\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^4} \, dx\) [1145]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=-\frac {\left (b^2-6 a c\right ) x}{32 c^3 d^4}+\frac {b x^2}{32 c^2 d^4}+\frac {x^3}{48 c d^4}+\frac {\left (b^2-4 a c\right )^3}{384 c^4 d^4 (b+2 c x)^3}-\frac {3 \left (b^2-4 a c\right )^2}{128 c^4 d^4 (b+2 c x)} \]

[Out]

-1/32*(-6*a*c+b^2)*x/c^3/d^4+1/32*b*x^2/c^2/d^4+1/48*x^3/c/d^4+1/384*(-4*a*c+b^2)^3/c^4/d^4/(2*c*x+b)^3-3/128*
(-4*a*c+b^2)^2/c^4/d^4/(2*c*x+b)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {697} \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=\frac {\left (b^2-4 a c\right )^3}{384 c^4 d^4 (b+2 c x)^3}-\frac {3 \left (b^2-4 a c\right )^2}{128 c^4 d^4 (b+2 c x)}-\frac {x \left (b^2-6 a c\right )}{32 c^3 d^4}+\frac {b x^2}{32 c^2 d^4}+\frac {x^3}{48 c d^4} \]

[In]

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^4,x]

[Out]

-1/32*((b^2 - 6*a*c)*x)/(c^3*d^4) + (b*x^2)/(32*c^2*d^4) + x^3/(48*c*d^4) + (b^2 - 4*a*c)^3/(384*c^4*d^4*(b +
2*c*x)^3) - (3*(b^2 - 4*a*c)^2)/(128*c^4*d^4*(b + 2*c*x))

Rule 697

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] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-b^2+6 a c}{32 c^3 d^4}+\frac {b x}{16 c^2 d^4}+\frac {x^2}{16 c d^4}+\frac {\left (-b^2+4 a c\right )^3}{64 c^3 d^4 (b+2 c x)^4}+\frac {3 \left (-b^2+4 a c\right )^2}{64 c^3 d^4 (b+2 c x)^2}\right ) \, dx \\ & = -\frac {\left (b^2-6 a c\right ) x}{32 c^3 d^4}+\frac {b x^2}{32 c^2 d^4}+\frac {x^3}{48 c d^4}+\frac {\left (b^2-4 a c\right )^3}{384 c^4 d^4 (b+2 c x)^3}-\frac {3 \left (b^2-4 a c\right )^2}{128 c^4 d^4 (b+2 c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=\frac {12 c \left (-b^2+6 a c\right ) x+12 b c^2 x^2+8 c^3 x^3+\frac {\left (b^2-4 a c\right )^3}{(b+2 c x)^3}-\frac {9 \left (b^2-4 a c\right )^2}{b+2 c x}}{384 c^4 d^4} \]

[In]

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^4,x]

[Out]

(12*c*(-b^2 + 6*a*c)*x + 12*b*c^2*x^2 + 8*c^3*x^3 + (b^2 - 4*a*c)^3/(b + 2*c*x)^3 - (9*(b^2 - 4*a*c)^2)/(b + 2
*c*x))/(384*c^4*d^4)

