\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^{9/2}} \, dx\) [106]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=\frac {\left (b^2-4 a c\right )^3}{448 c^4 d (b d+2 c d x)^{7/2}}-\frac {\left (b^2-4 a c\right )^2}{64 c^4 d^3 (b d+2 c d x)^{3/2}}-\frac {3 \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{64 c^4 d^5}+\frac {(b d+2 c d x)^{5/2}}{320 c^4 d^7} \] Output:

1/448*(-4*a*c+b^2)^3/c^4/d/(2*c*d*x+b*d)^(7/2)-1/64*(-4*a*c+b^2)^2/c^4/d^3 
/(2*c*d*x+b*d)^(3/2)-3/64*(-4*a*c+b^2)*(2*c*d*x+b*d)^(1/2)/c^4/d^5+1/320*( 
2*c*d*x+b*d)^(5/2)/c^4/d^7
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=\frac {5 b^6-60 a b^4 c+240 a^2 b^2 c^2-320 a^3 c^3-35 b^4 (b+2 c x)^2+280 a b^2 c (b+2 c x)^2-560 a^2 c^2 (b+2 c x)^2-105 b^2 (b+2 c x)^4+420 a c (b+2 c x)^4+7 (b+2 c x)^6}{2240 c^4 d (d (b+2 c x))^{7/2}} \] Input:

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(9/2),x]
 

Output:

(5*b^6 - 60*a*b^4*c + 240*a^2*b^2*c^2 - 320*a^3*c^3 - 35*b^4*(b + 2*c*x)^2 
 + 280*a*b^2*c*(b + 2*c*x)^2 - 560*a^2*c^2*(b + 2*c*x)^2 - 105*b^2*(b + 2* 
c*x)^4 + 420*a*c*(b + 2*c*x)^4 + 7*(b + 2*c*x)^6)/(2240*c^4*d*(d*(b + 2*c* 
x))^(7/2))
 

Rubi [A] (verified)

Time = 0.26 (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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx\)

\(\Big \downarrow \) 1107

\(\displaystyle \int \left (\frac {3 \left (4 a c-b^2\right )}{64 c^3 d^4 \sqrt {b d+2 c d x}}+\frac {3 \left (4 a c-b^2\right )^2}{64 c^3 d^2 (b d+2 c d x)^{5/2}}+\frac {\left (4 a c-b^2\right )^3}{64 c^3 (b d+2 c d x)^{9/2}}+\frac {(b d+2 c d x)^{3/2}}{64 c^3 d^6}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{64 c^4 d^5}-\frac {\left (b^2-4 a c\right )^2}{64 c^4 d^3 (b d+2 c d x)^{3/2}}+\frac {\left (b^2-4 a c\right )^3}{448 c^4 d (b d+2 c d x)^{7/2}}+\frac {(b d+2 c d x)^{5/2}}{320 c^4 d^7}\)

Input:

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(9/2),x]
 

Output:

(b^2 - 4*a*c)^3/(448*c^4*d*(b*d + 2*c*d*x)^(7/2)) - (b^2 - 4*a*c)^2/(64*c^ 
4*d^3*(b*d + 2*c*d*x)^(3/2)) - (3*(b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x])/(64*c 
^4*d^5) + (b*d + 2*c*d*x)^(5/2)/(320*c^4*d^7)
 

