3.744 \(\int \frac{\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=98 \[ \frac{3 \cos (c+d x)}{a^3 d}+\frac{3 \cot (c+d x)}{a^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{5 x}{2 a^3} \]

[Out]

(5*x)/(2*a^3) - (5*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (3*Cos[c + d*x])/(a^3*d) + (3*Cot[c + d*x])/(a^3*d) - (C
ot[c + d*x]*Csc[c + d*x])/(2*a^3*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d)

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Rubi [A]  time = 0.246939, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2872, 3770, 3767, 8, 3768, 2638, 2635} \[ \frac{3 \cos (c+d x)}{a^3 d}+\frac{3 \cot (c+d x)}{a^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{5 x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(5*x)/(2*a^3) - (5*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (3*Cos[c + d*x])/(a^3*d) + (3*Cot[c + d*x])/(a^3*d) - (C
ot[c + d*x]*Csc[c + d*x])/(2*a^3*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d)

Rule 2875

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

Rule 2872

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

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

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

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \cot ^2(c+d x) \csc (c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (2 a^5+2 a^5 \csc (c+d x)-3 a^5 \csc ^2(c+d x)+a^5 \csc ^3(c+d x)-3 a^5 \sin (c+d x)+a^5 \sin ^2(c+d x)\right ) \, dx}{a^8}\\ &=\frac{2 x}{a^3}+\frac{\int \csc ^3(c+d x) \, dx}{a^3}+\frac{\int \sin ^2(c+d x) \, dx}{a^3}+\frac{2 \int \csc (c+d x) \, dx}{a^3}-\frac{3 \int \csc ^2(c+d x) \, dx}{a^3}-\frac{3 \int \sin (c+d x) \, dx}{a^3}\\ &=\frac{2 x}{a^3}-\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{3 \cos (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{\int 1 \, dx}{2 a^3}+\frac{\int \csc (c+d x) \, dx}{2 a^3}+\frac{3 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{5 x}{2 a^3}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \cos (c+d x)}{a^3 d}+\frac{3 \cot (c+d x)}{a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.906118, size = 144, normalized size = 1.47 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (20 (c+d x)-2 \sin (2 (c+d x))+24 \cos (c+d x)-12 \tan \left (\frac{1}{2} (c+d x)\right )+12 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-20 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{8 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(20*(c + d*x) + 24*Cos[c + d*x] + 12*Cot[(c + d*x)/2] - Csc[(c + d*x)
/2]^2 - 20*Log[Cos[(c + d*x)/2]] + 20*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 2*Sin[2*(c + d*x)] - 12*Tan
[(c + d*x)/2]))/(8*d*(a + a*Sin[c + d*x])^3)

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Maple [B]  time = 0.157, size = 234, normalized size = 2.4 \begin{align*}{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{3}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+6\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{5}{2\,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)^8*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*tan(1/2*d*x+1/2*c)+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)^3+6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1
/2*c)+6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2+5/d/a^3*arctan(tan(1/2*d*x+1/2*c))-1/8/d/a^3/tan(1/2*d*x+1/2*c)^2+3/2
/d/a^3/tan(1/2*d*x+1/2*c)+5/2/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.56469, size = 360, normalized size = 3.67 \begin{align*} \frac{\frac{\frac{12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{46 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{47 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{20 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1}{\frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{\frac{12 \, \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^{3}} + \frac{40 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{20 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/8*((12*sin(d*x + c)/(cos(d*x + c) + 1) + 46*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 16*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 47*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1)/(a^3*sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6) - (12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 40*arctan(sin(d*x +
c)/(cos(d*x + c) + 1))/a^3 + 20*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.19672, size = 358, normalized size = 3.65 \begin{align*} \frac{10 \, d x \cos \left (d x + c\right )^{2} + 12 \, \cos \left (d x + c\right )^{3} - 10 \, d x - 5 \,{\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 ) + 5 \,{\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 ) - 2 \,{\left (\cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 10 \, \cos \left (d x + c\right )}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(10*d*x*cos(d*x + c)^2 + 12*cos(d*x + c)^3 - 10*d*x - 5*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) +
5*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) - 2*(cos(d*x + c)^3 + 5*cos(d*x + c))*sin(d*x + c) - 10*c
os(d*x + c))/(a^3*d*cos(d*x + c)^2 - a^3*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)**8*csc(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33558, size = 232, normalized size = 2.37 \begin{align*} \frac{\frac{20 \,{\left (d x + c\right )}}{a^{3}} + \frac{20 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, \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 ) + 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} a^{3}} + \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(20*(d*x + c)/a^3 + 20*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (10*tan(1/2*d*x + 1/2*c)^6 - 20*tan(1/2*d*x +
1/2*c)^5 - 27*tan(1/2*d*x + 1/2*c)^4 - 16*tan(1/2*d*x + 1/2*c)^3 - 36*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x
+ 1/2*c) + 1)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2*a^3) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*t
an(1/2*d*x + 1/2*c))/a^6)/d