x^{2}-14 a b \,c^{2} x +2 b^{3} c x +21 a^{2} c^{2}-14 c a \,b^{2}+2 b^{4}\right )}{21 c^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}\) | \(96\) |
trager | \(-\frac {\left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-14 a \,c^{3} x^{2}-b^{2} c^{2} x^{2}-14 a b \,c^{2} x +2 b^{3} c x +21 a^{2} c^{2}-14 c a \,b^{2}+2 b^{4}\right ) \sqrt {2 c d x +b d}}{21 d^{2} c^{3} \left (2 c x +b \right )}\) | \(101\) |
Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/21*(3*c^4*x^4+6*b*c^3*x^3+14*a*c^3*x^2+b^2*c^2*x^2+14*a*b*c^2*x-2*b^3*c* x-21*a^2*c^2+14*a*b^2*c-2*b^4)/d/(d*(2*c*x+b))^(1/2)/c^3
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {{\left (3 \, c^{4} x^{4} + 6 \, b c^{3} x^{3} - 2 \, b^{4} + 14 \, a b^{2} c - 21 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 14 \, a c^{3}\right )} x^{2} - 2 \, {\left (b^{3} c - 7 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{21 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(3/2),x, algorithm="fricas")
Output:
1/21*(3*c^4*x^4 + 6*b*c^3*x^3 - 2*b^4 + 14*a*b^2*c - 21*a^2*c^2 + (b^2*c^2 + 14*a*c^3)*x^2 - 2*(b^3*c - 7*a*b*c^2)*x)*sqrt(2*c*d*x + b*d)/(2*c^4*d^2 *x + b*c^3*d^2)
Time = 1.23 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{2}}{16 c^{2} \sqrt {b d + 2 c d x}} + \frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {3}{2}}}{24 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {7}{2}}}{112 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}}{\left (b d\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((c*x**2+b*x+a)**2/(2*c*d*x+b*d)**(3/2),x)
Output:
Piecewise(((-(4*a*c - b**2)**2/(16*c**2*sqrt(b*d + 2*c*d*x)) + (4*a*c - b* *2)*(b*d + 2*c*d*x)**(3/2)/(24*c**2*d**2) + (b*d + 2*c*d*x)**(7/2)/(112*c* *2*d**4))/(c*d), Ne(c*d, 0)), ((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)/(b*d)**(3/2), True))
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=-\frac {\frac {21 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{\sqrt {2 \, c d x + b d} c^{2}} + \frac {14 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 3 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}}{c^{2} d^{4}}}{336 \, c d} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(3/2),x, algorithm="maxima")
Output:
-1/336*(21*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(sqrt(2*c*d*x + b*d)*c^2) + (14* (2*c*d*x + b*d)^(3/2)*(b^2 - 4*a*c)*d^2 - 3*(2*c*d*x + b*d)^(7/2))/(c^2*d^ 4))/(c*d)
Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=-\frac {b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{16 \, \sqrt {2 \, c d x + b d} c^{3} d} - \frac {14 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{18} d^{32} - 56 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{19} d^{32} - 3 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{18} d^{30}}{336 \, c^{21} d^{35}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(3/2),x, algorithm="giac")
Output:
-1/16*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(sqrt(2*c*d*x + b*d)*c^3*d) - 1/336*( 14*(2*c*d*x + b*d)^(3/2)*b^2*c^18*d^32 - 56*(2*c*d*x + b*d)^(3/2)*a*c^19*d ^32 - 3*(2*c*d*x + b*d)^(7/2)*c^18*d^30)/(c^21*d^35)
Time = 5.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^4-21\,b^4\,d^4-14\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-336\,a^2\,c^2\,d^4+56\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+168\,a\,b^2\,c\,d^4}{336\,c^3\,d^5\,\sqrt {b\,d+2\,c\,d\,x}} \] Input:
int((a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(3/2),x)
Output:
(3*(b*d + 2*c*d*x)^4 - 21*b^4*d^4 - 14*b^2*d^2*(b*d + 2*c*d*x)^2 - 336*a^2 *c^2*d^4 + 56*a*c*d^2*(b*d + 2*c*d*x)^2 + 168*a*b^2*c*d^4)/(336*c^3*d^5*(b *d + 2*c*d*x)^(1/2))
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {\sqrt {d}\, \left (3 c^{4} x^{4}+6 b \,c^{3} x^{3}+14 a \,c^{3} x^{2}+b^{2} c^{2} x^{2}+14 a b \,c^{2} x -2 b^{3} c x -21 a^{2} c^{2}+14 a \,b^{2} c -2 b^{4}\right )}{21 \sqrt {2 c x +b}\, c^{3} d^{2}} \] Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(3/2),x)
Output:
(sqrt(d)*( - 21*a**2*c**2 + 14*a*b**2*c + 14*a*b*c**2*x + 14*a*c**3*x**2 - 2*b**4 - 2*b**3*c*x + b**2*c**2*x**2 + 6*b*c**3*x**3 + 3*c**4*x**4))/(21* sqrt(b + 2*c*x)*c**3*d**2)