\(\int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [347]

Optimal result

Integrand size = 23, antiderivative size = 62 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {x}{a}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a (a+b)^{3/2} f}-\frac {\cot (e+f x)}{(a+b) f} \]

[Out]

-x/a+b^(3/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a/(a+b)^(3/2)/f-cot(f*x+e)/(a+b)/f

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4226, 2000, 491, 536, 209, 211} \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a f (a+b)^{3/2}}-\frac {\cot (e+f x)}{f (a+b)}-\frac {x}{a} \]

[In]

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

[Out]

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

Rule 209

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

Rule 211

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

Rule 491

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 536

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 2000

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 4226

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x)}{(a+b) f}+\frac {\text {Subst}\left (\int \frac {-a-2 b-b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{(a+b) f} \\ & = -\frac {\cot (e+f x)}{(a+b) f}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a f}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a (a+b) f} \\ & = -\frac {x}{a}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a (a+b)^{3/2} f}-\frac {\cot (e+f x)}{(a+b) f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.25 (sec) , antiderivative size = 204, normalized size of antiderivative = 3.29 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (b^2 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4} ((a+b) f x-a \csc (e) \csc (e+f x) \sin (f x))\right )}{2 a (a+b)^{3/2} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]

[In]

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

[Out]

-1/2*((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(b^2*ArcTan[(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])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[
a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]*((a + b)*f*x - a*Csc[e]*Csc[e + f*x]*Sin[f*x])))/(a*(a + b)^(3/2)*f*(a +
b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right ) a \sqrt {\left (a +b \right ) b}}-\frac {1}{\left (a +b \right ) \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\)	\(68\)
default	\(\frac {\frac {b^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right ) a \sqrt {\left (a +b \right ) b}}-\frac {1}{\left (a +b \right ) \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\)	\(68\)
risch	\(-\frac {x}{a}-\frac {2 i}{f \left (a +b \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}-\frac {\sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{2} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{2} f a}\)	\(141\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (54) = 108\).

Time = 0.28 (sec) , antiderivative size = 310, normalized size of antiderivative = 5.00 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {4 \, {\left (a + b\right )} f x \sin \left (f x + e\right ) - b \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 4 \, a \cos \left (f x + e\right )}{4 \, {\left (a^{2} + a b\right )} f \sin \left (f x + e\right )}, -\frac {2 \, {\left (a + b\right )} f x \sin \left (f x + e\right ) + b \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, a \cos \left (f x + e\right )}{2 \, {\left (a^{2} + a b\right )} f \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[-1/4*(4*(a + b)*f*x*sin(f*x + e) - b*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)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*s
in(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*a*cos(f*x + e))/((a^2 +
a*b)*f*sin(f*x + e)), -1/2*(2*(a + b)*f*x*sin(f*x + e) + b*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*a*cos(f*x + e))/((a^2 + a*b)*f*sin(f*x
+ e))]

Sympy [F]

\[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{2}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {b^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} + a b\right )} \sqrt {{\left (a + b\right )} b}} - \frac {f x + e}{a} - \frac {1}{{\left (a + b\right )} \tan \left (f x + e\right )}}{f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.40 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} + a b\right )} \sqrt {a b + b^{2}}} - \frac {f x + e}{a} - \frac {1}{{\left (a + b\right )} \tan \left (f x + e\right )}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 21.80 (sec) , antiderivative size = 637, normalized size of antiderivative = 10.27 \[ \int \frac {\cot ^2(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a\,b^2+2\,a^2\,b+a^3+a^3\,\mathrm {tan}\left (e+f\,x\right )\,\mathrm {atan}\left (\mathrm {tan}\left (e+f\,x\right )\right )+b^3\,\mathrm {tan}\left (e+f\,x\right )\,\mathrm {atan}\left (\mathrm {tan}\left (e+f\,x\right )\right )+3\,a\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,\mathrm {atan}\left (\mathrm {tan}\left (e+f\,x\right )\right )+3\,a^2\,b\,\mathrm {tan}\left (e+f\,x\right )\,\mathrm {atan}\left (\mathrm {tan}\left (e+f\,x\right )\right )-\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6\right )}^{3/2}\,1{}\mathrm {i}+b\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6\right )}^{3/2}\,2{}\mathrm {i}+b^7\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,2{}\mathrm {i}+a\,b^6\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,10{}\mathrm {i}+a^6\,b\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,1{}\mathrm {i}+a^2\,b^5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,21{}\mathrm {i}+a^3\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,24{}\mathrm {i}+a^4\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,16{}\mathrm {i}+a^5\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,6{}\mathrm {i}}{a^8\,b^2+8\,a^7\,b^3+28\,a^6\,b^4+55\,a^5\,b^5+65\,a^4\,b^6+46\,a^3\,b^7+18\,a^2\,b^8+3\,a\,b^9}\right )\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}\,1{}\mathrm {i}}{f\,\mathrm {tan}\left (e+f\,x\right )\,a^4+3\,f\,\mathrm {tan}\left (e+f\,x\right )\,a^3\,b+3\,f\,\mathrm {tan}\left (e+f\,x\right )\,a^2\,b^2+f\,\mathrm {tan}\left (e+f\,x\right )\,a\,b^3} \]

[In]

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

[Out]

-(a*b^2 + 2*a^2*b + a^3 - atan((a*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(3/2)*1i + b*tan(e + f*
x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(3/2)*2i + b^7*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)
^(1/2)*2i + a*b^6*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*10i + a^6*b*tan(e + f*x)*(- 3*a*b
^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*1i + a^2*b^5*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*
21i + a^3*b^4*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*24i + a^4*b^3*tan(e + f*x)*(- 3*a*b^5
- b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*16i + a^5*b^2*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*6
i)/(3*a*b^9 + 18*a^2*b^8 + 46*a^3*b^7 + 65*a^4*b^6 + 55*a^5*b^5 + 28*a^6*b^4 + 8*a^7*b^3 + a^8*b^2))*tan(e + f
*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^3*b^3)^(1/2)*1i + a^3*tan(e + f*x)*atan(tan(e + f*x)) + b^3*tan(e + f*x)*
atan(tan(e + f*x)) + 3*a*b^2*tan(e + f*x)*atan(tan(e + f*x)) + 3*a^2*b*tan(e + f*x)*atan(tan(e + f*x)))/(a^4*f
*tan(e + f*x) + a*b^3*f*tan(e + f*x) + 3*a^3*b*f*tan(e + f*x) + 3*a^2*b^2*f*tan(e + f*x))