3.291 \(\int \frac{\cot (c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac{b^2 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )}+\frac{\log (1-\sec (c+d x))}{2 d (a+b)}+\frac{\log (\sec (c+d x)+1)}{2 d (a-b)}+\frac{\log (\cos (c+d x))}{a d} \]

[Out]

Log[Cos[c + d*x]]/(a*d) + Log[1 - Sec[c + d*x]]/(2*(a + b)*d) + Log[1 + Sec[c + d*x]]/(2*(a - b)*d) - (b^2*Log
[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)*d)

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Rubi [A]  time = 0.102287, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3885, 894} \[ -\frac{b^2 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )}+\frac{\log (1-\sec (c+d x))}{2 d (a+b)}+\frac{\log (\sec (c+d x)+1)}{2 d (a-b)}+\frac{\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + b*Sec[c + d*x]),x]

[Out]

Log[Cos[c + d*x]]/(a*d) + Log[1 - Sec[c + d*x]]/(2*(a + b)*d) + Log[1 + Sec[c + d*x]]/(2*(a - b)*d) - (b^2*Log
[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)*d)

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.105076, size = 70, normalized size = 0.74 \[ \frac{b^2 (-\log (a \cos (c+d x)+b))+a (a-b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+a (a+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + b*Sec[c + d*x]),x]

[Out]

(a*(a + b)*Log[Cos[(c + d*x)/2]] - b^2*Log[b + a*Cos[c + d*x]] + a*(a - b)*Log[Sin[(c + d*x)/2]])/(a*(a - b)*(
a + b)*d)

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Maple [A]  time = 0.058, size = 80, normalized size = 0.9 \begin{align*} -{\frac{{b}^{2}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ) a}}+{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{d \left ( 2\,a+2\,b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+b*sec(d*x+c)),x)

[Out]

-1/d*b^2/(a+b)/(a-b)/a*ln(b+a*cos(d*x+c))+1/d/(2*a-2*b)*ln(cos(d*x+c)+1)+1/d/(2*a+2*b)*ln(-1+cos(d*x+c))

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Maxima [A]  time = 0.956272, size = 92, normalized size = 0.98 \begin{align*} -\frac{\frac{2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{3} - a b^{2}} - \frac{\log \left (\cos \left (d x + c\right ) + 1\right )}{a - b} - \frac{\log \left (\cos \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(2*b^2*log(a*cos(d*x + c) + b)/(a^3 - a*b^2) - log(cos(d*x + c) + 1)/(a - b) - log(cos(d*x + c) - 1)/(a +
b))/d

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Fricas [A]  time = 1.07341, size = 190, normalized size = 2.02 \begin{align*} -\frac{2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (a^{2} + a b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a^{2} - a b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*log(a*cos(d*x + c) + b) - (a^2 + a*b)*log(1/2*cos(d*x + c) + 1/2) - (a^2 - a*b)*log(-1/2*cos(d*x +
c) + 1/2))/((a^3 - a*b^2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)/(a + b*sec(c + d*x)), x)

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Giac [A]  time = 1.29026, size = 180, normalized size = 1.91 \begin{align*} -\frac{\frac{2 \, b^{2} \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{3} - a b^{2}} - \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} + \frac{2 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(2*b^2*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
))/(a^3 - a*b^2) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) + 2*log(abs(-(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) + 1))/a)/d