### 3.1175 $$\int \frac{1}{(b d+2 c d x)^2 (a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=98 $-\frac{1}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )}-\frac{12 c}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac{12 c \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{5/2}}$

[Out]

(-12*c)/((b^2 - 4*a*c)^2*d^2*(b + 2*c*x)) - 1/((b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)) + (12*c*ArcTan
h[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0575797, antiderivative size = 98, 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 = {687, 693, 618, 206} $-\frac{1}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )}-\frac{12 c}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac{12 c \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-12*c)/((b^2 - 4*a*c)^2*d^2*(b + 2*c*x)) - 1/((b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)) + (12*c*ArcTan
h[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)*d^2)

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 618

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^2} \, dx &=-\frac{1}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac{(6 c) \int \frac{1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c}\\ &=-\frac{12 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x)}-\frac{1}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac{(6 c) \int \frac{1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{12 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x)}-\frac{1}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )}+\frac{(12 c) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac{12 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x)}-\frac{1}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )}+\frac{12 c \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d^2}\\ \end{align*}

Mathematica [A]  time = 0.121909, size = 84, normalized size = 0.86 $-\frac{\frac{12 c \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{b+2 c x}{a+x (b+c x)}+\frac{8 c}{b+2 c x}}{d^2 \left (b^2-4 a c\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(((8*c)/(b + 2*c*x) + (b + 2*c*x)/(a + x*(b + c*x)) + (12*c*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2
+ 4*a*c])/((b^2 - 4*a*c)^2*d^2))

________________________________________________________________________________________

Maple [A]  time = 0.158, size = 127, normalized size = 1.3 \begin{align*} -2\,{\frac{cx}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{b}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-12\,{\frac{c}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{5/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-8\,{\frac{c}{{d}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( 2\,cx+b \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*x*c-1/d^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b-12/d^2/(4*a*c-b^2)^(5/2)*c*arctan((
2*c*x+b)/(4*a*c-b^2)^(1/2))-8/d^2/(4*a*c-b^2)^2*c/(2*c*x+b)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.15535, size = 1384, normalized size = 14.12 \begin{align*} \left [-\frac{b^{4} + 4 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 6 \,{\left (2 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} + a b c +{\left (b^{2} c + 2 \, a c^{2}\right )} x\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 12 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x}{2 \,{\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} d^{2} x^{3} + 3 \,{\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} d^{2} x^{2} +{\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} d^{2} x +{\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} d^{2}}, -\frac{b^{4} + 4 \, a b^{2} c - 32 \, a^{2} c^{2} + 12 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 12 \,{\left (2 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} + a b c +{\left (b^{2} c + 2 \, a c^{2}\right )} x\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 12 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x}{2 \,{\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} d^{2} x^{3} + 3 \,{\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} d^{2} x^{2} +{\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} d^{2} x +{\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} d^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(b^4 + 4*a*b^2*c - 32*a^2*c^2 + 12*(b^2*c^2 - 4*a*c^3)*x^2 - 6*(2*c^3*x^3 + 3*b*c^2*x^2 + a*b*c + (b^2*c + 2
*a*c^2)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 +
b*x + a)) + 12*(b^3*c - 4*a*b*c^2)*x)/(2*(b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d^2*x^3 + 3*(b
^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*d^2*x^2 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2
*c^3 - 128*a^4*c^4)*d^2*x + (a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*d^2), -(b^4 + 4*a*b^2*c - 3
2*a^2*c^2 + 12*(b^2*c^2 - 4*a*c^3)*x^2 - 12*(2*c^3*x^3 + 3*b*c^2*x^2 + a*b*c + (b^2*c + 2*a*c^2)*x)*sqrt(-b^2
+ 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 12*(b^3*c - 4*a*b*c^2)*x)/(2*(b^6*c^2 - 12*a*
b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d^2*x^3 + 3*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*d^2*
x^2 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d^2*x + (a*b^7 - 12*a^2*b^5*c + 48*a^
3*b^3*c^2 - 64*a^4*b*c^3)*d^2)]

________________________________________________________________________________________

Sympy [B]  time = 2.676, size = 457, normalized size = 4.66 \begin{align*} \frac{6 c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{- 384 a^{3} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 288 a^{2} b^{2} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 72 a b^{4} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b^{6} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c}{12 c^{2}} \right )}}{d^{2}} - \frac{6 c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{384 a^{3} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 288 a^{2} b^{2} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 72 a b^{4} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 6 b^{6} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c}{12 c^{2}} \right )}}{d^{2}} - \frac{8 a c + b^{2} + 12 b c x + 12 c^{2} x^{2}}{16 a^{3} b c^{2} d^{2} - 8 a^{2} b^{3} c d^{2} + a b^{5} d^{2} + x^{3} \left (32 a^{2} c^{4} d^{2} - 16 a b^{2} c^{3} d^{2} + 2 b^{4} c^{2} d^{2}\right ) + x^{2} \left (48 a^{2} b c^{3} d^{2} - 24 a b^{3} c^{2} d^{2} + 3 b^{5} c d^{2}\right ) + x \left (32 a^{3} c^{3} d^{2} - 6 a b^{4} c d^{2} + b^{6} d^{2}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-384*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5) + 288*a**2*b**2*c**3*sqrt(-1
/(4*a*c - b**2)**5) - 72*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5) + 6*b**6*c*sqrt(-1/(4*a*c - b**2)**5) + 6*b*c)
/(12*c**2))/d**2 - 6*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (384*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5) - 288*a**2
*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5) + 72*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5) - 6*b**6*c*sqrt(-1/(4*a*c -
b**2)**5) + 6*b*c)/(12*c**2))/d**2 - (8*a*c + b**2 + 12*b*c*x + 12*c**2*x**2)/(16*a**3*b*c**2*d**2 - 8*a**2*b*
*3*c*d**2 + a*b**5*d**2 + x**3*(32*a**2*c**4*d**2 - 16*a*b**2*c**3*d**2 + 2*b**4*c**2*d**2) + x**2*(48*a**2*b*
c**3*d**2 - 24*a*b**3*c**2*d**2 + 3*b**5*c*d**2) + x*(32*a**3*c**3*d**2 - 6*a*b**4*c*d**2 + b**6*d**2))

________________________________________________________________________________________

Giac [B]  time = 1.23668, size = 297, normalized size = 3.03 \begin{align*} -\frac{8 \, c^{5} d^{7}}{{\left (b^{4} c^{4} d^{8} - 8 \, a b^{2} c^{5} d^{8} + 16 \, a^{2} c^{6} d^{8}\right )}{\left (2 \, c d x + b d\right )}} - \frac{12 \, c \arctan \left (\frac{\frac{b^{2} d}{2 \, c d x + b d} - \frac{4 \, a c d}{2 \, c d x + b d}}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c} d^{2}} + \frac{4 \, c}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}{\left (2 \, c d x + b d\right )}{\left (\frac{b^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac{4 \, a c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - 1\right )} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*c^5*d^7/((b^4*c^4*d^8 - 8*a*b^2*c^5*d^8 + 16*a^2*c^6*d^8)*(2*c*d*x + b*d)) - 12*c*arctan((b^2*d/(2*c*d*x +
b*d) - 4*a*c*d/(2*c*d*x + b*d))/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)*d^2) +
4*c/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)*(b^2*d^2/(2*c*d*x + b*d)^2 - 4*a*c*d^2/(2*c*d*x + b*d)^2 -
1)*d)