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((a^2-b^2)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b))/a/(a^2-b^2)^(3/2)/d-a*cot(d*x+c)/(a^2-b^2)/d+b*cs
c(d*x+c)/(a^2-b^2)/d

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Rubi [A]  time = 0.24, 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 8

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

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]

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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.42, 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]

-1/2*(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])))/(a*(a - b)*(a + b)*Sq
rt[a^2 - b^2]*d)

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fricas [A]  time = 0.51, size = 362, normalized size = 3.42 \[ \left [-\frac {\sqrt {a^{2} - b^{2}} b^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, a^{3} b + 2 \, a b^{3} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \sin \left (d x + c\right ) + 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sin \left (d x + c\right )}, -\frac {\sqrt {-a^{2} + b^{2}} b^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - a^{3} b + a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \sin \left (d x + c\right ) + {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \sin \left (d x + c\right )}\right ] \]

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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giac [B]  time = 0.45, size = 582, normalized size = 5.49 \[ -\frac {\frac {2 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4} - 2 \, b^{5} - a^{2} {\left | -a^{3} + a b^{2} \right |} + a b {\left | -a^{3} + a b^{2} \right |} + b^{2} {\left | -a^{3} + a b^{2} \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a^{2} b - b^{3} + \sqrt {{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} + {\left (a^{2} b - b^{3}\right )}^{2}}}{a^{3} - a^{2} b - a b^{2} + b^{3}}}}\right )\right )}}{a^{2} b {\left | -a^{3} + a b^{2} \right |} - b^{3} {\left | -a^{3} + a b^{2} \right |} + {\left (a^{3} - a b^{2}\right )}^{2}} + \frac {2 \, {\left ({\left (a^{2} - a b - b^{2}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a^{3} + a b^{2} \right |} {\left | -a + b \right |} + {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4} - 2 \, b^{5}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a^{2} b - b^{3} - \sqrt {{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} + {\left (a^{2} b - b^{3}\right )}^{2}}}{a^{3} - a^{2} b - a b^{2} + b^{3}}}}\right )\right )}}{{\left (a^{3} - a b^{2}\right )}^{2} {\left (a^{2} - 2 \, a b + b^{2}\right )} - {\left (a^{4} b - 2 \, a^{3} b^{2} + 2 \, a b^{4} - b^{5}\right )} {\left | -a^{3} + a b^{2} \right |}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a - b} + \frac {1}{{\left (a + b\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

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*(2*(a^5 - a^4*b - 2*a^3*b^2 + 3*a^2*b^3 + a*b^4 - 2*b^5 - a^2*abs(-a^3 + a*b^2) + a*b*abs(-a^3 + a*b^2) +
b^2*abs(-a^3 + a*b^2))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^2*b - b^3 + s
qrt((a^3 + a^2*b - a*b^2 - b^3)*(a^3 - a^2*b - a*b^2 + b^3) + (a^2*b - b^3)^2))/(a^3 - a^2*b - a*b^2 + b^3))))
/(a^2*b*abs(-a^3 + a*b^2) - b^3*abs(-a^3 + a*b^2) + (a^3 - a*b^2)^2) + 2*((a^2 - a*b - b^2)*sqrt(-a^2 + b^2)*a
bs(-a^3 + a*b^2)*abs(-a + b) + (a^5 - a^4*b - 2*a^3*b^2 + 3*a^2*b^3 + a*b^4 - 2*b^5)*sqrt(-a^2 + b^2)*abs(-a +
b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^2*b - b^3 - sqrt((a^3 + a^2*b -
a*b^2 - b^3)*(a^3 - a^2*b - a*b^2 + b^3) + (a^2*b - b^3)^2))/(a^3 - a^2*b - a*b^2 + b^3))))/((a^3 - a*b^2)^2*(
a^2 - 2*a*b + b^2) - (a^4*b - 2*a^3*b^2 + 2*a*b^4 - b^5)*abs(-a^3 + a*b^2)) - tan(1/2*d*x + 1/2*c)/(a - b) + 1
/((a + b)*tan(1/2*d*x + 1/2*c)))/d

