Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{8 c^3 d^3}+\frac {(b d+2 c d x)^{5/2}}{80 c^3 d^5} \] Output:
-1/48*(-4*a*c+b^2)^2/c^3/d/(2*c*d*x+b*d)^(3/2)-1/8*(-4*a*c+b^2)*(2*c*d*x+b *d)^(1/2)/c^3/d^3+1/80*(2*c*d*x+b*d)^(5/2)/c^3/d^5
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {-5 b^4+40 a b^2 c-80 a^2 c^2-30 b^2 (b+2 c x)^2+120 a c (b+2 c x)^2+3 (b+2 c x)^4}{240 c^3 d (d (b+2 c x))^{3/2}} \] Input:
Integrate[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(5/2),x]
Output:
(-5*b^4 + 40*a*b^2*c - 80*a^2*c^2 - 30*b^2*(b + 2*c*x)^2 + 120*a*c*(b + 2* c*x)^2 + 3*(b + 2*c*x)^4)/(240*c^3*d*(d*(b + 2*c*x))^(3/2))
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {4 a c-b^2}{8 c^2 d^2 \sqrt {b d+2 c d x}}+\frac {\left (4 a c-b^2\right )^2}{16 c^2 (b d+2 c d x)^{5/2}}+\frac {(b d+2 c d x)^{3/2}}{16 c^2 d^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{8 c^3 d^3}-\frac {\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}+\frac {(b d+2 c d x)^{5/2}}{80 c^3 d^5}\) |
Input:
Int[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(5/2),x]
Output:
-1/48*(b^2 - 4*a*c)^2/(c^3*d*(b*d + 2*c*d*x)^(3/2)) - ((b^2 - 4*a*c)*Sqrt[ b*d + 2*c*d*x])/(8*c^3*d^3) + (b*d + 2*c*d*x)^(5/2)/(80*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 = 95, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {8 a c \,d^{2} \sqrt {2 c d x +b d}-2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{16 c^{3} d^{5}}\) | \(95\) |
default | \(\frac {8 a c \,d^{2} \sqrt {2 c d x +b d}-2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{16 c^{3} d^{5}}\) | \(95\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-30 a \,c^{3} x^{2}+3 b^{2} c^{2} x^{2}-30 a b \,c^{2} x +6 b^{3} c x +5 a^{2} c^{2}-10 c a \,b^{2}+2 b^{4}\right )}{15 c^{3} \left (2 c d x +b d \right )^{\frac {5}{2}}}\) | \(96\) |
orering | \(-\frac {\left (2 c x +b \right ) \left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-30 a \,c^{3} x^{2}+3 b^{2} c^{2} x^{2}-30 a b \,c^{2} x +6 b^{3} c x +5 a^{2} c^{2}-10 c a \,b^{2}+2 b^{4}\right )}{15 c^{3} \left (2 c d x +b d \right )^{\frac {5}{2}}}\) | \(96\) |
pseudoelliptic | \(\frac {3 c^{4} x^{4}+6 b \,c^{3} x^{3}+30 a \,c^{3} x^{2}-3 b^{2} c^{2} x^{2}+30 a b \,c^{2} x -6 b^{3} c x -5 a^{2} c^{2}+10 c a \,b^{2}-2 b^{4}}{15 d^{2} \left (2 c x +b \right ) \sqrt {d \left (2 c x +b \right )}\, c^{3}}\) | \(100\) |
trager | \(-\frac {\left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-30 a \,c^{3} x^{2}+3 b^{2} c^{2} x^{2}-30 a b \,c^{2} x +6 b^{3} c x +5 a^{2} c^{2}-10 c a \,b^{2}+2 b^{4}\right ) \sqrt {2 c d x +b d}}{15 d^{3} c^{3} \left (2 c x +b \right )^{2}}\) | \(101\) |
Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/16/c^3/d^5*(8*a*c*d^2*(2*c*d*x+b*d)^(1/2)-2*b^2*d^2*(2*c*d*x+b*d)^(1/2)+ 1/5*(2*c*d*x+b*d)^(5/2)-1/3*d^4*(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)^( 3/2))
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {{\left (3 \, c^{4} x^{4} + 6 \, b c^{3} x^{3} - 2 \, b^{4} + 10 \, a b^{2} c - 5 \, a^{2} c^{2} - 3 \, {\left (b^{2} c^{2} - 10 \, a c^{3}\right )} x^{2} - 6 \, {\left (b^{3} c - 5 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{15 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(5/2),x, algorithm="fricas")
