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

Optimal. Leaf size=97 \[ \frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^2}+\frac{b \cot (c+d x)}{2 a d (a-b) \left (a+b \cot ^2(c+d x)\right )}+\frac{x}{(a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.100319, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3661, 414, 522, 203, 205} \[ \frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^2}+\frac{b \cot (c+d x)}{2 a d (a-b) \left (a+b \cot ^2(c+d x)\right )}+\frac{x}{(a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x]^2)^(-2),x]

[Out]

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

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

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

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

Mathematica [A]  time = 0.877054, size = 90, normalized size = 0.93 \[ \frac{\frac{\sqrt{b} (3 a-b) \tan ^{-1}\left (\frac{\sqrt{b} \cot (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{b (a-b) \cot (c+d x)}{a \left (a+b \cot ^2(c+d x)\right )}-2 \tan ^{-1}(\cot (c+d x))}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[c + d*x]^2)^(-2),x]

[Out]

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

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Maple [B]  time = 0.028, size = 173, normalized size = 1.8 \begin{align*} -{\frac{\pi }{2\,d \left ( a-b \right ) ^{2}}}+{\frac{{\rm arccot} \left (\cot \left ( dx+c \right ) \right )}{d \left ( a-b \right ) ^{2}}}+{\frac{b\cot \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{2} \left ( a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{2}\cot \left ( dx+c \right ) }{2\,d \left ( a-b \right ) ^{2}a \left ( a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{3\,b}{2\,d \left ( a-b \right ) ^{2}}\arctan \left ({b\cot \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{2}}{2\,d \left ( a-b \right ) ^{2}a}\arctan \left ({b\cot \left ( dx+c \right ){\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*cot(d*x+c)^2)^2,x)

[Out]

-1/2/d/(a-b)^2*Pi+1/d/(a-b)^2*arccot(cot(d*x+c))+1/2/d/(a-b)^2*b*cot(d*x+c)/(a+b*cot(d*x+c)^2)-1/2/d/(a-b)^2*b
^2/a*cot(d*x+c)/(a+b*cot(d*x+c)^2)+3/2/d/(a-b)^2*b/(a*b)^(1/2)*arctan(cot(d*x+c)*b/(a*b)^(1/2))-1/2/d/(a-b)^2*
b^2/a/(a*b)^(1/2)*arctan(cot(d*x+c)*b/(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*cot(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89343, size = 1184, normalized size = 12.21 \begin{align*} \left [\frac{8 \,{\left (a^{2} - a b\right )} d x \cos \left (2 \, d x + 2 \, c\right ) - 8 \,{\left (a^{2} + a b\right )} d x +{\left (3 \, a^{2} + 2 \, a b - b^{2} -{\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + 4 \,{\left (a^{2} - a b -{\left (a^{2} + a b\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt{-\frac{b}{a}} \sin \left (2 \, d x + 2 \, c\right ) + a^{2} - 6 \, a b + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + a^{2} + 2 \, a b + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}\right ) - 4 \,{\left (a b - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{8 \,{\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) -{\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac{4 \,{\left (a^{2} - a b\right )} d x \cos \left (2 \, d x + 2 \, c\right ) - 4 \,{\left (a^{2} + a b\right )} d x -{\left (3 \, a^{2} + 2 \, a b - b^{2} -{\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a + b\right )} \sqrt{\frac{b}{a}}}{2 \, b \sin \left (2 \, d x + 2 \, c\right )}\right ) - 2 \,{\left (a b - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \,{\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) -{\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(8*(a^2 - a*b)*d*x*cos(2*d*x + 2*c) - 8*(a^2 + a*b)*d*x + (3*a^2 + 2*a*b - b^2 - (3*a^2 - 4*a*b + b^2)*co
s(2*d*x + 2*c))*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(2*d*x + 2*c)^2 + 4*(a^2 - a*b - (a^2 + a*b)*cos(2*d*x
+ 2*c))*sqrt(-b/a)*sin(2*d*x + 2*c) + a^2 - 6*a*b + b^2 - 2*(a^2 - b^2)*cos(2*d*x + 2*c))/((a^2 - 2*a*b + b^2)
*cos(2*d*x + 2*c)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - b^2)*cos(2*d*x + 2*c))) - 4*(a*b - b^2)*sin(2*d*x + 2*c))/(
(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c) - (a^4 - a^3*b - a^2*b^2 + a*b^3)*d), 1/4*(4*(a^2 - a*b
)*d*x*cos(2*d*x + 2*c) - 4*(a^2 + a*b)*d*x - (3*a^2 + 2*a*b - b^2 - (3*a^2 - 4*a*b + b^2)*cos(2*d*x + 2*c))*sq
rt(b/a)*arctan(1/2*((a + b)*cos(2*d*x + 2*c) - a + b)*sqrt(b/a)/(b*sin(2*d*x + 2*c))) - 2*(a*b - b^2)*sin(2*d*
x + 2*c))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c) - (a^4 - a^3*b - a^2*b^2 + a*b^3)*d)]

