\(\int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx\) [1116]

Optimal result

Integrand size = 22, antiderivative size = 45 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=\frac {b^2-4 a c}{24 c^2 d^4 (b+2 c x)^3}-\frac {1}{8 c^2 d^4 (b+2 c x)} \]

[Out]

1/24*(-4*a*c+b^2)/c^2/d^4/(2*c*x+b)^3-1/8/c^2/d^4/(2*c*x+b)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {697} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=\frac {b^2-4 a c}{24 c^2 d^4 (b+2 c x)^3}-\frac {1}{8 c^2 d^4 (b+2 c x)} \]

[In]

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

[Out]

(b^2 - 4*a*c)/(24*c^2*d^4*(b + 2*c*x)^3) - 1/(8*c^2*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+4 a c}{4 c d^4 (b+2 c x)^4}+\frac {1}{4 c d^4 (b+2 c x)^2}\right ) \, dx \\ & = \frac {b^2-4 a c}{24 c^2 d^4 (b+2 c x)^3}-\frac {1}{8 c^2 d^4 (b+2 c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=-\frac {b^2+6 b c x+2 c \left (a+3 c x^2\right )}{12 c^2 d^4 (b+2 c x)^3} \]

[In]

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

[Out]

-1/12*(b^2 + 6*b*c*x + 2*c*(a + 3*c*x^2))/(c^2*d^4*(b + 2*c*x)^3)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84

method	result	size
gosper	\(-\frac {6 c^{2} x^{2}+6 b c x +2 a c +b^{2}}{12 c^{2} \left (2 c x +b \right )^{3} d^{4}}\)	\(38\)
risch	\(\frac {-\frac {x^{2}}{2}-\frac {b x}{2 c}-\frac {2 a c +b^{2}}{12 c^{2}}}{d^{4} \left (2 c x +b \right )^{3}}\)	\(39\)
default	\(\frac {-\frac {4 a c -b^{2}}{24 c^{2} \left (2 c x +b \right )^{3}}-\frac {1}{8 c^{2} \left (2 c x +b \right )}}{d^{4}}\)	\(42\)
parallelrisch	\(\frac {8 a \,c^{2} x^{3}+4 b^{2} c \,x^{3}+12 a b c \,x^{2}+3 b^{3} x^{2}+6 a \,b^{2} x}{6 b^{3} d^{4} \left (2 c x +b \right )^{3}}\)	\(59\)
norman	\(\frac {\frac {a x}{b d}+\frac {\left (4 a c +b^{2}\right ) x^{2}}{2 b^{2} d}+\frac {2 c \left (2 a c +b^{2}\right ) x^{3}}{3 b^{3} d}}{d^{3} \left (2 c x +b \right )^{3}}\)	\(62\)

[In]

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

[Out]

-1/12/c^2*(6*c^2*x^2+6*b*c*x+2*a*c+b^2)/(2*c*x+b)^3/d^4

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=-\frac {6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c}{12 \, {\left (8 \, c^{5} d^{4} x^{3} + 12 \, b c^{4} d^{4} x^{2} + 6 \, b^{2} c^{3} d^{4} x + b^{3} c^{2} d^{4}\right )}} \]

[In]

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

[Out]

-1/12*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)/(8*c^5*d^4*x^3 + 12*b*c^4*d^4*x^2 + 6*b^2*c^3*d^4*x + b^3*c^2*d^4)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.67 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=\frac {- 2 a c - b^{2} - 6 b c x - 6 c^{2} x^{2}}{12 b^{3} c^{2} d^{4} + 72 b^{2} c^{3} d^{4} x + 144 b c^{4} d^{4} x^{2} + 96 c^{5} d^{4} x^{3}} \]

[In]

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

[Out]

(-2*a*c - b**2 - 6*b*c*x - 6*c**2*x**2)/(12*b**3*c**2*d**4 + 72*b**2*c**3*d**4*x + 144*b*c**4*d**4*x**2 + 96*c
**5*d**4*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=-\frac {6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c}{12 \, {\left (8 \, c^{5} d^{4} x^{3} + 12 \, b c^{4} d^{4} x^{2} + 6 \, b^{2} c^{3} d^{4} x + b^{3} c^{2} d^{4}\right )}} \]

[In]

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

[Out]

-1/12*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)/(8*c^5*d^4*x^3 + 12*b*c^4*d^4*x^2 + 6*b^2*c^3*d^4*x + b^3*c^2*d^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=-\frac {6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c}{12 \, {\left (2 \, c x + b\right )}^{3} c^{2} d^{4}} \]

[In]

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

[Out]

-1/12*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)/((2*c*x + b)^3*c^2*d^4)

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^4} \, dx=-\frac {\frac {b^2+2\,a\,c}{12\,c^2}+\frac {x^2}{2}+\frac {b\,x}{2\,c}}{b^3\,d^4+6\,b^2\,c\,d^4\,x+12\,b\,c^2\,d^4\,x^2+8\,c^3\,d^4\,x^3} \]

[In]

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

[Out]

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