3.221 \(\int \frac{\cot ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=64 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} f (a-b)}-\frac{x}{a-b}-\frac{\cot (e+f x)}{a f} \]

[Out]

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

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Rubi [A]  time = 0.11064, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3670, 480, 522, 203, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} f (a-b)}-\frac{x}{a-b}-\frac{\cot (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2),x]

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(((d*ff*x)/c)^m*(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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, 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{\cot ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x)}{a f}+\frac{\operatorname{Subst}\left (\int \frac{-a-b-b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{\cot (e+f x)}{a f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a (a-b) f}\\ &=-\frac{x}{a-b}+\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} (a-b) f}-\frac{\cot (e+f x)}{a f}\\ \end{align*}

Mathematica [A]  time = 0.23702, size = 68, normalized size = 1.06 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )-\sqrt{a} ((a-b) \cot (e+f x)+a (e+f x))}{a^{3/2} f (a-b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2),x]

[Out]

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

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Maple [A]  time = 0.067, size = 73, normalized size = 1.1 \begin{align*} -{\frac{1}{fa\tan \left ( fx+e \right ) }}+{\frac{{b}^{2}}{fa \left ( a-b \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2/(a+b*tan(f*x+e)^2),x)

[Out]

-1/f/a/tan(f*x+e)+1/f/a*b^2/(a-b)/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/f/(a-b)*arctan(tan(f*x+e))

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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(cot(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.17911, size = 568, normalized size = 8.88 \begin{align*} \left [-\frac{4 \, a f x \tan \left (f x + e\right ) + b \sqrt{-\frac{b}{a}} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \,{\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt{-\frac{b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) \tan \left (f x + e\right ) + 4 \, a - 4 \, b}{4 \,{\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}, -\frac{2 \, a f x \tan \left (f x + e\right ) - b \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt{\frac{b}{a}}}{2 \, b \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right ) + 2 \, a - 2 \, b}{2 \,{\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/4*(4*a*f*x*tan(f*x + e) + b*sqrt(-b/a)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 - 4*(a*b*tan(f
*x + e)^3 - a^2*tan(f*x + e))*sqrt(-b/a))/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2))*tan(f*x + e) + 4*
a - 4*b)/((a^2 - a*b)*f*tan(f*x + e)), -1/2*(2*a*f*x*tan(f*x + e) - b*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 -
a)*sqrt(b/a)/(b*tan(f*x + e)))*tan(f*x + e) + 2*a - 2*b)/((a^2 - a*b)*f*tan(f*x + e))]

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Sympy [A]  time = 22.1719, size = 570, normalized size = 8.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2/(a+b*tan(f*x+e)**2),x)

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(e, 0) & Eq(f, 0)), ((x + 1/(f*tan(e + f*x)) - 1/(3*f*tan(e + f*x)**
3))/b, Eq(a, 0)), (-3*f*x*tan(e + f*x)**3/(2*b*f*tan(e + f*x)**3 + 2*b*f*tan(e + f*x)) - 3*f*x*tan(e + f*x)/(2
*b*f*tan(e + f*x)**3 + 2*b*f*tan(e + f*x)) - 3*tan(e + f*x)**2/(2*b*f*tan(e + f*x)**3 + 2*b*f*tan(e + f*x)) -
2/(2*b*f*tan(e + f*x)**3 + 2*b*f*tan(e + f*x)), Eq(a, b)), (zoo*x/a, Eq(e, -f*x)), (x*cot(e)**2/(a + b*tan(e)*
*2), Eq(f, 0)), ((-x - cot(e + f*x)/f)/a, Eq(b, 0)), (-2*I*a**(3/2)*f*x*sqrt(1/b)*tan(e + f*x)/(2*I*a**(5/2)*f
*sqrt(1/b)*tan(e + f*x) - 2*I*a**(3/2)*b*f*sqrt(1/b)*tan(e + f*x)) - 2*I*a**(3/2)*sqrt(1/b)/(2*I*a**(5/2)*f*sq
rt(1/b)*tan(e + f*x) - 2*I*a**(3/2)*b*f*sqrt(1/b)*tan(e + f*x)) + 2*I*sqrt(a)*b*sqrt(1/b)/(2*I*a**(5/2)*f*sqrt
(1/b)*tan(e + f*x) - 2*I*a**(3/2)*b*f*sqrt(1/b)*tan(e + f*x)) + b*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan
(e + f*x)/(2*I*a**(5/2)*f*sqrt(1/b)*tan(e + f*x) - 2*I*a**(3/2)*b*f*sqrt(1/b)*tan(e + f*x)) - b*log(I*sqrt(a)*
sqrt(1/b) + tan(e + f*x))*tan(e + f*x)/(2*I*a**(5/2)*f*sqrt(1/b)*tan(e + f*x) - 2*I*a**(3/2)*b*f*sqrt(1/b)*tan
(e + f*x)), True))

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Giac [B]  time = 1.44382, size = 390, normalized size = 6.09 \begin{align*} \frac{\frac{{\left (a^{2} b + a b^{2} - b{\left | -a^{2} + a b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \tan \left (f x + e\right )}{\sqrt{\frac{2 \, a^{2} + 2 \, a b + \sqrt{-16 \, a^{3} b + 4 \,{\left (a^{2} + a b\right )}^{2}}}{a b}}}\right )\right )}}{a^{2}{\left | -a^{2} + a b \right |} + a b{\left | -a^{2} + a b \right |} +{\left (a^{2} - a b\right )}^{2}} + \frac{{\left ({\left (a^{2} + a b\right )} \sqrt{a b}{\left | b \right |} + \sqrt{a b}{\left | -a^{2} + a b \right |}{\left | b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \tan \left (f x + e\right )}{\sqrt{\frac{2 \, a^{2} + 2 \, a b - \sqrt{-16 \, a^{3} b + 4 \,{\left (a^{2} + a b\right )}^{2}}}{a b}}}\right )\right )}}{{\left (a^{2} - a b\right )}^{2} b -{\left (a^{2} b + a b^{2}\right )}{\left | -a^{2} + a b \right |}} - \frac{1}{a \tan \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

((a^2*b + a*b^2 - b*abs(-a^2 + a*b))*(pi*floor((f*x + e)/pi + 1/2) + arctan(2*tan(f*x + e)/sqrt((2*a^2 + 2*a*b
+ sqrt(-16*a^3*b + 4*(a^2 + a*b)^2))/(a*b))))/(a^2*abs(-a^2 + a*b) + a*b*abs(-a^2 + a*b) + (a^2 - a*b)^2) + (
(a^2 + a*b)*sqrt(a*b)*abs(b) + sqrt(a*b)*abs(-a^2 + a*b)*abs(b))*(pi*floor((f*x + e)/pi + 1/2) + arctan(2*tan(
f*x + e)/sqrt((2*a^2 + 2*a*b - sqrt(-16*a^3*b + 4*(a^2 + a*b)^2))/(a*b))))/((a^2 - a*b)^2*b - (a^2*b + a*b^2)*
abs(-a^2 + a*b)) - 1/(a*tan(f*x + e)))/f