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

Optimal. Leaf size=63 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} d}+\frac{c+d x}{2 a d \left (a+b (c+d x)^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0329842, antiderivative size = 63, normalized size of antiderivative = 1., 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 = {247, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} d}+\frac{c+d x}{2 a d \left (a+b (c+d x)^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c + d*x)^2)^(-2),x]

[Out]

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

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.0247364, size = 60, normalized size = 0.95 \[ \frac{\frac{\sqrt{a} (c+d x)}{a+b (c+d x)^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} (c+d x)}{\sqrt{a}}\right )}{\sqrt{b}}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c + d*x)^2)^(-2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 86, normalized size = 1.4 \begin{align*}{\frac{2\,b{d}^{2}x+2\,bcd}{4\,ab{d}^{2} \left ( b{d}^{2}{x}^{2}+2\,bcdx+{c}^{2}b+a \right ) }}+{\frac{1}{2\,ad}\arctan \left ({\frac{2\,b{d}^{2}x+2\,bcd}{2\,d}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8116, size = 555, normalized size = 8.81 \begin{align*} \left [\frac{2 \, a b d x + 2 \, a b c -{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )} \sqrt{-a b} \log \left (\frac{b d^{2} x^{2} + 2 \, b c d x + b c^{2} - 2 \, \sqrt{-a b}{\left (d x + c\right )} - a}{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}\right )}{4 \,{\left (a^{2} b^{2} d^{3} x^{2} + 2 \, a^{2} b^{2} c d^{2} x +{\left (a^{2} b^{2} c^{2} + a^{3} b\right )} d\right )}}, \frac{a b d x + a b c +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}{\left (d x + c\right )}}{a}\right )}{2 \,{\left (a^{2} b^{2} d^{3} x^{2} + 2 \, a^{2} b^{2} c d^{2} x +{\left (a^{2} b^{2} c^{2} + a^{3} b\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.727365, size = 117, normalized size = 1.86 \begin{align*} \frac{c + d x}{2 a^{2} d + 2 a b c^{2} d + 4 a b c d^{2} x + 2 a b d^{3} x^{2}} + \frac{- \frac{\sqrt{- \frac{1}{a^{3} b}} \log{\left (x + \frac{- a^{2} \sqrt{- \frac{1}{a^{3} b}} + c}{d} \right )}}{4} + \frac{\sqrt{- \frac{1}{a^{3} b}} \log{\left (x + \frac{a^{2} \sqrt{- \frac{1}{a^{3} b}} + c}{d} \right )}}{4}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x)/(2*a**2*d + 2*a*b*c**2*d + 4*a*b*c*d**2*x + 2*a*b*d**3*x**2) + (-sqrt(-1/(a**3*b))*log(x + (-a**2*sq
rt(-1/(a**3*b)) + c)/d)/4 + sqrt(-1/(a**3*b))*log(x + (a**2*sqrt(-1/(a**3*b)) + c)/d)/4)/d

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Giac [A]  time = 1.11284, size = 88, normalized size = 1.4 \begin{align*} \frac{\arctan \left (\frac{b d x + b c}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a d} + \frac{d x + c}{2 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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