3.305 \(\int \frac{\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=95 \[ \frac{\cot ^3(c+d x)}{3 a d}+\frac{\cot (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(8*a*d) + Cot[c + d*x]/(a*d) + Cot[c + d*x]^3/(3*a*d) - (3*Cot[c + d*x]*Csc[c + d*x
])/(8*a*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d)

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Rubi [A]  time = 0.13442, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2839, 3768, 3770, 3767} \[ \frac{\cot ^3(c+d x)}{3 a d}+\frac{\cot (c+d x)}{a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(8*a*d) + Cot[c + d*x]/(a*d) + Cot[c + d*x]^3/(3*a*d) - (3*Cot[c + d*x]*Csc[c + d*x
])/(8*a*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*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 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]

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]

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \csc ^4(c+d x) \, dx}{a}+\frac{\int \csc ^5(c+d x) \, dx}{a}\\ &=-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac{3 \int \csc ^3(c+d x) \, dx}{4 a}+\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac{\cot (c+d x)}{a d}+\frac{\cot ^3(c+d x)}{3 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac{3 \int \csc (c+d x) \, dx}{8 a}\\ &=-\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{\cot (c+d x)}{a d}+\frac{\cot ^3(c+d x)}{3 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 1.07876, size = 125, normalized size = 1.32 \[ -\frac{\csc ^4(c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-48 \sin (2 (c+d x))+66 \cos (c+d x)+2 (16 \sin (c+d x)-9) \cos (3 (c+d x))+72 \sin ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{192 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-(Csc[c + d*x]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(66*Cos[c + d*x] + 72*(Log[Cos[(c + d*x)/2]] - Log[Si
n[(c + d*x)/2]])*Sin[c + d*x]^4 + 2*Cos[3*(c + d*x)]*(-9 + 16*Sin[c + d*x]) - 48*Sin[2*(c + d*x)]))/(192*a*d*(
1 + Sin[c + d*x]))

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Maple [A]  time = 0.134, size = 170, normalized size = 1.8 \begin{align*}{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\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}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{3}{8\,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)^5/(a+a*sin(d*x+c)),x)

[Out]

1/64/d/a*tan(1/2*d*x+1/2*c)^4-1/24/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)-1/64/d/a/tan(1/2*d*x+1/2*c)^4+3/8/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 = 1.1745, size = 263, normalized size = 2.77 \begin{align*} -\frac{\frac{\frac{72 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a} - \frac{72 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{{\left (\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{72 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 3\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a \sin \left (d x + c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/192*((72*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a - 72*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (8*sin
(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 72*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
- 3)*(cos(d*x + c) + 1)^4/(a*sin(d*x + c)^4))/d

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Fricas [A]  time = 1.70459, size = 396, normalized size = 4.17 \begin{align*} \frac{18 \, \cos \left (d x + c\right )^{3} - 9 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 16 \,{\left (2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )}{48 \,{\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(18*cos(d*x + c)^3 - 9*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 9*(cos(d*x +
c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 16*(2*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x +
c) - 30*cos(d*x + c))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)

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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)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.38365, size = 212, normalized size = 2.23 \begin{align*} \frac{\frac{72 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} - \frac{150 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 72 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, \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 ) + 3}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*(72*log(abs(tan(1/2*d*x + 1/2*c)))/a + (3*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/2*c)^3 + 24
*a^3*tan(1/2*d*x + 1/2*c)^2 - 72*a^3*tan(1/2*d*x + 1/2*c))/a^4 - (150*tan(1/2*d*x + 1/2*c)^4 - 72*tan(1/2*d*x
+ 1/2*c)^3 + 24*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 3)/(a*tan(1/2*d*x + 1/2*c)^4))/d