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\frac {\frac {2}{3} c^{2} x^{3}+c b \,x^{2}+6 a c x -b^{2} x}{32 c^{3}}-\frac {64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}{384 c^{4} \left (2 c x +b \right )^{3}}-\frac {48 a^{2} c^{2}-24 a \,b^{2} c +3 b^{4}}{128 c^{4} \left (2 c x +b \right )}}{d^{4}}\)	\(116\)
gosper	\(-\frac {-4 c^{5} x^{6}-12 b \,c^{4} x^{5}-36 a \,c^{4} x^{4}-6 b^{2} c^{3} x^{4}+36 a^{2} c^{3} x^{2}+36 a \,b^{2} c^{2} x^{2}+36 a^{2} b \,c^{2} x +18 a \,b^{3} c x +4 a^{3} c^{2}+6 a^{2} b^{2} c +3 a \,b^{4}}{24 \left (2 c x +b \right )^{3} d^{4} c^{3}}\)	\(119\)
parallelrisch	\(\frac {4 c^{5} x^{6}+12 b \,c^{4} x^{5}+36 a \,c^{4} x^{4}+6 b^{2} c^{3} x^{4}-36 a^{2} c^{3} x^{2}-36 a \,b^{2} c^{2} x^{2}-36 a^{2} b \,c^{2} x -18 a \,b^{3} c x -4 a^{3} c^{2}-6 a^{2} b^{2} c -3 a \,b^{4}}{24 c^{3} d^{4} \left (2 c x +b \right )^{3}}\)	\(119\)
norman	\(\frac {\frac {a^{3} x}{b d}+\frac {c^{2} x^{6}}{6 d}+\frac {\left (6 a c +b^{2}\right ) x^{4}}{4 d}+\frac {b c \,x^{5}}{2 d}+\frac {\left (4 c \,a^{3}+3 a^{2} b^{2}\right ) x^{2}}{2 b^{2} d}+\frac {\left (4 a^{3} c^{2}+6 a^{2} b^{2} c +3 a \,b^{4}\right ) x^{3}}{3 b^{3} d}}{d^{3} \left (2 c x +b \right )^{3}}\)	\(123\)
risch	\(\frac {x^{3}}{48 c \,d^{4}}+\frac {b \,x^{2}}{32 c^{2} d^{4}}+\frac {3 a x}{16 d^{4} c^{2}}-\frac {b^{2} x}{32 d^{4} c^{3}}+\frac {\left (-48 a^{2} d^{4} c^{3}+24 a \,b^{2} d^{4} c^{2}-3 b^{4} d^{4} c \right ) x^{2}-3 b \,d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) x -\frac {d^{4} \left (16 c^{3} a^{3}+24 a^{2} b^{2} c^{2}-15 a \,b^{4} c +2 b^{6}\right )}{3 c}}{32 d^{8} c^{3} \left (2 c x +b \right )^{3}}\)	\(167\)

[In]

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^4,x,method=_RETURNVERBOSE)

[Out]

1/d^4*(1/32/c^3*(2/3*c^2*x^3+c*b*x^2+6*a*c*x-b^2*x)-1/384*(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/c^4/(2*c*
x+b)^3-1/128*(48*a^2*c^2-24*a*b^2*c+3*b^4)/c^4/(2*c*x+b))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (93) = 186\).

Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=\frac {16 \, c^{6} x^{6} + 48 \, b c^{5} x^{5} - 2 \, b^{6} + 15 \, a b^{4} c - 24 \, a^{2} b^{2} c^{2} - 16 \, a^{3} c^{3} + 24 \, {\left (b^{2} c^{4} + 6 \, a c^{5}\right )} x^{4} - 8 \, {\left (2 \, b^{3} c^{3} - 27 \, a b c^{4}\right )} x^{3} - 12 \, {\left (2 \, b^{4} c^{2} - 15 \, a b^{2} c^{3} + 12 \, a^{2} c^{4}\right )} x^{2} - 6 \, {\left (2 \, b^{5} c - 15 \, a b^{3} c^{2} + 24 \, a^{2} b c^{3}\right )} x}{96 \, {\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^4,x, algorithm="fricas")

[Out]

1/96*(16*c^6*x^6 + 48*b*c^5*x^5 - 2*b^6 + 15*a*b^4*c - 24*a^2*b^2*c^2 - 16*a^3*c^3 + 24*(b^2*c^4 + 6*a*c^5)*x^
4 - 8*(2*b^3*c^3 - 27*a*b*c^4)*x^3 - 12*(2*b^4*c^2 - 15*a*b^2*c^3 + 12*a^2*c^4)*x^2 - 6*(2*b^5*c - 15*a*b^3*c^
2 + 24*a^2*b*c^3)*x)/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2*c^5*d^4*x + b^3*c^4*d^4)

Sympy [A] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.86 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=\frac {b x^{2}}{32 c^{2} d^{4}} + x \left (\frac {3 a}{16 c^{2} d^{4}} - \frac {b^{2}}{32 c^{3} d^{4}}\right ) + \frac {- 16 a^{3} c^{3} - 24 a^{2} b^{2} c^{2} + 15 a b^{4} c - 2 b^{6} + x^{2} \left (- 144 a^{2} c^{4} + 72 a b^{2} c^{3} - 9 b^{4} c^{2}\right ) + x \left (- 144 a^{2} b c^{3} + 72 a b^{3} c^{2} - 9 b^{5} c\right )}{96 b^{3} c^{4} d^{4} + 576 b^{2} c^{5} d^{4} x + 1152 b c^{6} d^{4} x^{2} + 768 c^{7} d^{4} x^{3}} + \frac {x^{3}}{48 c d^{4}} \]

[In]

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**4,x)

