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

Optimal. Leaf size=72 \[ -\frac{\cot ^3(c+d x)}{3 a d}-\frac{\cot (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

ArcTanh[Cos[c + d*x]]/(2*a*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*a*d) + (Cot[c + d*x]*Csc[c + d*x])/(2*a
*d)

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Rubi [A]  time = 0.123105, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2839, 3767, 3768, 3770} \[ -\frac{\cot ^3(c+d x)}{3 a d}-\frac{\cot (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(2*a*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*a*d) + (Cot[c + d*x]*Csc[c + d*x])/(2*a
*d)

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

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 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 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{\cot ^2(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \csc ^3(c+d x) \, dx}{a}+\frac{\int \csc ^4(c+d x) \, dx}{a}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{2 a d}-\frac{\int \csc (c+d x) \, dx}{2 a}-\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\cot (c+d x)}{a d}-\frac{\cot ^3(c+d x)}{3 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.574997, size = 126, normalized size = 1.75 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\csc \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-12 (\sin (c+d x)-1) \cos (c+d x)-4 \left (\cos (3 (c+d x))+3 \sin ^3(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )\right )}{192 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(-12*Cos[c + d*x]*(-1 + Sin[c + d*
x]) - 4*(Cos[3*(c + d*x)] + 3*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3)))/(192*a*d*(1 +
Sin[c + d*x]))

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Maple [A]  time = 0.135, size = 132, normalized size = 1.8 \begin{align*}{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{3}{8\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/8/d/a*tan(1/2*d*x+1/2*c)^2+3/8/d/a*tan(1/2*d*x+1/2*c)-3/8/d/a/tan(1/2*d*x+1/2*
c)-1/2/d/a*ln(tan(1/2*d*x+1/2*c))-1/24/d/a/tan(1/2*d*x+1/2*c)^3+1/8/d/a/tan(1/2*d*x+1/2*c)^2

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Maxima [B]  time = 1.10887, size = 207, normalized size = 2.88 \begin{align*} \frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{{\left (\frac{3 \, \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}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/24*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c
) + 1)^3)/a - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a + (3*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

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Fricas [A]  time = 1.65169, size = 336, normalized size = 4.67 \begin{align*} -\frac{8 \, \cos \left (d x + c\right )^{3} - 3 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 3 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 12 \, \cos \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(8*cos(d*x + c)^3 - 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(cos(d*x + c)^2
- 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*cos(d*x + c)*sin(d*x + c) - 12*cos(d*x + c))/((a*d*cos(d*x
+ c)^2 - a*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)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30829, size = 173, normalized size = 2.4 \begin{align*} -\frac{\frac{12 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{22 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/24*(12*log(abs(tan(1/2*d*x + 1/2*c)))/a - (a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 + 9*a^
2*tan(1/2*d*x + 1/2*c))/a^3 - (22*tan(1/2*d*x + 1/2*c)^3 - 9*tan(1/2*d*x + 1/2*c)^2 + 3*tan(1/2*d*x + 1/2*c) -
1)/(a*tan(1/2*d*x + 1/2*c)^3))/d