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

Optimal. Leaf size=46 \[ \frac{\log (\sin (e+f x))}{f (a+b)}+\frac{b \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)} \]

[Out]

(b*Log[b + a*Cos[e + f*x]^2])/(2*a*(a + b)*f) + Log[Sin[e + f*x]]/((a + b)*f)

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Rubi [A]  time = 0.0803005, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4138, 446, 72} \[ \frac{\log (\sin (e+f x))}{f (a+b)}+\frac{b \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Log[b + a*Cos[e + f*x]^2])/(2*a*(a + b)*f) + Log[Sin[e + f*x]]/((a + b)*f)

Rule 4138

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

Rule 446

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x) (b+a x)} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(-a-b) (-1+x)}-\frac{b}{(a+b) (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{b \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b) f}+\frac{\log (\sin (e+f x))}{(a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.105223, size = 43, normalized size = 0.93 \[ \frac{b \log \left (-a \sin ^2(e+f x)+a+b\right )+2 a \log (\sin (e+f x))}{2 a^2 f+2 a b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Log[Sin[e + f*x]] + b*Log[a + b - a*Sin[e + f*x]^2])/(2*a^2*f + 2*a*b*f)

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Maple [A]  time = 0.078, size = 73, normalized size = 1.6 \begin{align*}{\frac{b\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\, \left ( a+b \right ) af}}+{\frac{\ln \left ( 1+\cos \left ( fx+e \right ) \right ) }{f \left ( 2\,a+2\,b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( fx+e \right ) \right ) }{f \left ( 2\,a+2\,b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*ln(b+a*cos(f*x+e)^2)/a/(a+b)/f+1/f/(2*a+2*b)*ln(1+cos(f*x+e))+1/f/(2*a+2*b)*ln(-1+cos(f*x+e))

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Maxima [A]  time = 0.980614, size = 68, normalized size = 1.48 \begin{align*} \frac{\frac{b \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{2} + a b} + \frac{\log \left (\sin \left (f x + e\right )^{2}\right )}{a + b}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*log(a*sin(f*x + e)^2 - a - b)/(a^2 + a*b) + log(sin(f*x + e)^2)/(a + b))/f

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Fricas [A]  time = 0.663496, size = 107, normalized size = 2.33 \begin{align*} \frac{b \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 2 \, a \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{2 \,{\left (a^{2} + a b\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*log(a*cos(f*x + e)^2 + b) + 2*a*log(1/2*sin(f*x + e)))/((a^2 + a*b)*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.36048, size = 235, normalized size = 5.11 \begin{align*} \frac{\frac{b \log \left (a + b + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{2 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{2} + a b} - \frac{2 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} + \frac{\log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{a + b}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(
cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^2 + a*b) - 2*log(-(
cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a + log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/(a + b))/f