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

Optimal. Leaf size=78 \[ \frac{2 \cot (c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (2*C
os[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.220133, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2874, 2966, 3770, 3767, 8, 3768, 2648} \[ \frac{2 \cot (c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (2*C
os[c + d*x])/(a^2*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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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{\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \frac{\csc ^3(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{\int \left (2 \csc (c+d x)-2 \csc ^2(c+d x)+\csc ^3(c+d x)-\frac{2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=\frac{\int \csc ^3(c+d x) \, dx}{a^2}+\frac{2 \int \csc (c+d x) \, dx}{a^2}-\frac{2 \int \csc ^2(c+d x) \, dx}{a^2}-\frac{2 \int \frac{1}{1+\sin (c+d x)} \, dx}{a^2}\\ &=-\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}+\frac{\int \csc (c+d x) \, dx}{2 a^2}+\frac{2 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}\\ &=-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.750099, size = 364, normalized size = 4.67 \[ -\frac{4 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}{d (a \sin (c+d x)+a)^2}-\frac{5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{2 d (a \sin (c+d x)+a)^2}+\frac{5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{2 d (a \sin (c+d x)+a)^2}-\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{d (a \sin (c+d x)+a)^2}+\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{d (a \sin (c+d x)+a)^2}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{8 d (a \sin (c+d x)+a)^2}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{8 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)/(d*(a + a*Sin[c + d*x])^2) + (Cot[(c + d*x)/2]*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(d*(a + a*Sin[c + d*x])^2) - (Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x])^2) - (5*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^4)/(2*d*(a + a*Sin[c + d*x])^2) + (5*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(2*d*(
a + a*Sin[c + d*x])^2) + (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(8*d*(a + a*Sin[c + d*x]
)^2) - ((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/2])/(d*(a + a*Sin[c + d*x])^2)

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Maple [A]  time = 0.16, size = 114, normalized size = 1.5 \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{5}{2\,d{a}^{2}}\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)^3/(a+a*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)+4/d/a^2/(tan(1/2*d*x+1/2*c)+1)-1/8/d/a^2/tan(1/2*d*x
+1/2*c)^2+1/d/a^2/tan(1/2*d*x+1/2*c)+5/2/d/a^2*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.19286, size = 217, normalized size = 2.78 \begin{align*} \frac{\frac{\frac{7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac{\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac{20 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((7*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a^2*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) - (8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c
)^2/(cos(d*x + c) + 1)^2)/a^2 + 20*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [B]  time = 1.703, size = 659, normalized size = 8.45 \begin{align*} \frac{16 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} - 5 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 5 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (8 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 14 \, \cos \left (d x + c\right ) - 8}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d +{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} 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)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(16*cos(d*x + c)^3 + 10*cos(d*x + c)^2 - 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x
+ c) - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 -
1)*sin(d*x + c) - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 2*(8*cos(d*x + c)^2 + 3*cos(d*x + c) - 4)*
sin(d*x + c) - 14*cos(d*x + c) - 8)/(a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - a^2*d
+ (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.36502, size = 157, normalized size = 2.01 \begin{align*} \frac{\frac{20 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} + \frac{32}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} - \frac{30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(20*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + 3
2/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) - (30*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x
+ 1/2*c)^2))/d