3.318 \(\int \frac{\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=68 \[ \frac{5 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac{2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^3*d)) + (2*Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x])^2) + (5*Cos[c + d*x])/(3*a^3*d
*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.193275, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2874, 2966, 3770, 2650, 2648} \[ \frac{5 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)}+\frac{2 \cos (c+d x)}{3 a^3 d (\sin (c+d x)+1)^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^3*d)) + (2*Cos[c + d*x])/(3*a^3*d*(1 + Sin[c + d*x])^2) + (5*Cos[c + d*x])/(3*a^3*d
*(1 + Sin[c + d*x]))

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]

Rule 2650

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

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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \frac{\csc (c+d x) (a-a \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=\frac{\int \left (\frac{\csc (c+d x)}{a}-\frac{2}{a (1+\sin (c+d x))^2}-\frac{1}{a (1+\sin (c+d x))}\right ) \, dx}{a^2}\\ &=\frac{\int \csc (c+d x) \, dx}{a^3}-\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}-\frac{2 \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac{\cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\frac{2 \int \frac{1}{1+\sin (c+d x)} \, dx}{3 a^3}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac{5 \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.37831, size = 185, normalized size = 2.72 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (-4 \sin \left (\frac{1}{2} (c+d x)\right )-10 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3+3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3\right )}{3 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-4*Sin[(c + d*x)/2] + 2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 10*S
in[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 3*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^3 + 3*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3))/(3*d*(a + a*Sin[c + d*x])^3)

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Maple [A]  time = 0.147, size = 82, normalized size = 1.2 \begin{align*}{\frac{8}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+6\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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))^3,x)

[Out]

8/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+6/d/a^3/(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*ln(
tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.16581, size = 193, normalized size = 2.84 \begin{align*} \frac{\frac{2 \,{\left (\frac{12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 7\right )}}{a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, d} \end{align*}

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))^3,x, algorithm="maxima")

[Out]

1/3*(2*(12*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 7)/(a^3 + 3*a^3*sin(d*x +
c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
+ 3*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [B]  time = 1.71214, size = 528, normalized size = 7.76 \begin{align*} -\frac{10 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (5 \, \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 14 \, \cos \left (d x + c\right ) + 4}{6 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d -{\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

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))^3,x, algorithm="fricas")

[Out]

-1/6*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*log(1/2*cos(
d*x + c) + 1/2) - 3*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*log(-1/2*cos(d*x + c
) + 1/2) + 2*(5*cos(d*x + c) - 2)*sin(d*x + c) + 14*cos(d*x + c) + 4)/(a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x +
c) - 2*a^3*d - (a^3*d*cos(d*x + c) + 2*a^3*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{2}{\left (c + d x \right )} \csc{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

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))**3,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.32223, size = 89, normalized size = 1.31 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}}}{3 \, d} \end{align*}

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))^3,x, algorithm="giac")

[Out]

1/3*(3*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 2*(9*tan(1/2*d*x + 1/2*c)^2 + 12*tan(1/2*d*x + 1/2*c) + 7)/(a^3*(t
an(1/2*d*x + 1/2*c) + 1)^3))/d