3.1081 \(\int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=92 \[ -\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \sqrt {a^2-b^2}}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2889, 3056, 12, 2747, 3770, 2660, 618, 204} \[ -\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \sqrt {a^2-b^2}}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2747

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

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac {\csc (c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx\\ &=\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \frac {\left (a^2-b^2\right ) \csc (c+d x)}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {b \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}\\ &=-\frac {2 b \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 97, normalized size = 1.05 \[ \frac {-\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a \cos (c+d x)}{a+b \sin (c+d x)}+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.02, size = 483, normalized size = 5.25 \[ \left [-\frac {{\left (b^{2} \sin \left (d x + c\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - a^{3} b^{2}\right )} d\right )}}, \frac {2 \, {\left (b^{2} \sin \left (d x + c\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - a^{3} b^{2}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 130, normalized size = 1.41 \[ -\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.68, size = 153, normalized size = 1.66 \[ \frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2}{d a \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(cos(d*x+c)^2*csc(d*x+c)/(a+b*sin(d*x+c))^2,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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 10.16, size = 523, normalized size = 5.68 \[ \frac {a^3\,\cos \left (c+d\,x\right )-a\,b^2-b^3\,\sin \left (c+d\,x\right )+a^3+a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-a\,b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-b^3\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-a\,b^2\,\cos \left (c+d\,x\right )+a^2\,b\,\sin \left (c+d\,x\right )+a^2\,b\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+b^2\,\sin \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}+a\,b\,\mathrm {atan}\left (\frac {-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{a^2\,d\,\left (a^2-b^2\right )\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*cos(c + d*x) - a*b^2 - b^3*sin(c + d*x) + a^3 + a^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - a*b^2*lo
g(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - b^3*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - a*b^2
*cos(c + d*x) + a^2*b*sin(c + d*x) + b^2*sin(c + d*x)*atan((b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^2*
sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + a*b*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*2i)/(a^3*cos(c/2 + (d*x)/2)
- 4*b^3*sin(c/2 + (d*x)/2) - 2*a*b^2*cos(c/2 + (d*x)/2) + 3*a^2*b*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*2i +
a*b*atan((b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + a*b*cos
(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*2i)/(a^3*cos(c/2 + (d*x)/2) - 4*b^3*sin(c/2 + (d*x)/2) - 2*a*b^2*cos(c/2 + (
d*x)/2) + 3*a^2*b*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*2i + a^2*b*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2)))/(a^2*d*(a^2 - b^2)*(a + b*sin(c + d*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)/(a + b*sin(c + d*x))**2, x)

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