Output:
1/15*(3*c^4*x^4 + 6*b*c^3*x^3 - 2*b^4 + 10*a*b^2*c - 5*a^2*c^2 - 3*(b^2*c^ 2 - 10*a*c^3)*x^2 - 6*(b^3*c - 5*a*b*c^2)*x)*sqrt(2*c*d*x + b*d)/(4*c^5*d^ 3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3)
Time = 1.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{2}}{48 c^{2} \left (b d + 2 c d x\right )^{\frac {3}{2}}} + \frac {\left (4 a c - b^{2}\right ) \sqrt {b d + 2 c d x}}{8 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{80 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}}{\left (b d\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((c*x**2+b*x+a)**2/(2*c*d*x+b*d)**(5/2),x)
Output:
Piecewise(((-(4*a*c - b**2)**2/(48*c**2*(b*d + 2*c*d*x)**(3/2)) + (4*a*c - b**2)*sqrt(b*d + 2*c*d*x)/(8*c**2*d**2) + (b*d + 2*c*d*x)**(5/2)/(80*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)**(5/2), True))
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\frac {5 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{2}} + \frac {3 \, {\left (10 \, \sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )}}{c^{2} d^{4}}}{240 \, c d} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(5/2),x, algorithm="maxima")
Output:
-1/240*(5*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/((2*c*d*x + b*d)^(3/2)*c^2) + 3*( 10*sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*d^2 - (2*c*d*x + b*d)^(5/2))/(c^2*d^4 ))/(c*d)
Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{48 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{3} d} - \frac {10 \, \sqrt {2 \, c d x + b d} b^{2} c^{12} d^{22} - 40 \, \sqrt {2 \, c d x + b d} a c^{13} d^{22} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{12} d^{20}}{80 \, c^{15} d^{25}} \] Input:
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(5/2),x, algorithm="giac")
Output:
-1/48*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/((2*c*d*x + b*d)^(3/2)*c^3*d) - 1/80* (10*sqrt(2*c*d*x + b*d)*b^2*c^12*d^22 - 40*sqrt(2*c*d*x + b*d)*a*c^13*d^22 - (2*c*d*x + b*d)^(5/2)*c^12*d^20)/(c^15*d^25)
Time = 5.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^4-5\,b^4\,d^4-30\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-80\,a^2\,c^2\,d^4+120\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+40\,a\,b^2\,c\,d^4}{240\,c^3\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}} \] Input:
int((a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^(5/2),x)
Output:
(3*(b*d + 2*c*d*x)^4 - 5*b^4*d^4 - 30*b^2*d^2*(b*d + 2*c*d*x)^2 - 80*a^2*c ^2*d^4 + 120*a*c*d^2*(b*d + 2*c*d*x)^2 + 40*a*b^2*c*d^4)/(240*c^3*d^5*(b*d + 2*c*d*x)^(3/2))
Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {\sqrt {d}\, \left (3 c^{4} x^{4}+6 b \,c^{3} x^{3}+30 a \,c^{3} x^{2}-3 b^{2} c^{2} x^{2}+30 a b \,c^{2} x -6 b^{3} c x -5 a^{2} c^{2}+10 a \,b^{2} c -2 b^{4}\right )}{15 \sqrt {2 c x +b}\, c^{3} d^{3} \left (2 c x +b \right )} \] Input:
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^(5/2),x)
Output:
(sqrt(d)*( - 5*a**2*c**2 + 10*a*b**2*c + 30*a*b*c**2*x + 30*a*c**3*x**2 - 2*b**4 - 6*b**3*c*x - 3*b**2*c**2*x**2 + 6*b*c**3*x**3 + 3*c**4*x**4))/(15 *sqrt(b + 2*c*x)*c**3*d**3*(b + 2*c*x))