3.247 \(\int \frac{\cot ^2(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=189 \[ \frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac{b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]

[Out]

-(x/(a - b)^3) + (b^(3/2)*(35*a^2 - 42*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(7/2)*(a - b
)^3*f) - ((8*a^2 - 27*a*b + 15*b^2)*Cot[e + f*x])/(8*a^3*(a - b)^2*f) - (b*Cot[e + f*x])/(4*a*(a - b)*f*(a + b
*Tan[e + f*x]^2)^2) - ((9*a - 5*b)*b*Cot[e + f*x])/(8*a^2*(a - b)^2*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.290321, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3670, 472, 579, 583, 522, 203, 205} \[ \frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac{b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac{b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-(x/(a - b)^3) + (b^(3/2)*(35*a^2 - 42*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(7/2)*(a - b
)^3*f) - ((8*a^2 - 27*a*b + 15*b^2)*Cot[e + f*x])/(8*a^3*(a - b)^2*f) - (b*Cot[e + f*x])/(4*a*(a - b)*f*(a + b
*Tan[e + f*x]^2)^2) - ((9*a - 5*b)*b*Cot[e + f*x])/(8*a^2*(a - b)^2*f*(a + b*Tan[e + f*x]^2))

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 472

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

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{4 a-5 b-5 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a-b) f}\\ &=-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2-27 a b+15 b^2-3 (9 a-5 b) b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a-b)^2 f}\\ &=-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{8 a^3+8 a^2 b-27 a b^2+15 b^3+b \left (8 a^2-27 a b+15 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^2 f}\\ &=-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}+\frac{\left (b^2 \left (35 a^2-42 a b+15 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^3 f}\\ &=-\frac{x}{(a-b)^3}+\frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} (a-b)^3 f}-\frac{\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac{b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.01344, size = 174, normalized size = 0.92 \[ \frac{\frac{b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} (a-b)^3}-\frac{4 b^3 \sin (2 (e+f x))}{a^2 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}+\frac{b^2 (13 a-7 b) \sin (2 (e+f x))}{a^3 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)}-\frac{8 \cot (e+f x)}{a^3}+\frac{8 (e+f x)}{(b-a)^3}}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

((8*(e + f*x))/(-a + b)^3 + (b^(3/2)*(35*a^2 - 42*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(7/
2)*(a - b)^3) - (8*Cot[e + f*x])/a^3 - (4*b^3*Sin[2*(e + f*x)])/(a^2*(a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x
)])^2) + ((13*a - 7*b)*b^2*Sin[2*(e + f*x)])/(a^3*(a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])))/(8*f)

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Maple [B]  time = 0.09, size = 379, normalized size = 2. \begin{align*} -{\frac{1}{f{a}^{3}\tan \left ( fx+e \right ) }}+{\frac{11\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}-{\frac{9\,{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}{a}^{2}}}+{\frac{7\,{b}^{5} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{13\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{11\,{b}^{3}\tan \left ( fx+e \right ) }{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}+{\frac{9\,{b}^{4}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}{a}^{2}}}+{\frac{35\,{b}^{2}}{8\,f \left ( a-b \right ) ^{3}a}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{21\,{b}^{3}}{4\,f{a}^{2} \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{15\,{b}^{4}}{8\,f{a}^{3} \left ( a-b \right ) ^{3}}\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 ) ^{3}}} \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)^3,x)

[Out]

-1/f/a^3/tan(f*x+e)+11/8/f*b^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2/a*tan(f*x+e)^3-9/4/f*b^4/(a-b)^3/(a+b*tan(f*x+e)^2
)^2/a^2*tan(f*x+e)^3+7/8/f*b^5/a^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3+13/8/f*b^2/(a-b)^3/(a+b*tan(f*x+e
)^2)^2*tan(f*x+e)-11/4/f*b^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2/a*tan(f*x+e)+9/8/f*b^4/a^2/(a-b)^3/(a+b*tan(f*x+e)^2
)^2*tan(f*x+e)+35/8/f*b^2/(a-b)^3/a/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-21/4/f*b^3/(a-b)^3/a^2/(a*b)^
(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+15/8/f*b^4/a^3/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/f
/(a-b)^3*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.93673, size = 1989, normalized size = 10.52 \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)^3,x, algorithm="fricas")

[Out]

[-1/32*(32*a^3*b^2*f*x*tan(f*x + e)^5 + 64*a^4*b*f*x*tan(f*x + e)^3 + 32*a^5*f*x*tan(f*x + e) + 32*a^5 - 96*a^
4*b + 96*a^3*b^2 - 32*a^2*b^3 + 4*(8*a^3*b^2 - 35*a^2*b^3 + 42*a*b^4 - 15*b^5)*tan(f*x + e)^4 + 4*(16*a^4*b -
61*a^3*b^2 + 70*a^2*b^3 - 25*a*b^4)*tan(f*x + e)^2 + ((35*a^2*b^3 - 42*a*b^4 + 15*b^5)*tan(f*x + e)^5 + 2*(35*
a^3*b^2 - 42*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^3 + (35*a^4*b - 42*a^3*b^2 + 15*a^2*b^3)*tan(f*x + e))*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)))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x +
e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f*
tan(f*x + e)), -1/16*(16*a^3*b^2*f*x*tan(f*x + e)^5 + 32*a^4*b*f*x*tan(f*x + e)^3 + 16*a^5*f*x*tan(f*x + e) +
16*a^5 - 48*a^4*b + 48*a^3*b^2 - 16*a^2*b^3 + 2*(8*a^3*b^2 - 35*a^2*b^3 + 42*a*b^4 - 15*b^5)*tan(f*x + e)^4 +
2*(16*a^4*b - 61*a^3*b^2 + 70*a^2*b^3 - 25*a*b^4)*tan(f*x + e)^2 - ((35*a^2*b^3 - 42*a*b^4 + 15*b^5)*tan(f*x +
e)^5 + 2*(35*a^3*b^2 - 42*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^3 + (35*a^4*b - 42*a^3*b^2 + 15*a^2*b^3)*tan(f*x +
e))*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(b/a)/(b*tan(f*x + e))))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^
4 - a^3*b^5)*f*tan(f*x + e)^5 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^3 + (a^8 - 3*a^7*b
+ 3*a^6*b^2 - a^5*b^3)*f*tan(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.6405, size = 312, normalized size = 1.65 \begin{align*} \frac{\frac{{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt{a b}} - \frac{8 \,{\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{11 \, a b^{3} \tan \left (f x + e\right )^{3} - 7 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) - 9 \, a b^{3} \tan \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}} - \frac{8}{a^{3} \tan \left (f x + e\right )}}{8 \, 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)^3,x, algorithm="giac")

[Out]

1/8*((35*a^2*b^2 - 42*a*b^3 + 15*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))
/((a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*sqrt(a*b)) - 8*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (11*a*b^3*t
an(f*x + e)^3 - 7*b^4*tan(f*x + e)^3 + 13*a^2*b^2*tan(f*x + e) - 9*a*b^3*tan(f*x + e))/((a^5 - 2*a^4*b + a^3*b
^2)*(b*tan(f*x + e)^2 + a)^2) - 8/(a^3*tan(f*x + e)))/f