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

Optimal. Leaf size=121 \[ \frac{b^{3/2} (5 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 f (a+b)^{5/2}}-\frac{x}{a^2}-\frac{(2 a-b) \cot (e+f x)}{2 a f (a+b)^2}-\frac{b \cot (e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

-(x/a^2) + (b^(3/2)*(5*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(5/2)*f) - ((2*a -
b)*Cot[e + f*x])/(2*a*(a + b)^2*f) - (b*Cot[e + f*x])/(2*a*(a + b)*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.253772, antiderivative size = 121, 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 = {4141, 1975, 472, 583, 522, 203, 205} \[ \frac{b^{3/2} (5 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 f (a+b)^{5/2}}-\frac{x}{a^2}-\frac{(2 a-b) \cot (e+f x)}{2 a f (a+b)^2}-\frac{b \cot (e+f x)}{2 a f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) + (b^(3/2)*(5*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(5/2)*f) - ((2*a -
b)*Cot[e + f*x])/(2*a*(a + b)^2*f) - (b*Cot[e + f*x])/(2*a*(a + b)*f*(a + b + b*Tan[e + f*x]^2))

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

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 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 \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot (e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a-b-3 b x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a+b) f}\\ &=-\frac{(2 a-b) \cot (e+f x)}{2 a (a+b)^2 f}-\frac{b \cot (e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{2 a^2+6 a b+b^2+(2 a-b) b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a+b)^2 f}\\ &=-\frac{(2 a-b) \cot (e+f x)}{2 a (a+b)^2 f}-\frac{b \cot (e+f x)}{2 a (a+b) f \left (a+b+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^2 f}+\frac{\left (b^2 (5 a+2 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 (a+b)^2 f}\\ &=-\frac{x}{a^2}+\frac{b^{3/2} (5 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 (a+b)^{5/2} f}-\frac{(2 a-b) \cot (e+f x)}{2 a (a+b)^2 f}-\frac{b \cot (e+f x)}{2 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.92856, size = 288, normalized size = 2.38 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{b^2 (a \sin (2 f x)-(a+2 b) \sin (2 e))}{a^2 f (a+b)^2 (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac{b^2 (5 a+2 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{a^2 f (a+b)^{5/2} \sqrt{b (\cos (e)-i \sin (e))^4}}-\frac{2 x (a \cos (2 (e+f x))+a+2 b)}{a^2}+\frac{2 \csc (e) \sin (f x) \csc (e+f x) (a \cos (2 (e+f x))+a+2 b)}{f (a+b)^2}\right )}{8 \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((-2*x*(a + 2*b + a*Cos[2*(e + f*x)]))/a^2 - (b^2*(5*a + 2*b)*A
rcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos
[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/(a^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[
e] - I*Sin[e])^4]) + (2*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[e]*Csc[e + f*x]*Sin[f*x])/((a + b)^2*f) + (b^2*(-((
a + 2*b)*Sin[2*e]) + a*Sin[2*f*x]))/(a^2*(a + b)^2*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8*(a + b*Sec[e +
f*x]^2)^2)

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Maple [A]  time = 0.104, size = 149, normalized size = 1.2 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{2}}}-{\frac{1}{f \left ( a+b \right ) ^{2}\tan \left ( fx+e \right ) }}+{\frac{{b}^{2}\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{2}a \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{5\,{b}^{2}}{2\,f \left ( a+b \right ) ^{2}a}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{{b}^{3}}{f \left ( a+b \right ) ^{2}{a}^{2}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/a^2*arctan(tan(f*x+e))-1/f/(a+b)^2/tan(f*x+e)+1/2/f*b^2/(a+b)^2/a*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+5/2/f*b
^2/(a+b)^2/a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/f*b^3/(a+b)^2/a^2/((a+b)*b)^(1/2)*arctan(t
an(f*x+e)*b/((a+b)*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(cot(f*x+e)^2/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.724258, size = 1384, normalized size = 11.44 \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*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/8*(4*(2*a^3 + a*b^2)*cos(f*x + e)^3 - (5*a*b^2 + 2*b^3 + (5*a^2*b + 2*a*b^2)*cos(f*x + e)^2)*sqrt(-b/(a +
b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*co
s(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(
f*x + e)^2 + b^2))*sin(f*x + e) + 4*(2*a^2*b - a*b^2)*cos(f*x + e) + 8*((a^3 + 2*a^2*b + a*b^2)*f*x*cos(f*x +
e)^2 + (a^2*b + 2*a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^5 + 2*a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^
3*b^2 + a^2*b^3)*f)*sin(f*x + e)), -1/4*(2*(2*a^3 + a*b^2)*cos(f*x + e)^3 + (5*a*b^2 + 2*b^3 + (5*a^2*b + 2*a*
b^2)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)
*sin(f*x + e)))*sin(f*x + e) + 2*(2*a^2*b - a*b^2)*cos(f*x + e) + 4*((a^3 + 2*a^2*b + a*b^2)*f*x*cos(f*x + e)^
2 + (a^2*b + 2*a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^5 + 2*a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^4*b + 2*a^3*b
^2 + a^2*b^3)*f)*sin(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*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.37533, size = 247, normalized size = 2.04 \begin{align*} \frac{\frac{{\left (5 \, a b^{2} + 2 \, b^{3}\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 + b^{2}}}\right )\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt{a b + b^{2}}} - \frac{2 \, a b \tan \left (f x + e\right )^{2} - b^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{2} + 2 \, a b}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )\right )}{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}} - \frac{2 \,{\left (f x + e\right )}}{a^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2 + 2*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^4 +
2*a^3*b + a^2*b^2)*sqrt(a*b + b^2)) - (2*a*b*tan(f*x + e)^2 - b^2*tan(f*x + e)^2 + 2*a^2 + 2*a*b)/((b*tan(f*x
+ e)^3 + a*tan(f*x + e) + b*tan(f*x + e))*(a^3 + 2*a^2*b + a*b^2)) - 2*(f*x + e)/a^2)/f