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maple [A]  time = 0.62, size = 123, normalized size = 1.16 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \left (a -b \right )}-\frac {2 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \left (a -b \right ) \left (a +b \right ) a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 d \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 3.94, size = 1002, normalized size = 9.45 \[ -\frac {1{}\mathrm {i}\,\cos \left (c+d\,x\right )\,a^6-a^5\,b\,1{}\mathrm {i}-2{}\mathrm {i}\,\cos \left (c+d\,x\right )\,a^4\,b^2+a^3\,b^3\,2{}\mathrm {i}+1{}\mathrm {i}\,\cos \left (c+d\,x\right )\,a^2\,b^4-a\,b^5\,1{}\mathrm {i}}{1{}\mathrm {i}\,d\,\sin \left (c+d\,x\right )\,a^7-3{}\mathrm {i}\,d\,\sin \left (c+d\,x\right )\,a^5\,b^2+3{}\mathrm {i}\,d\,\sin \left (c+d\,x\right )\,a^3\,b^4-1{}\mathrm {i}\,d\,\sin \left (c+d\,x\right )\,a\,b^6}+\frac {-a^6\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+b^6\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-a^2\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}+a^4\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}+b^3\,\mathrm {atanh}\left (\frac {2\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}^{3/2}-a^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+2\,b^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-9\,a^2\,b^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+3\,a^3\,b^{10}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+18\,a^4\,b^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-12\,a^5\,b^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-21\,a^6\,b^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+19\,a^7\,b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+15\,a^8\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-15\,a^9\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-6\,a^{10}\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+6\,a^{11}\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+a^{12}\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^{16}-8\,a^{14}\,b^2+28\,a^{12}\,b^4-55\,a^{10}\,b^6+65\,a^8\,b^8-46\,a^6\,b^{10}+18\,a^4\,b^{12}-3\,a^2\,b^{14}\right )}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}\,2{}\mathrm {i}}{1{}\mathrm {i}\,d\,a^7-3{}\mathrm {i}\,d\,a^5\,b^2+3{}\mathrm {i}\,d\,a^3\,b^4-1{}\mathrm {i}\,d\,a\,b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^6*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i - a^6*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*2i - a^2
*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*6i + a^4*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*6i +
b^3*atanh((2*b^7*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - a^13*sin(c/2 + (d*x)/2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 2*b^13*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 9
*a^2*b^11*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 3*a^3*b^10*sin(c/2 + (d*x)/2)*(a^6 -
b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 18*a^4*b^9*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) -
12*a^5*b^8*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 21*a^6*b^7*sin(c/2 + (d*x)/2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 19*a^7*b^6*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2)
+ 15*a^8*b^5*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 15*a^9*b^4*sin(c/2 + (d*x)/2)*(a^
6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 6*a^10*b^3*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/
2) + 6*a^11*b^2*sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + a^12*b*sin(c/2 + (d*x)/2)*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a^16 - 3*a^2*b^14 + 18*a^4*b^12 - 46*a^6*b^10 + 65*
a^8*b^8 - 55*a^10*b^6 + 28*a^12*b^4 - 8*a^14*b^2)))*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2)*2i)/(a^7*d*1i +
a^3*b^4*d*3i - a^5*b^2*d*3i - a*b^6*d*1i) - (a^6*cos(c + d*x)*1i - a*b^5*1i - a^5*b*1i + a^3*b^3*2i + a^2*b^4*
cos(c + d*x)*1i - a^4*b^2*cos(c + d*x)*2i)/(a^7*d*sin(c + d*x)*1i - a*b^6*d*sin(c + d*x)*1i + a^3*b^4*d*sin(c
+ d*x)*3i - a^5*b^2*d*sin(c + d*x)*3i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

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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