### 3.1185 $$\int \frac{(b d+2 c d x)^2}{(a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=100 $\frac{4 c^2 d^2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{c d^2 (b+2 c x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{d^2 (b+2 c x)}{2 \left (a+b x+c x^2\right )^2}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0528953, antiderivative size = 100, 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 = {686, 614, 618, 206} $\frac{4 c^2 d^2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{c d^2 (b+2 c x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{d^2 (b+2 c x)}{2 \left (a+b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 686

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

Rule 614

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

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

Mathematica [A]  time = 0.0910937, size = 98, normalized size = 0.98 $d^2 \left (\frac{4 c^2 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}-\frac{(b+2 c x) \left (2 c \left (c x^2-a\right )+b^2+2 b c x\right )}{2 \left (b^2-4 a c\right ) (a+x (b+c x))^2}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.155, size = 245, normalized size = 2.5 \begin{align*} 2\,{\frac{{d}^{2}{c}^{3}{x}^{3}}{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}+3\,{\frac{{d}^{2}b{c}^{2}{x}^{2}}{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}-2\,{\frac{{c}^{2}{d}^{2}xa}{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{c{d}^{2}x{b}^{2}}{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{d}^{2}bac}{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{{d}^{2}{b}^{3}}{2\, \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( 4\,ac-{b}^{2} \right ) }}+4\,{\frac{{c}^{2}{d}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.15957, size = 1485, normalized size = 14.85 \begin{align*} \left [-\frac{4 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{2} x^{3} + 6 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{2} x^{2} + 4 \,{\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} d^{2} x +{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} d^{2} + 4 \,{\left (c^{4} d^{2} x^{4} + 2 \, b c^{3} d^{2} x^{3} + 2 \, a b c^{2} d^{2} x + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} + 2 \, a c^{3}\right )} d^{2} x^{2}\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 )}{2 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}, -\frac{4 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{2} x^{3} + 6 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{2} x^{2} + 4 \,{\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} d^{2} x +{\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} d^{2} - 8 \,{\left (c^{4} d^{2} x^{4} + 2 \, b c^{3} d^{2} x^{3} + 2 \, a b c^{2} d^{2} x + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} + 2 \, a c^{3}\right )} d^{2} x^{2}\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 )}{2 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(4*(b^2*c^3 - 4*a*c^4)*d^2*x^3 + 6*(b^3*c^2 - 4*a*b*c^3)*d^2*x^2 + 4*(b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*d
^2*x + (b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*d^2 + 4*(c^4*d^2*x^4 + 2*b*c^3*d^2*x^3 + 2*a*b*c^2*d^2*x + a^2*c^2*d^2
+ (b^2*c^2 + 2*a*c^3)*d^2*x^2)*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)))/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4
+ 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 1
6*a^3*b*c^2)*x), -1/2*(4*(b^2*c^3 - 4*a*c^4)*d^2*x^3 + 6*(b^3*c^2 - 4*a*b*c^3)*d^2*x^2 + 4*(b^4*c - 5*a*b^2*c^
2 + 4*a^2*c^3)*d^2*x + (b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*d^2 - 8*(c^4*d^2*x^4 + 2*b*c^3*d^2*x^3 + 2*a*b*c^2*d^2*
x + a^2*c^2*d^2 + (b^2*c^2 + 2*a*c^3)*d^2*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2
- 4*a*c)))/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3
*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)]

________________________________________________________________________________________

Sympy [B]  time = 2.7708, size = 430, normalized size = 4.3 \begin{align*} - 2 c^{2} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{- 32 a^{2} c^{4} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 16 a b^{2} c^{3} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 2 b^{4} c^{2} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b c^{2} d^{2}}{4 c^{3} d^{2}} \right )} + 2 c^{2} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{32 a^{2} c^{4} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 16 a b^{2} c^{3} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b^{4} c^{2} d^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b c^{2} d^{2}}{4 c^{3} d^{2}} \right )} + \frac{- 2 a b c d^{2} + b^{3} d^{2} + 6 b c^{2} d^{2} x^{2} + 4 c^{3} d^{2} x^{3} + x \left (- 4 a c^{2} d^{2} + 4 b^{2} c d^{2}\right )}{8 a^{3} c - 2 a^{2} b^{2} + x^{4} \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{3} \left (16 a b c^{2} - 4 b^{3} c\right ) + x^{2} \left (16 a^{2} c^{2} + 4 a b^{2} c - 2 b^{4}\right ) + x \left (16 a^{2} b c - 4 a b^{3}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c**2*d**2*sqrt(-1/(4*a*c - b**2)**3)*log(x + (-32*a**2*c**4*d**2*sqrt(-1/(4*a*c - b**2)**3) + 16*a*b**2*c**
3*d**2*sqrt(-1/(4*a*c - b**2)**3) - 2*b**4*c**2*d**2*sqrt(-1/(4*a*c - b**2)**3) + 2*b*c**2*d**2)/(4*c**3*d**2)
) + 2*c**2*d**2*sqrt(-1/(4*a*c - b**2)**3)*log(x + (32*a**2*c**4*d**2*sqrt(-1/(4*a*c - b**2)**3) - 16*a*b**2*c
**3*d**2*sqrt(-1/(4*a*c - b**2)**3) + 2*b**4*c**2*d**2*sqrt(-1/(4*a*c - b**2)**3) + 2*b*c**2*d**2)/(4*c**3*d**
2)) + (-2*a*b*c*d**2 + b**3*d**2 + 6*b*c**2*d**2*x**2 + 4*c**3*d**2*x**3 + x*(-4*a*c**2*d**2 + 4*b**2*c*d**2))
/(8*a**3*c - 2*a**2*b**2 + x**4*(8*a*c**3 - 2*b**2*c**2) + x**3*(16*a*b*c**2 - 4*b**3*c) + x**2*(16*a**2*c**2
+ 4*a*b**2*c - 2*b**4) + x*(16*a**2*b*c - 4*a*b**3))

________________________________________________________________________________________

Giac [A]  time = 1.25046, size = 181, normalized size = 1.81 \begin{align*} -\frac{4 \, c^{2} d^{2} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{4 \, c^{3} d^{2} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b^{2} c d^{2} x - 4 \, a c^{2} d^{2} x + b^{3} d^{2} - 2 \, a b c d^{2}}{2 \,{\left (c x^{2} + b x + a\right )}^{2}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c^2*d^2*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(4*c^3*d^2*x^3 + 6*
b*c^2*d^2*x^2 + 4*b^2*c*d^2*x - 4*a*c^2*d^2*x + b^3*d^2 - 2*a*b*c*d^2)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c))