3.2195 \(\int \frac {1}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=66 \[ \frac {4 c \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {614, 618, 206} \[ \frac {4 c \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(-2),x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 70, normalized size = 1.06 \[ -\frac {\frac {4 c \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {b+2 c x}{a+x (b+c x)}}{b^2-4 a c} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(-2),x]

[Out]

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

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fricas [B]  time = 0.89, size = 341, normalized size = 5.17 \[ \left [-\frac {b^{3} - 4 \, a b c + 2 \, {\left (c^{2} x^{2} + b c x + a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}, -\frac {b^{3} - 4 \, a b c - 4 \, {\left (c^{2} x^{2} + b c x + a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 76, normalized size = 1.15 \[ -\frac {4 \, c \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c x + b}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) - (2*c*x + b)/((c*x^2 + b*x + a
)*(b^2 - 4*a*c))

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maple [A]  time = 0.07, size = 68, normalized size = 1.03 \[ \frac {4 c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {2 c x +b}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.15, size = 119, normalized size = 1.80 \[ \frac {\frac {b}{4\,a\,c-b^2}+\frac {2\,c\,x}{4\,a\,c-b^2}}{c\,x^2+b\,x+a}-\frac {4\,c\,\mathrm {atan}\left (\frac {\left (\frac {2\,c\,\left (b^3-4\,a\,b\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {4\,c^2\,x}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )\,\left (4\,a\,c-b^2\right )}{2\,c}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x + c*x^2)^2,x)

[Out]

(b/(4*a*c - b^2) + (2*c*x)/(4*a*c - b^2))/(a + b*x + c*x^2) - (4*c*atan((((2*c*(b^3 - 4*a*b*c))/(4*a*c - b^2)^
(5/2) - (4*c^2*x)/(4*a*c - b^2)^(3/2))*(4*a*c - b^2))/(2*c)))/(4*a*c - b^2)^(3/2)

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sympy [B]  time = 0.58, size = 265, normalized size = 4.02 \[ - 2 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {- 32 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 16 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 2 b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b c}{4 c^{2}} \right )} + 2 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {32 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 16 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 b c}{4 c^{2}} \right )} + \frac {b + 2 c x}{4 a^{2} c - a b^{2} + x^{2} \left (4 a c^{2} - b^{2} c\right ) + x \left (4 a b c - b^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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