3.1.95 \(\int \frac {1}{(a+c x+b x^2)^2} \, dx\) [95]

Optimal. Leaf size=71 \[ \frac {c+2 b x}{\left (4 a b-c^2\right ) \left (a+c x+b x^2\right )}+\frac {4 b \tan ^{-1}\left (\frac {c+2 b x}{\sqrt {4 a b-c^2}}\right )}{\left (4 a b-c^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 71, 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 = {628, 632, 210} \begin {gather*} \frac {4 b \text {ArcTan}\left (\frac {2 b x+c}{\sqrt {4 a b-c^2}}\right )}{\left (4 a b-c^2\right )^{3/2}}+\frac {2 b x+c}{\left (4 a b-c^2\right ) \left (a+b x^2+c x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x + b*x^2)^(-2),x]

[Out]

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

Rule 210

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

Rule 628

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 632

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

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Mathematica [A]
time = 0.04, size = 70, normalized size = 0.99 \begin {gather*} \frac {c+2 b x}{\left (4 a b-c^2\right ) (a+x (c+b x))}+\frac {4 b \tan ^{-1}\left (\frac {c+2 b x}{\sqrt {4 a b-c^2}}\right )}{\left (4 a b-c^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x + b*x^2)^(-2),x]

[Out]

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

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Maple [A]
time = 0.61, size = 68, normalized size = 0.96

method	result	size
default	\(\frac {2 b x +c}{\left (4 a b -c^{2}\right ) \left (b \,x^{2}+c x +a \right )}+\frac {4 b \arctan \left (\frac {2 b x +c}{\sqrt {4 a b -c^{2}}}\right )}{\left (4 a b -c^{2}\right )^{\frac {3}{2}}}\)	\(68\)
risch	\(\frac {\frac {2 b x}{4 a b -c^{2}}+\frac {c}{4 a b -c^{2}}}{b \,x^{2}+c x +a}+\frac {2 b \ln \left (\left (-8 a \,b^{2}+2 b \,c^{2}\right ) x +\left (-4 a b +c^{2}\right )^{\frac {3}{2}}-4 a b c +c^{3}\right )}{\left (-4 a b +c^{2}\right )^{\frac {3}{2}}}-\frac {2 b \ln \left (\left (8 a \,b^{2}-2 b \,c^{2}\right ) x +\left (-4 a b +c^{2}\right )^{\frac {3}{2}}+4 a b c -c^{3}\right )}{\left (-4 a b +c^{2}\right )^{\frac {3}{2}}}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+c*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(c^2-4*a*b>0)', see `assume?` f
or more deta

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (67) = 134\).
time = 2.17, size = 334, normalized size = 4.70 \begin {gather*} \left [\frac {4 \, a b c - c^{3} + 2 \, {\left (b^{2} x^{2} + b c x + a b\right )} \sqrt {-4 \, a b + c^{2}} \log \left (\frac {2 \, b^{2} x^{2} + 2 \, b c x - 2 \, a b + c^{2} + \sqrt {-4 \, a b + c^{2}} {\left (2 \, b x + c\right )}}{b x^{2} + c x + a}\right ) + 2 \, {\left (4 \, a b^{2} - b c^{2}\right )} x}{16 \, a^{3} b^{2} - 8 \, a^{2} b c^{2} + a c^{4} + {\left (16 \, a^{2} b^{3} - 8 \, a b^{2} c^{2} + b c^{4}\right )} x^{2} + {\left (16 \, a^{2} b^{2} c - 8 \, a b c^{3} + c^{5}\right )} x}, \frac {4 \, a b c - c^{3} - 4 \, {\left (b^{2} x^{2} + b c x + a b\right )} \sqrt {4 \, a b - c^{2}} \arctan \left (-\frac {2 \, b x + c}{\sqrt {4 \, a b - c^{2}}}\right ) + 2 \, {\left (4 \, a b^{2} - b c^{2}\right )} x}{16 \, a^{3} b^{2} - 8 \, a^{2} b c^{2} + a c^{4} + {\left (16 \, a^{2} b^{3} - 8 \, a b^{2} c^{2} + b c^{4}\right )} x^{2} + {\left (16 \, a^{2} b^{2} c - 8 \, a b c^{3} + c^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (60) = 120\).
time = 0.30, size = 265, normalized size = 3.73 \begin {gather*} - 2 b \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} \log {\left (x + \frac {- 32 a^{2} b^{3} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} + 16 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} - 2 b c^{4} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} + 2 b c}{4 b^{2}} \right )} + 2 b \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} \log {\left (x + \frac {32 a^{2} b^{3} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} - 16 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} + 2 b c^{4} \sqrt {- \frac {1}{\left (4 a b - c^{2}\right )^{3}}} + 2 b c}{4 b^{2}} \right )} + \frac {2 b x + c}{4 a^{2} b - a c^{2} + x^{2} \cdot \left (4 a b^{2} - b c^{2}\right ) + x \left (4 a b c - c^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.83, size = 67, normalized size = 0.94 \begin {gather*} \frac {4 \, b \arctan \left (\frac {2 \, b x + c}{\sqrt {4 \, a b - c^{2}}}\right )}{{\left (4 \, a b - c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, b x + c}{{\left (b x^{2} + c x + a\right )} {\left (4 \, a b - c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.17, size = 119, normalized size = 1.68 \begin {gather*} \frac {\frac {c}{4\,a\,b-c^2}+\frac {2\,b\,x}{4\,a\,b-c^2}}{b\,x^2+c\,x+a}-\frac {4\,b\,\mathrm {atan}\left (\frac {\left (\frac {2\,b\,\left (c^3-4\,a\,b\,c\right )}{{\left (4\,a\,b-c^2\right )}^{5/2}}-\frac {4\,b^2\,x}{{\left (4\,a\,b-c^2\right )}^{3/2}}\right )\,\left (4\,a\,b-c^2\right )}{2\,b}\right )}{{\left (4\,a\,b-c^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + c*x + b*x^2)^2,x)

[Out]

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

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