### 3.1117 $$\int \frac{a+b x+c x^2}{(b d+2 c d x)^5} \, dx$$

Optimal. Leaf size=37 $\frac{\left (a+b x+c x^2\right )^2}{2 d^5 \left (b^2-4 a c\right ) (b+2 c x)^4}$

[Out]

(a + b*x + c*x^2)^2/(2*(b^2 - 4*a*c)*d^5*(b + 2*c*x)^4)

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Rubi [A]  time = 0.010164, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.045, Rules used = {682} $\frac{\left (a+b x+c x^2\right )^2}{2 d^5 \left (b^2-4 a c\right ) (b+2 c x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^5,x]

[Out]

(a + b*x + c*x^2)^2/(2*(b^2 - 4*a*c)*d^5*(b + 2*c*x)^4)

Rule 682

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{(b d+2 c d x)^5} \, dx &=\frac{\left (a+b x+c x^2\right )^2}{2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0148447, size = 38, normalized size = 1.03 $-\frac{4 c \left (a+2 c x^2\right )+b^2+8 b c x}{32 c^2 d^5 (b+2 c x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^5,x]

[Out]

-(b^2 + 8*b*c*x + 4*c*(a + 2*c*x^2))/(32*c^2*d^5*(b + 2*c*x)^4)

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Maple [A]  time = 0.046, size = 42, normalized size = 1.1 \begin{align*}{\frac{1}{{d}^{5}} \left ( -{\frac{1}{16\,{c}^{2} \left ( 2\,cx+b \right ) ^{2}}}-{\frac{4\,ac-{b}^{2}}{32\,{c}^{2} \left ( 2\,cx+b \right ) ^{4}}} \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)^5,x)

[Out]

1/d^5*(-1/16/c^2/(2*c*x+b)^2-1/32*(4*a*c-b^2)/c^2/(2*c*x+b)^4)

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Maxima [B]  time = 1.22513, size = 115, normalized size = 3.11 \begin{align*} -\frac{8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{32 \,{\left (16 \, c^{6} d^{5} x^{4} + 32 \, b c^{5} d^{5} x^{3} + 24 \, b^{2} c^{4} d^{5} x^{2} + 8 \, b^{3} c^{3} d^{5} x + b^{4} c^{2} d^{5}\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)^5,x, algorithm="maxima")

[Out]

-1/32*(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)/(16*c^6*d^5*x^4 + 32*b*c^5*d^5*x^3 + 24*b^2*c^4*d^5*x^2 + 8*b^3*c^3*
d^5*x + b^4*c^2*d^5)

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Fricas [B]  time = 2.4989, size = 178, normalized size = 4.81 \begin{align*} -\frac{8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{32 \,{\left (16 \, c^{6} d^{5} x^{4} + 32 \, b c^{5} d^{5} x^{3} + 24 \, b^{2} c^{4} d^{5} x^{2} + 8 \, b^{3} c^{3} d^{5} x + b^{4} c^{2} d^{5}\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)^5,x, algorithm="fricas")

[Out]

-1/32*(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)/(16*c^6*d^5*x^4 + 32*b*c^5*d^5*x^3 + 24*b^2*c^4*d^5*x^2 + 8*b^3*c^3*
d^5*x + b^4*c^2*d^5)

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Sympy [B]  time = 0.869247, size = 90, normalized size = 2.43 \begin{align*} - \frac{4 a c + b^{2} + 8 b c x + 8 c^{2} x^{2}}{32 b^{4} c^{2} d^{5} + 256 b^{3} c^{3} d^{5} x + 768 b^{2} c^{4} d^{5} x^{2} + 1024 b c^{5} d^{5} x^{3} + 512 c^{6} d^{5} x^{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)**5,x)

[Out]

-(4*a*c + b**2 + 8*b*c*x + 8*c**2*x**2)/(32*b**4*c**2*d**5 + 256*b**3*c**3*d**5*x + 768*b**2*c**4*d**5*x**2 +
1024*b*c**5*d**5*x**3 + 512*c**6*d**5*x**4)

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Giac [A]  time = 1.18428, size = 81, normalized size = 2.19 \begin{align*} \frac{\frac{b^{2}}{{\left (2 \, c d x + b d\right )}^{4} c^{2}} - \frac{4 \, a}{{\left (2 \, c d x + b d\right )}^{4} c} - \frac{2}{{\left (2 \, c d x + b d\right )}^{2} c^{2} d^{2}}}{32 \, d} \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)^5,x, algorithm="giac")

[Out]

1/32*(b^2/((2*c*d*x + b*d)^4*c^2) - 4*a/((2*c*d*x + b*d)^4*c) - 2/((2*c*d*x + b*d)^2*c^2*d^2))/d