Defintions of rubi rules used

rule 1107

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])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(-\frac {-\frac {7 x^{6} c^{6}}{5}-21 x^{4} \left (\frac {b x}{5}+a \right ) c^{5}+7 a \,x^{2} \left (-6 b x +a \right ) c^{4}+\left (7 b^{3} x^{3}-35 a \,b^{2} x^{2}+7 a^{2} b x +a^{3}\right ) c^{3}+b^{2} \left (7 b^{2} x^{2}-14 a b x +a^{2}\right ) c^{2}-2 \left (-\frac {7 b x}{5}+a \right ) b^{4} c +\frac {2 b^{6}}{5}}{7 \sqrt {d \left (2 c x +b \right )}\, d^{4} \left (2 c x +b \right )^{3} c^{4}}\)	\(138\)
derivativedivides	\(\frac {12 a c \,d^{2} \sqrt {2 c d x +b d}-3 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{\left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {d^{6} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{7 \left (2 c d x +b d \right )^{\frac {7}{2}}}}{64 c^{4} d^{7}}\)	\(143\)
default	\(\frac {12 a c \,d^{2} \sqrt {2 c d x +b d}-3 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}{\left (2 c d x +b d \right )^{\frac {3}{2}}}-\frac {d^{6} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{7 \left (2 c d x +b d \right )^{\frac {7}{2}}}}{64 c^{4} d^{7}}\)	\(143\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-7 x^{6} c^{6}-21 x^{5} b \,c^{5}-105 a \,c^{5} x^{4}-210 a b \,c^{4} x^{3}+35 b^{3} c^{3} x^{3}+35 a^{2} c^{4} x^{2}-175 a \,b^{2} c^{3} x^{2}+35 c^{2} x^{2} b^{4}+35 a^{2} b \,c^{3} x -70 x a \,b^{3} c^{2}+14 x c \,b^{5}+5 a^{3} c^{3}+5 a^{2} b^{2} c^{2}-10 a \,b^{4} c +2 b^{6}\right )}{35 c^{4} \left (2 c d x +b d \right )^{\frac {9}{2}}}\)	\(163\)
orering	\(-\frac {\left (2 c x +b \right ) \left (-7 x^{6} c^{6}-21 x^{5} b \,c^{5}-105 a \,c^{5} x^{4}-210 a b \,c^{4} x^{3}+35 b^{3} c^{3} x^{3}+35 a^{2} c^{4} x^{2}-175 a \,b^{2} c^{3} x^{2}+35 c^{2} x^{2} b^{4}+35 a^{2} b \,c^{3} x -70 x a \,b^{3} c^{2}+14 x c \,b^{5}+5 a^{3} c^{3}+5 a^{2} b^{2} c^{2}-10 a \,b^{4} c +2 b^{6}\right )}{35 c^{4} \left (2 c d x +b d \right )^{\frac {9}{2}}}\)	\(163\)
trager	\(-\frac {\left (-7 x^{6} c^{6}-21 x^{5} b \,c^{5}-105 a \,c^{5} x^{4}-210 a b \,c^{4} x^{3}+35 b^{3} c^{3} x^{3}+35 a^{2} c^{4} x^{2}-175 a \,b^{2} c^{3} x^{2}+35 c^{2} x^{2} b^{4}+35 a^{2} b \,c^{3} x -70 x a \,b^{3} c^{2}+14 x c \,b^{5}+5 a^{3} c^{3}+5 a^{2} b^{2} c^{2}-10 a \,b^{4} c +2 b^{6}\right ) \sqrt {2 c d x +b d}}{35 d^{5} c^{4} \left (2 c x +b \right )^{4}}\)	\(168\)

Input:

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/7/(d*(2*c*x+b))^(1/2)*(-7/5*x^6*c^6-21*x^4*(1/5*b*x+a)*c^5+7*a*x^2*(-6* 
b*x+a)*c^4+(7*b^3*x^3-35*a*b^2*x^2+7*a^2*b*x+a^3)*c^3+b^2*(7*b^2*x^2-14*a* 
b*x+a^2)*c^2-2*(-7/5*b*x+a)*b^4*c+2/5*b^6)/d^4/(2*c*x+b)^3/c^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=\frac {{\left (7 \, c^{6} x^{6} + 21 \, b c^{5} x^{5} + 105 \, a c^{5} x^{4} - 2 \, b^{6} + 10 \, a b^{4} c - 5 \, a^{2} b^{2} c^{2} - 5 \, a^{3} c^{3} - 35 \, {\left (b^{3} c^{3} - 6 \, a b c^{4}\right )} x^{3} - 35 \, {\left (b^{4} c^{2} - 5 \, a b^{2} c^{3} + a^{2} c^{4}\right )} x^{2} - 7 \, {\left (2 \, b^{5} c - 10 \, a b^{3} c^{2} + 5 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{35 \, {\left (16 \, c^{8} d^{5} x^{4} + 32 \, b c^{7} d^{5} x^{3} + 24 \, b^{2} c^{6} d^{5} x^{2} + 8 \, b^{3} c^{5} d^{5} x + b^{4} c^{4} d^{5}\right )}} \] Input:

