### 3.1164 $$\int \frac{1}{(b d+2 c d x)^3 (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=70 $\frac{\log \left (a+b x+c x^2\right )}{d^3 \left (b^2-4 a c\right )^2}+\frac{1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}-\frac{2 \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^2}$

[Out]

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

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Rubi [A]  time = 0.0403733, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {693, 681, 31, 628} $\frac{\log \left (a+b x+c x^2\right )}{d^3 \left (b^2-4 a c\right )^2}+\frac{1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}-\frac{2 \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]

[Out]

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

Rule 693

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

Rule 681

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.0391309, size = 65, normalized size = 0.93 $\frac{\frac{\log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2}+\frac{1}{\left (b^2-4 a c\right ) (b+2 c x)^2}-\frac{2 \log (b+2 c x)}{\left (b^2-4 a c\right )^2}}{d^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.046, size = 78, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( c{x}^{2}+bx+a \right ) }{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( 2\,cx+b \right ) }{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2}}}-{\frac{1}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) \left ( 2\,cx+b \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.20086, size = 174, normalized size = 2.49 \begin{align*} \frac{1}{4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{3} x^{2} + 4 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} d^{3} x +{\left (b^{4} - 4 \, a b^{2} c\right )} d^{3}} + \frac{\log \left (c x^{2} + b x + a\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} - \frac{2 \, \log \left (2 \, c x + b\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.99394, size = 338, normalized size = 4.83 \begin{align*} \frac{b^{2} - 4 \, a c +{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (c x^{2} + b x + a\right ) - 2 \,{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (2 \, c x + b\right )}{4 \,{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{3} x^{2} + 4 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{3} x +{\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.34491, size = 119, normalized size = 1.7 \begin{align*} - \frac{1}{4 a b^{2} c d^{3} - b^{4} d^{3} + x^{2} \left (16 a c^{3} d^{3} - 4 b^{2} c^{2} d^{3}\right ) + x \left (16 a b c^{2} d^{3} - 4 b^{3} c d^{3}\right )} - \frac{2 \log{\left (\frac{b}{2 c} + x \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} + \frac{\log{\left (\frac{a}{c} + \frac{b x}{c} + x^{2} \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a),x)

[Out]

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

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Giac [A]  time = 1.16075, size = 150, normalized size = 2.14 \begin{align*} -\frac{2 \, c \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{4} c d^{3} - 8 \, a b^{2} c^{2} d^{3} + 16 \, a^{2} c^{3} d^{3}} + \frac{\log \left (c x^{2} + b x + a\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \frac{1}{{\left (b^{2} - 4 \, a c\right )}{\left (2 \, c x + b\right )}^{2} d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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