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

Optimal. Leaf size=72 \[ \frac {b \tanh ^{-1}\left (\frac {a+b x}{\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}}-\frac {a+b x}{2 \left (a^2-b^2\right ) \left (2 a x+b x^2+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {614, 618, 206} \[ \frac {b \tanh ^{-1}\left (\frac {a+b x}{\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2}}-\frac {a+b x}{2 \left (a^2-b^2\right ) \left (2 a x+b x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(b + 2*a*x + b*x^2)^(-2),x]

[Out]

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

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Mathematica [A]  time = 0.04, size = 72, normalized size = 1.00 \[ \frac {a+b x}{2 \left (b^2-a^2\right ) \left (2 a x+b x^2+b\right )}+\frac {b \tan ^{-1}\left (\frac {a+b x}{\sqrt {b^2-a^2}}\right )}{2 \left (b^2-a^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b + 2*a*x + b*x^2)^(-2),x]

[Out]

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

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fricas [B]  time = 0.95, size = 317, normalized size = 4.40 \[ \left [-\frac {2 \, a^{3} - 2 \, a b^{2} + {\left (b^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} x^{2} + 2 \, a b x + 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b x + a\right )}}{b x^{2} + 2 \, a x + b}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} x}{4 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x\right )}}, -\frac {a^{3} - a b^{2} - {\left (b^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b x + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} b - b^{3}\right )} x}{2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} x^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a^3 - 2*a*b^2 + (b^2*x^2 + 2*a*b*x + b^2)*sqrt(a^2 - b^2)*log((b^2*x^2 + 2*a*b*x + 2*a^2 - b^2 - 2*sq
rt(a^2 - b^2)*(b*x + a))/(b*x^2 + 2*a*x + b)) + 2*(a^2*b - b^3)*x)/(a^4*b - 2*a^2*b^3 + b^5 + (a^4*b - 2*a^2*b
^3 + b^5)*x^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*x), -1/2*(a^3 - a*b^2 - (b^2*x^2 + 2*a*b*x + b^2)*sqrt(-a^2 + b^2)
*arctan(-sqrt(-a^2 + b^2)*(b*x + a)/(a^2 - b^2)) + (a^2*b - b^3)*x)/(a^4*b - 2*a^2*b^3 + b^5 + (a^4*b - 2*a^2*
b^3 + b^5)*x^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*x)]

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giac [A]  time = 0.51, size = 75, normalized size = 1.04 \[ -\frac {b \arctan \left (\frac {b x + a}{\sqrt {-a^{2} + b^{2}}}\right )}{2 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {b x + a}{2 \, {\left (b x^{2} + 2 \, a x + b\right )} {\left (a^{2} - b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*arctan((b*x + a)/sqrt(-a^2 + b^2))/((a^2 - b^2)*sqrt(-a^2 + b^2)) - 1/2*(b*x + a)/((b*x^2 + 2*a*x + b)*
(a^2 - b^2))

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maple [A]  time = 0.05, size = 86, normalized size = 1.19 \[ \frac {2 b \arctan \left (\frac {2 b x +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-4 a^{2}+4 b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {2 b x +2 a}{\left (-4 a^{2}+4 b^{2}\right ) \left (b \,x^{2}+2 a x +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*x+2*a)/(-4*a^2+4*b^2)/(b*x^2+2*a*x+b)+2*b/(-4*a^2+4*b^2)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*x+2*a)/(-a^2+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/(b*x^2+2*a*x+b)^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^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 0.32, size = 107, normalized size = 1.49 \[ -\frac {\frac {a}{2\,\left (a^2-b^2\right )}+\frac {b\,x}{2\,\left (a^2-b^2\right )}}{b\,x^2+2\,a\,x+b}+\frac {b\,\mathrm {atan}\left (\frac {-a^3\,1{}\mathrm {i}-1{}\mathrm {i}\,x\,a^2\,b+a\,b^2\,1{}\mathrm {i}+1{}\mathrm {i}\,x\,b^3}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,1{}\mathrm {i}}{2\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b + 2*a*x + b*x^2)^2,x)

[Out]

(b*atan((a*b^2*1i + b^3*x*1i - a^3*1i - a^2*b*x*1i)/((a + b)^(3/2)*(a - b)^(3/2)))*1i)/(2*(a + b)^(3/2)*(a - b
)^(3/2)) - (a/(2*(a^2 - b^2)) + (b*x)/(2*(a^2 - b^2)))/(b + 2*a*x + b*x^2)

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sympy [B]  time = 0.55, size = 230, normalized size = 3.19 \[ - \frac {b \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} \log {\left (x + \frac {- a^{4} b \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} + 2 a^{2} b^{3} \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} + a b - b^{5} \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}}}{b^{2}} \right )}}{4} + \frac {b \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} \log {\left (x + \frac {a^{4} b \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} - 2 a^{2} b^{3} \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}} + a b + b^{5} \sqrt {\frac {1}{\left (a - b\right )^{3} \left (a + b\right )^{3}}}}{b^{2}} \right )}}{4} + \frac {- a - b x}{2 a^{2} b - 2 b^{3} + x^{2} \left (2 a^{2} b - 2 b^{3}\right ) + x \left (4 a^{3} - 4 a b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*sqrt(1/((a - b)**3*(a + b)**3))*log(x + (-a**4*b*sqrt(1/((a - b)**3*(a + b)**3)) + 2*a**2*b**3*sqrt(1/((a -
b)**3*(a + b)**3)) + a*b - b**5*sqrt(1/((a - b)**3*(a + b)**3)))/b**2)/4 + b*sqrt(1/((a - b)**3*(a + b)**3))*
log(x + (a**4*b*sqrt(1/((a - b)**3*(a + b)**3)) - 2*a**2*b**3*sqrt(1/((a - b)**3*(a + b)**3)) + a*b + b**5*sqr
t(1/((a - b)**3*(a + b)**3)))/b**2)/4 + (-a - b*x)/(2*a**2*b - 2*b**3 + x**2*(2*a**2*b - 2*b**3) + x*(4*a**3 -
4*a*b**2))

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