3.1115 \(\int \frac{a+b x+c x^2}{(b d+2 c d x)^3} \, dx\)

Optimal. Leaf size=44 \[ \frac{b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac{\log (b+2 c x)}{8 c^2 d^3} \]

[Out]

(b^2 - 4*a*c)/(16*c^2*d^3*(b + 2*c*x)^2) + Log[b + 2*c*x]/(8*c^2*d^3)

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Rubi [A]  time = 0.0326059, antiderivative size = 44, 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} \[ \frac{b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac{\log (b+2 c x)}{8 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2 - 4*a*c)/(16*c^2*d^3*(b + 2*c*x)^2) + Log[b + 2*c*x]/(8*c^2*d^3)

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)^3} \, dx &=\int \left (\frac{-b^2+4 a c}{4 c d^3 (b+2 c x)^3}+\frac{1}{4 c d^3 (b+2 c x)}\right ) \, dx\\ &=\frac{b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac{\log (b+2 c x)}{8 c^2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0154983, size = 37, normalized size = 0.84 \[ \frac{\frac{b^2-4 a c}{(b+2 c x)^2}+2 \log (b+2 c x)}{16 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)/(b + 2*c*x)^2 + 2*Log[b + 2*c*x])/(16*c^2*d^3)

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Maple [A]  time = 0.044, size = 53, normalized size = 1.2 \begin{align*} -{\frac{a}{4\,c{d}^{3} \left ( 2\,cx+b \right ) ^{2}}}+{\frac{{b}^{2}}{16\,{c}^{2}{d}^{3} \left ( 2\,cx+b \right ) ^{2}}}+{\frac{\ln \left ( 2\,cx+b \right ) }{8\,{c}^{2}{d}^{3}}} \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)^3,x)

[Out]

-1/4/d^3/c/(2*c*x+b)^2*a+1/16/d^3/c^2/(2*c*x+b)^2*b^2+1/8*ln(2*c*x+b)/c^2/d^3

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Maxima [A]  time = 1.18744, size = 81, normalized size = 1.84 \begin{align*} \frac{b^{2} - 4 \, a c}{16 \,{\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}} + \frac{\log \left (2 \, c x + b\right )}{8 \, c^{2} d^{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)^3,x, algorithm="maxima")

[Out]

1/16*(b^2 - 4*a*c)/(4*c^4*d^3*x^2 + 4*b*c^3*d^3*x + b^2*c^2*d^3) + 1/8*log(2*c*x + b)/(c^2*d^3)

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Fricas [A]  time = 2.33354, size = 153, normalized size = 3.48 \begin{align*} \frac{b^{2} - 4 \, a c + 2 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (2 \, c x + b\right )}{16 \,{\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\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)^3,x, algorithm="fricas")

[Out]

1/16*(b^2 - 4*a*c + 2*(4*c^2*x^2 + 4*b*c*x + b^2)*log(2*c*x + b))/(4*c^4*d^3*x^2 + 4*b*c^3*d^3*x + b^2*c^2*d^3
)

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Sympy [A]  time = 0.621888, size = 60, normalized size = 1.36 \begin{align*} - \frac{4 a c - b^{2}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac{\log{\left (b + 2 c x \right )}}{8 c^{2} d^{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)**3,x)

[Out]

-(4*a*c - b**2)/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*x**2) + log(b + 2*c*x)/(8*c**2*d**3)

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Giac [A]  time = 1.22163, size = 55, normalized size = 1.25 \begin{align*} \frac{\log \left ({\left | 2 \, c x + b \right |}\right )}{8 \, c^{2} d^{3}} + \frac{b^{2} - 4 \, a c}{16 \,{\left (2 \, c x + b\right )}^{2} c^{2} d^{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)^3,x, algorithm="giac")

[Out]

1/8*log(abs(2*c*x + b))/(c^2*d^3) + 1/16*(b^2 - 4*a*c)/((2*c*x + b)^2*c^2*d^3)