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Sympy [A]  time = 151.394, size = 2086, normalized size = 21.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/cot(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x - 1/(d*cot(c + d*x)) + 1/(3*d*cot(c + d*x)**3
))/b**2, Eq(a, 0)), (x/(a + b*cot(c)**2)**2, Eq(d, 0)), (x/a**2, Eq(b, 0)), (4*I*a**(5/2)*d*x*sqrt(1/b)/(4*I*a
**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b
**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**
2) + 4*I*a**(3/2)*b*d*x*sqrt(1/b)*cot(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c
+ d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*
sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) + 2*I*a**(3/2)*b*sqrt(1/b)*cot(c + d*x)/(4*I*a**(9/
2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d
*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) -
2*I*sqrt(a)*b**2*sqrt(1/b)*cot(c + d*x)/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2
- 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b)
+ 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) + 3*a**2*log(-I*sqrt(a)*sqrt(1/b) + cot(c + d*x))/(4*I*a**(9/
2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d
*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) -
3*a**2*log(I*sqrt(a)*sqrt(1/b) + cot(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c +
d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqr
t(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) + 3*a*b*log(-I*sqrt(a)*sqrt(1/b) + cot(c + d*x))*cot(c
+ d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b)
- 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/
b)*cot(c + d*x)**2) - a*b*log(-I*sqrt(a)*sqrt(1/b) + cot(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*
d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I
*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) - 3*a*b*log(I*sqrt(a)*sqrt(1/b) +
cot(c + d*x))*cot(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**
(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(
3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) + a*b*log(I*sqrt(a)*sqrt(1/b) + cot(c + d*x))/(4*I*a**(9/2)*d*sqrt(1/b)
+ 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot
(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) - b**2*log(-I*sq
rt(a)*sqrt(1/b) + cot(c + d*x))*cot(c + d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c +
d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b) - 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sq
rt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b)*cot(c + d*x)**2) + b**2*log(I*sqrt(a)*sqrt(1/b) + cot(c + d*x))*cot(c
+ d*x)**2/(4*I*a**(9/2)*d*sqrt(1/b) + 4*I*a**(7/2)*b*d*sqrt(1/b)*cot(c + d*x)**2 - 8*I*a**(7/2)*b*d*sqrt(1/b)
- 8*I*a**(5/2)*b**2*d*sqrt(1/b)*cot(c + d*x)**2 + 4*I*a**(5/2)*b**2*d*sqrt(1/b) + 4*I*a**(3/2)*b**3*d*sqrt(1/b
)*cot(c + d*x)**2), True))

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Giac [A]  time = 1.23714, size = 166, normalized size = 1.71 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (d x + c\right )}{\sqrt{a b}}\right )\right )}{\left (3 \, a b - b^{2}\right )}}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt{a b}} - \frac{2 \,{\left (d x + c\right )}}{a^{2} - 2 \, a b + b^{2}} - \frac{b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b\right )}{\left (a^{2} - a b\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(a) + arctan(a*tan(d*x + c)/sqrt(a*b)))*(3*a*b - b^2)/((a^3 - 2*a^2*b +
a*b^2)*sqrt(a*b)) - 2*(d*x + c)/(a^2 - 2*a*b + b^2) - b*tan(d*x + c)/((a*tan(d*x + c)^2 + b)*(a^2 - a*b)))/d