[Out]

b*x**2/(32*c**2*d**4) + x*(3*a/(16*c**2*d**4) - b**2/(32*c**3*d**4)) + (-16*a**3*c**3 - 24*a**2*b**2*c**2 + 15
*a*b**4*c - 2*b**6 + x**2*(-144*a**2*c**4 + 72*a*b**2*c**3 - 9*b**4*c**2) + x*(-144*a**2*b*c**3 + 72*a*b**3*c*
*2 - 9*b**5*c))/(96*b**3*c**4*d**4 + 576*b**2*c**5*d**4*x + 1152*b*c**6*d**4*x**2 + 768*c**7*d**4*x**3) + x**3
/(48*c*d**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=-\frac {2 \, b^{6} - 15 \, a b^{4} c + 24 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + 9 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + 9 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}{96 \, {\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}} + \frac {2 \, c^{2} x^{3} + 3 \, b c x^{2} - 3 \, {\left (b^{2} - 6 \, a c\right )} x}{96 \, c^{3} d^{4}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^4,x, algorithm="maxima")

[Out]

-1/96*(2*b^6 - 15*a*b^4*c + 24*a^2*b^2*c^2 + 16*a^3*c^3 + 9*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^2 + 9*(b^5*
c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x)/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2*c^5*d^4*x + b^3*c^4*d^4) + 1/96*(
2*c^2*x^3 + 3*b*c*x^2 - 3*(b^2 - 6*a*c)*x)/(c^3*d^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=-\frac {9 \, b^{4} c^{2} x^{2} - 72 \, a b^{2} c^{3} x^{2} + 144 \, a^{2} c^{4} x^{2} + 9 \, b^{5} c x - 72 \, a b^{3} c^{2} x + 144 \, a^{2} b c^{3} x + 2 \, b^{6} - 15 \, a b^{4} c + 24 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3}}{96 \, {\left (2 \, c x + b\right )}^{3} c^{4} d^{4}} + \frac {2 \, c^{11} d^{8} x^{3} + 3 \, b c^{10} d^{8} x^{2} - 3 \, b^{2} c^{9} d^{8} x + 18 \, a c^{10} d^{8} x}{96 \, c^{12} d^{12}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^4,x, algorithm="giac")

[Out]

-1/96*(9*b^4*c^2*x^2 - 72*a*b^2*c^3*x^2 + 144*a^2*c^4*x^2 + 9*b^5*c*x - 72*a*b^3*c^2*x + 144*a^2*b*c^3*x + 2*b
^6 - 15*a*b^4*c + 24*a^2*b^2*c^2 + 16*a^3*c^3)/((2*c*x + b)^3*c^4*d^4) + 1/96*(2*c^11*d^8*x^3 + 3*b*c^10*d^8*x
^2 - 3*b^2*c^9*d^8*x + 18*a*c^10*d^8*x)/(c^12*d^12)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.88 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^4} \, dx=x\,\left (\frac {3\,\left (b^2+a\,c\right )}{16\,c^3\,d^4}-\frac {7\,b^2}{32\,c^3\,d^4}\right )-\frac {\frac {16\,a^3\,c^3+24\,a^2\,b^2\,c^2-15\,a\,b^4\,c+2\,b^6}{3\,c}+x^2\,\left (48\,a^2\,c^3-24\,a\,b^2\,c^2+3\,b^4\,c\right )+x\,\left (48\,a^2\,b\,c^2-24\,a\,b^3\,c+3\,b^5\right )}{32\,b^3\,c^3\,d^4+192\,b^2\,c^4\,d^4\,x+384\,b\,c^5\,d^4\,x^2+256\,c^6\,d^4\,x^3}+\frac {x^3}{48\,c\,d^4}+\frac {b\,x^2}{32\,c^2\,d^4} \]

[In]

int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^4,x)

[Out]

x*((3*(a*c + b^2))/(16*c^3*d^4) - (7*b^2)/(32*c^3*d^4)) - ((2*b^6 + 16*a^3*c^3 + 24*a^2*b^2*c^2 - 15*a*b^4*c)/
(3*c) + x^2*(3*b^4*c + 48*a^2*c^3 - 24*a*b^2*c^2) + x*(3*b^5 + 48*a^2*b*c^2 - 24*a*b^3*c))/(32*b^3*c^3*d^4 + 2
56*c^6*d^4*x^3 + 192*b^2*c^4*d^4*x + 384*b*c^5*d^4*x^2) + x^3/(48*c*d^4) + (b*x^2)/(32*c^2*d^4)