### 3.1176 $$\int \frac{1}{(b d+2 c d x)^3 (a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=110 $-\frac{1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac{8 c \log \left (a+b x+c x^2\right )}{d^3 \left (b^2-4 a c\right )^3}-\frac{8 c}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{16 c \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^3}$

[Out]

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

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Rubi [A]  time = 0.0621794, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.208, Rules used = {687, 693, 681, 31, 628} $-\frac{1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac{8 c \log \left (a+b x+c x^2\right )}{d^3 \left (b^2-4 a c\right )^3}-\frac{8 c}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{16 c \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 687

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] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

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 )^2} \, dx &=-\frac{1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac{(8 c) \int \frac{1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{8 c}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac{(8 c) \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 )^2 d^2}\\ &=-\frac{8 c}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )}-\frac{(8 c) \int \frac{b d+2 c d x}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^4}+\frac{\left (32 c^2\right ) \int \frac{1}{b+2 c x} \, dx}{\left (b^2-4 a c\right )^3 d^3}\\ &=-\frac{8 c}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac{1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )}+\frac{16 c \log (b+2 c x)}{\left (b^2-4 a c\right )^3 d^3}-\frac{8 c \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^3 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0902598, size = 79, normalized size = 0.72 $\frac{-\frac{4 c \left (b^2-4 a c\right )}{(b+2 c x)^2}+\frac{4 a c-b^2}{a+x (b+c x)}-8 c \log (a+x (b+c x))+16 c \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 144, normalized size = 1.3 \begin{align*} -4\,{\frac{ac}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) }}+{\frac{{b}^{2}}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) }}+8\,{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) }{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}}}-16\,{\frac{c\ln \left ( 2\,cx+b \right ) }{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3}}}-4\,{\frac{c}{{d}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{2} \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)^2,x)

[Out]

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

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Maxima [B]  time = 1.12979, size = 400, normalized size = 3.64 \begin{align*} -\frac{8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{4 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \,{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} +{\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} +{\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x +{\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}} - \frac{8 \, c \log \left (c x^{2} + b x + a\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{3}} + \frac{16 \, c \log \left (2 \, c x + b\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\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)^2,x, algorithm="maxima")

[Out]

-(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^
3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c
- 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3) - 8*c*log(c*x^2 + b*x + a
)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^3) + 16*c*log(2*c*x + b)/((b^6 - 12*a*b^4*c + 48*a^2*b^2
*c^2 - 64*a^3*c^3)*d^3)

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Fricas [B]  time = 2.05137, size = 864, normalized size = 7.85 \begin{align*} -\frac{b^{4} - 16 \, a^{2} c^{2} + 8 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 8 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x + 8 \,{\left (4 \, c^{4} x^{4} + 8 \, b c^{3} x^{3} + a b^{2} c +{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} +{\left (b^{3} c + 4 \, a b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 16 \,{\left (4 \, c^{4} x^{4} + 8 \, b c^{3} x^{3} + a b^{2} c +{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} +{\left (b^{3} c + 4 \, a b c^{2}\right )} x\right )} \log \left (2 \, c x + b\right )}{4 \,{\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{3} x^{4} + 8 \,{\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{3} x^{3} +{\left (5 \, b^{8} c - 56 \, a b^{6} c^{2} + 192 \, a^{2} b^{4} c^{3} - 128 \, a^{3} b^{2} c^{4} - 256 \, a^{4} c^{5}\right )} d^{3} x^{2} +{\left (b^{9} - 8 \, a b^{7} c + 128 \, a^{3} b^{3} c^{3} - 256 \, a^{4} b c^{4}\right )} d^{3} x +{\left (a b^{8} - 12 \, a^{2} b^{6} c + 48 \, a^{3} b^{4} c^{2} - 64 \, a^{4} b^{2} c^{3}\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)^2,x, algorithm="fricas")

[Out]

