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

Optimal. Leaf size=40 \[ \frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2874, 2966, 3770, 2648} \[ \frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^2*d)) + (2*Cos[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2874

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

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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+a \sin (c+d x))^2} \, dx &=\frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {\int \left (\csc (c+d x)-\frac {2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^2}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 0.16, size = 115, normalized size = 2.88 \[ -\frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sin \left (\frac {1}{2} (c+d x)\right ) \left (-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4\right )\right )}{a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) +
(4 + Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[(c + d*x)/2]))/(a^2*d*(1 + Sin[c + d*x])^2))

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fricas [B]  time = 0.51, size = 103, normalized size = 2.58 \[ -\frac {{\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \cos \left (d x + c\right ) + 4 \, \sin \left (d x + c\right ) - 4}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((cos(d*x + c) + sin(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - (cos(d*x + c) + sin(d*x + c) + 1)*log(-1
/2*cos(d*x + c) + 1/2) - 4*cos(d*x + c) + 4*sin(d*x + c) - 4)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d
)

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giac [A]  time = 0.16, size = 38, normalized size = 0.95 \[ \frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {4}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 4/(a^2*(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.59, size = 40, normalized size = 1.00 \[ \frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {4}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^2*ln(tan(1/2*d*x+1/2*c))+4/d/a^2/(tan(1/2*d*x+1/2*c)+1)

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maxima [A]  time = 0.32, size = 55, normalized size = 1.38 \[ \frac {\frac {4}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1)) + log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 8.68, size = 39, normalized size = 0.98 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {4}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a^2*d) + 4/(a^2*d*(tan(c/2 + (d*x)/2) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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