3.297 \(\int \frac{\cot ^2(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{a \cot (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \csc (c+d x)}{d \left (a^2-b^2\right )}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a^2-b^2} \tan \left (\frac{1}{2} (c+d x)\right )}{a+b}\right )}{a d \left (a^2-b^2\right )^{3/2}}-\frac{x}{a} \]

[Out]

-(x/a) - (2*b^3*ArcTanh[(Sqrt[a^2 - b^2]*Tan[(c + d*x)/2])/(a + b)])/(a*(a^2 - b^2)^(3/2)*d) - (a*Cot[c + d*x]
)/((a^2 - b^2)*d) + (b*Csc[c + d*x])/((a^2 - b^2)*d)

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Rubi [A]  time = 0.243004, antiderivative size = 135, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3898, 2902, 2606, 8, 3473, 2735, 2659, 208} \[ -\frac{a \cot (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \csc (c+d x)}{d \left (a^2-b^2\right )}+\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{a x}{a^2-b^2}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x)/(a^2 - b^2)) + (b^2*x)/(a*(a^2 - b^2)) - (2*b^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(
a*(a - b)^(3/2)*(a + b)^(3/2)*d) - (a*Cot[c + d*x])/((a^2 - b^2)*d) + (b*Csc[c + d*x])/((a^2 - b^2)*d)

Rule 3898

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x)}{a+b \sec (c+d x)} \, dx &=\int \frac{\cos (c+d x) \cot ^2(c+d x)}{b+a \cos (c+d x)} \, dx\\ &=\frac{a \int \cot ^2(c+d x) \, dx}{a^2-b^2}-\frac{b \int \cot (c+d x) \csc (c+d x) \, dx}{a^2-b^2}+\frac{b^2 \int \frac{\cos (c+d x)}{b+a \cos (c+d x)} \, dx}{a^2-b^2}\\ &=\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{a \cot (c+d x)}{\left (a^2-b^2\right ) d}-\frac{a \int 1 \, dx}{a^2-b^2}-\frac{b^3 \int \frac{1}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}+\frac{b \operatorname{Subst}(\int 1 \, dx,x,\csc (c+d x))}{\left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{a \cot (c+d x)}{\left (a^2-b^2\right ) d}+\frac{b \csc (c+d x)}{\left (a^2-b^2\right ) d}-\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a (a-b)^{3/2} (a+b)^{3/2} d}-\frac{a \cot (c+d x)}{\left (a^2-b^2\right ) d}+\frac{b \csc (c+d x)}{\left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.409352, size = 147, normalized size = 1.39 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{a^2-b^2} \left (\left (a^2-b^2\right ) (c+d x) \sin (c+d x)+a^2 \cos (c+d x)-a b\right )-2 b^3 \sin (c+d x) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )\right )}{2 a d (a-b) (a+b) \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-2*b^3*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*Sin[c + d*x]
+ Sqrt[a^2 - b^2]*(-(a*b) + a^2*Cos[c + d*x] + (a^2 - b^2)*(c + d*x)*Sin[c + d*x])))/(2*a*(a - b)*(a + b)*Sqrt
[a^2 - b^2]*d)

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Maple [A]  time = 0.074, size = 123, normalized size = 1.2 \begin{align*}{\frac{1}{2\,d \left ( a-b \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}-2\,{\frac{{b}^{3}}{d \left ( a-b \right ) \left ( a+b \right ) a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{2\,d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(a+b*sec(d*x+c)),x)

[Out]

1/2/d/(a-b)*tan(1/2*d*x+1/2*c)-2/d/a*arctan(tan(1/2*d*x+1/2*c))-2/d/(a-b)/(a+b)*b^3/a/((a+b)*(a-b))^(1/2)*arct
anh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/2/d/(a+b)/tan(1/2*d*x+1/2*c)

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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(d*x+c)^2/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.969172, size = 815, normalized size = 7.69 \begin{align*} \left [-\frac{\sqrt{a^{2} - b^{2}} b^{3} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, a^{3} b + 2 \, a b^{3} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \sin \left (d x + c\right ) + 2 \,{\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )}{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sin \left (d x + c\right )}, -\frac{\sqrt{-a^{2} + b^{2}} b^{3} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - a^{3} b + a b^{3} +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \sin \left (d x + c\right ) +{\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(a^2 - b^2)*b^3*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d
*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 2*a^3
*b + 2*a*b^3 + 2*(a^4 - 2*a^2*b^2 + b^4)*d*x*sin(d*x + c) + 2*(a^4 - a^2*b^2)*cos(d*x + c))/((a^5 - 2*a^3*b^2
+ a*b^4)*d*sin(d*x + c)), -(sqrt(-a^2 + b^2)*b^3*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*si
n(d*x + c)))*sin(d*x + c) - a^3*b + a*b^3 + (a^4 - 2*a^2*b^2 + b^4)*d*x*sin(d*x + c) + (a^4 - a^2*b^2)*cos(d*x
+ c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\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)**2/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**2/(a + b*sec(c + d*x)), x)

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Giac [A]  time = 1.34272, size = 190, normalized size = 1.79 \begin{align*} -\frac{\frac{4 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} - a b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (d x + c\right )}}{a} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a - b} + \frac{1}{{\left (a + b\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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