-(b^4 - 16*a^2*c^2 + 8*(b^2*c^2 - 4*a*c^3)*x^2 + 8*(b^3*c - 4*a*b*c^2)*x + 8*(4*c^4*x^4 + 8*b*c^3*x^3 + a*b^2*
c + (5*b^2*c^2 + 4*a*c^3)*x^2 + (b^3*c + 4*a*b*c^2)*x)*log(c*x^2 + b*x + a) - 16*(4*c^4*x^4 + 8*b*c^3*x^3 + a*
b^2*c + (5*b^2*c^2 + 4*a*c^3)*x^2 + (b^3*c + 4*a*b*c^2)*x)*log(2*c*x + b))/(4*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2
*b^2*c^5 - 64*a^3*c^6)*d^3*x^4 + 8*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^3*x^3 + (5*b^8*c
- 56*a*b^6*c^2 + 192*a^2*b^4*c^3 - 128*a^3*b^2*c^4 - 256*a^4*c^5)*d^3*x^2 + (b^9 - 8*a*b^7*c + 128*a^3*b^3*c^
3 - 256*a^4*b*c^4)*d^3*x + (a*b^8 - 12*a^2*b^6*c + 48*a^3*b^4*c^2 - 64*a^4*b^2*c^3)*d^3)

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Sympy [B]  time = 6.44781, size = 303, normalized size = 2.75 \begin{align*} - \frac{16 c \log{\left (\frac{b}{2 c} + x \right )}}{d^{3} \left (4 a c - b^{2}\right )^{3}} + \frac{8 c \log{\left (\frac{a}{c} + \frac{b x}{c} + x^{2} \right )}}{d^{3} \left (4 a c - b^{2}\right )^{3}} - \frac{4 a c + b^{2} + 8 b c x + 8 c^{2} x^{2}}{16 a^{3} b^{2} c^{2} d^{3} - 8 a^{2} b^{4} c d^{3} + a b^{6} d^{3} + x^{4} \left (64 a^{2} c^{5} d^{3} - 32 a b^{2} c^{4} d^{3} + 4 b^{4} c^{3} d^{3}\right ) + x^{3} \left (128 a^{2} b c^{4} d^{3} - 64 a b^{3} c^{3} d^{3} + 8 b^{5} c^{2} d^{3}\right ) + x^{2} \left (64 a^{3} c^{4} d^{3} + 48 a^{2} b^{2} c^{3} d^{3} - 36 a b^{4} c^{2} d^{3} + 5 b^{6} c d^{3}\right ) + x \left (64 a^{3} b c^{3} d^{3} - 16 a^{2} b^{3} c^{2} d^{3} - 4 a b^{5} c d^{3} + b^{7} d^{3}\right )} \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)**2,x)

[Out]

-16*c*log(b/(2*c) + x)/(d**3*(4*a*c - b**2)**3) + 8*c*log(a/c + b*x/c + x**2)/(d**3*(4*a*c - b**2)**3) - (4*a*
c + b**2 + 8*b*c*x + 8*c**2*x**2)/(16*a**3*b**2*c**2*d**3 - 8*a**2*b**4*c*d**3 + a*b**6*d**3 + x**4*(64*a**2*c
**5*d**3 - 32*a*b**2*c**4*d**3 + 4*b**4*c**3*d**3) + x**3*(128*a**2*b*c**4*d**3 - 64*a*b**3*c**3*d**3 + 8*b**5
*c**2*d**3) + x**2*(64*a**3*c**4*d**3 + 48*a**2*b**2*c**3*d**3 - 36*a*b**4*c**2*d**3 + 5*b**6*c*d**3) + x*(64*
a**3*b*c**3*d**3 - 16*a**2*b**3*c**2*d**3 - 4*a*b**5*c*d**3 + b**7*d**3))

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Giac [A]  time = 1.17857, size = 274, normalized size = 2.49 \begin{align*} \frac{16 \, c^{2} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{6} c d^{3} - 12 \, a b^{4} c^{2} d^{3} + 48 \, a^{2} b^{2} c^{3} d^{3} - 64 \, a^{3} c^{4} d^{3}} - \frac{8 \, c \log \left (c x^{2} + b x + a\right )}{b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}} - \frac{b^{4} - 16 \, a^{2} c^{2} + 8 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 8 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x}{{\left (c x^{2} + b x + a\right )}{\left (b^{2} - 4 \, a c\right )}^{3}{\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)^2,x, algorithm="giac")

[Out]

16*c^2*log(abs(2*c*x + b))/(b^6*c*d^3 - 12*a*b^4*c^2*d^3 + 48*a^2*b^2*c^3*d^3 - 64*a^3*c^4*d^3) - 8*c*log(c*x^
2 + b*x + a)/(b^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3*d^3) - (b^4 - 16*a^2*c^2 + 8*(b^2*c^2
- 4*a*c^3)*x^2 + 8*(b^3*c - 4*a*b*c^2)*x)/((c*x^2 + b*x + a)*(b^2 - 4*a*c)^3*(2*c*x + b)^2*d^3)