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(9/2),x, algorithm="fricas")
 

Output:

1/35*(7*c^6*x^6 + 21*b*c^5*x^5 + 105*a*c^5*x^4 - 2*b^6 + 10*a*b^4*c - 5*a^ 
2*b^2*c^2 - 5*a^3*c^3 - 35*(b^3*c^3 - 6*a*b*c^4)*x^3 - 35*(b^4*c^2 - 5*a*b 
^2*c^3 + a^2*c^4)*x^2 - 7*(2*b^5*c - 10*a*b^3*c^2 + 5*a^2*b*c^3)*x)*sqrt(2 
*c*d*x + b*d)/(16*c^8*d^5*x^4 + 32*b*c^7*d^5*x^3 + 24*b^2*c^6*d^5*x^2 + 8* 
b^3*c^5*d^5*x + b^4*c^4*d^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1394 vs. \(2 (117) = 234\).

Time = 0.74 (sec) , antiderivative size = 1394, normalized size of antiderivative = 11.52 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx =\text {Too large to display} \] Input:

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**(9/2),x)
 

Output:

Piecewise((-5*a**3*c**3*sqrt(b*d + 2*c*d*x)/(35*b**4*c**4*d**5 + 280*b**3* 
c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d**5*x**3 + 560*c**8*d 
**5*x**4) - 5*a**2*b**2*c**2*sqrt(b*d + 2*c*d*x)/(35*b**4*c**4*d**5 + 280* 
b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d**5*x**3 + 560*c 
**8*d**5*x**4) - 35*a**2*b*c**3*x*sqrt(b*d + 2*c*d*x)/(35*b**4*c**4*d**5 + 
 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d**5*x**3 + 
560*c**8*d**5*x**4) - 35*a**2*c**4*x**2*sqrt(b*d + 2*c*d*x)/(35*b**4*c**4* 
d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d**5*x 
**3 + 560*c**8*d**5*x**4) + 10*a*b**4*c*sqrt(b*d + 2*c*d*x)/(35*b**4*c**4* 
d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d**5*x 
**3 + 560*c**8*d**5*x**4) + 70*a*b**3*c**2*x*sqrt(b*d + 2*c*d*x)/(35*b**4* 
c**4*d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120*b*c**7*d 
**5*x**3 + 560*c**8*d**5*x**4) + 175*a*b**2*c**3*x**2*sqrt(b*d + 2*c*d*x)/ 
(35*b**4*c**4*d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 + 1120 
*b*c**7*d**5*x**3 + 560*c**8*d**5*x**4) + 210*a*b*c**4*x**3*sqrt(b*d + 2*c 
*d*x)/(35*b**4*c**4*d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**2 
+ 1120*b*c**7*d**5*x**3 + 560*c**8*d**5*x**4) + 105*a*c**5*x**4*sqrt(b*d + 
 2*c*d*x)/(35*b**4*c**4*d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x 
**2 + 1120*b*c**7*d**5*x**3 + 560*c**8*d**5*x**4) - 2*b**6*sqrt(b*d + 2*c* 
d*x)/(35*b**4*c**4*d**5 + 280*b**3*c**5*d**5*x + 840*b**2*c**6*d**5*x**...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=-\frac {\frac {5 \, {\left (7 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{2} - {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{3} d^{2}} + \frac {7 \, {\left (15 \, \sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )}}{c^{3} d^{6}}}{2240 \, c d} \] Input:

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(9/2),x, algorithm="maxima")
 

Output:

-1/2240*(5*(7*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^2 - (b^6 - 12 
*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^2)/((2*c*d*x + b*d)^(7/2)*c^3*d^ 
2) + 7*(15*sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)*d^2 - (2*c*d*x + b*d)^(5/2))/ 
(c^3*d^6))/(c*d)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=\frac {b^{6} d^{2} - 12 \, a b^{4} c d^{2} + 48 \, a^{2} b^{2} c^{2} d^{2} - 64 \, a^{3} c^{3} d^{2} - 7 \, {\left (2 \, c d x + b d\right )}^{2} b^{4} + 56 \, {\left (2 \, c d x + b d\right )}^{2} a b^{2} c - 112 \, {\left (2 \, c d x + b d\right )}^{2} a^{2} c^{2}}{448 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{4} d^{3}} - \frac {15 \, \sqrt {2 \, c d x + b d} b^{2} c^{16} d^{30} - 60 \, \sqrt {2 \, c d x + b d} a c^{17} d^{30} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{16} d^{28}}{320 \, c^{20} d^{35}} \] Input:

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(9/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/448*(b^6*d^2 - 12*a*b^4*c*d^2 + 48*a^2*b^2*c^2*d^2 - 64*a^3*c^3*d^2 - 7* 
(2*c*d*x + b*d)^2*b^4 + 56*(2*c*d*x + b*d)^2*a*b^2*c - 112*(2*c*d*x + b*d) 
^2*a^2*c^2)/((2*c*d*x + b*d)^(7/2)*c^4*d^3) - 1/320*(15*sqrt(2*c*d*x + b*d 
)*b^2*c^16*d^30 - 60*sqrt(2*c*d*x + b*d)*a*c^17*d^30 - (2*c*d*x + b*d)^(5/ 
2)*c^16*d^28)/(c^20*d^35)
 

Mupad [B] (verification not implemented)

Time = 5.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=-\frac {5\,a^3\,c^3+5\,a^2\,b^2\,c^2+35\,a^2\,b\,c^3\,x+35\,a^2\,c^4\,x^2-10\,a\,b^4\,c-70\,a\,b^3\,c^2\,x-175\,a\,b^2\,c^3\,x^2-210\,a\,b\,c^4\,x^3-105\,a\,c^5\,x^4+2\,b^6+14\,b^5\,c\,x+35\,b^4\,c^2\,x^2+35\,b^3\,c^3\,x^3-21\,b\,c^5\,x^5-7\,c^6\,x^6}{35\,c^4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}} \] Input:

int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(9/2),x)
 

Output:

-(2*b^6 + 5*a^3*c^3 - 7*c^6*x^6 - 105*a*c^5*x^4 - 21*b*c^5*x^5 + 5*a^2*b^2 
*c^2 + 35*a^2*c^4*x^2 + 35*b^4*c^2*x^2 + 35*b^3*c^3*x^3 - 10*a*b^4*c + 14* 
b^5*c*x - 175*a*b^2*c^3*x^2 - 70*a*b^3*c^2*x + 35*a^2*b*c^3*x - 210*a*b*c^ 
4*x^3)/(35*c^4*d*(b*d + 2*c*d*x)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{9/2}} \, dx=\frac {\sqrt {d}\, \left (7 c^{6} x^{6}+21 b \,c^{5} x^{5}+105 a \,c^{5} x^{4}+210 a b \,c^{4} x^{3}-35 b^{3} c^{3} x^{3}-35 a^{2} c^{4} x^{2}+175 a \,b^{2} c^{3} x^{2}-35 b^{4} c^{2} x^{2}-35 a^{2} b \,c^{3} x +70 a \,b^{3} c^{2} x -14 b^{5} c x -5 a^{3} c^{3}-5 a^{2} b^{2} c^{2}+10 a \,b^{4} c -2 b^{6}\right )}{35 \sqrt {2 c x +b}\, c^{4} d^{5} \left (8 c^{3} x^{3}+12 b \,c^{2} x^{2}+6 b^{2} c x +b^{3}\right )} \] Input:

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(9/2),x)
 

Output:

(sqrt(d)*( - 5*a**3*c**3 - 5*a**2*b**2*c**2 - 35*a**2*b*c**3*x - 35*a**2*c 
**4*x**2 + 10*a*b**4*c + 70*a*b**3*c**2*x + 175*a*b**2*c**3*x**2 + 210*a*b 
*c**4*x**3 + 105*a*c**5*x**4 - 2*b**6 - 14*b**5*c*x - 35*b**4*c**2*x**2 - 
35*b**3*c**3*x**3 + 21*b*c**5*x**5 + 7*c**6*x**6))/(35*sqrt(b + 2*c*x)*c** 
4*d**5*(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3))