### 3.1116 $$\int \frac{a+b x+c x^2}{(b d+2 c d x)^4} \, dx$$

Optimal. Leaf size=45 $\frac{b^2-4 a c}{24 c^2 d^4 (b+2 c x)^3}-\frac{1}{8 c^2 d^4 (b+2 c x)}$

[Out]

(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))

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Rubi [A]  time = 0.0317319, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.045, Rules used = {683} $\frac{b^2-4 a c}{24 c^2 d^4 (b+2 c x)^3}-\frac{1}{8 c^2 d^4 (b+2 c x)}$

Antiderivative was successfully veriﬁed.

[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 683

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*} \int \frac{a+b x+c x^2}{(b d+2 c d x)^4} \, dx &=\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]  time = 0.0156746, size = 38, normalized size = 0.84 $-\frac{2 c \left (a+3 c x^2\right )+b^2+6 b c x}{12 c^2 d^4 (b+2 c x)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 42, normalized size = 0.9 \begin{align*}{\frac{1}{{d}^{4}} \left ( -{\frac{4\,ac-{b}^{2}}{24\,{c}^{2} \left ( 2\,cx+b \right ) ^{3}}}-{\frac{1}{8\,{c}^{2} \left ( 2\,cx+b \right ) }} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16272, size = 96, normalized size = 2.13 \begin{align*} -\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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.35558, size = 149, normalized size = 3.31 \begin{align*} -\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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.651829, size = 75, normalized size = 1.67 \begin{align*} - \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.25477, size = 50, normalized size = 1.11 \begin{align*} -\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)