3.729 \(\int \frac{\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=97 \[ \frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{3 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.252508, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2875, 2872, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ \frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{3 x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2633

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

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \cos (c+d x) \cot ^3(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac{\int \left (4 a^6-a^6 \csc (c+d x)-2 a^6 \csc ^2(c+d x)+a^6 \csc ^3(c+d x)-a^6 \sin (c+d x)-2 a^6 \sin ^2(c+d x)+a^6 \sin ^3(c+d x)\right ) \, dx}{a^8}\\ &=\frac{4 x}{a^2}-\frac{\int \csc (c+d x) \, dx}{a^2}+\frac{\int \csc ^3(c+d x) \, dx}{a^2}-\frac{\int \sin (c+d x) \, dx}{a^2}+\frac{\int \sin ^3(c+d x) \, dx}{a^2}-\frac{2 \int \csc ^2(c+d x) \, dx}{a^2}-\frac{2 \int \sin ^2(c+d x) \, dx}{a^2}\\ &=\frac{4 x}{a^2}+\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac{\cos (c+d x) \sin (c+d x)}{a^2 d}+\frac{\int \csc (c+d x) \, dx}{2 a^2}-\frac{\int 1 \, dx}{a^2}-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}+\frac{2 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}\\ &=\frac{3 x}{a^2}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac{\cos ^3(c+d x)}{3 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{\cos (c+d x) \sin (c+d x)}{a^2 d}\\ \end{align*}

Mathematica [A]  time = 2.04388, size = 158, normalized size = 1.63 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (6 \cos (c+d x)+2 \cos (3 (c+d x))+3 \left (4 \sin (2 (c+d x))-8 \tan \left (\frac{1}{2} (c+d x)\right )+8 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 c+24 d x\right )\right )}{24 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(6*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + 3*(24*c + 24*d*x + 8*Cot[(c +
d*x)/2] - Csc[(c + d*x)/2]^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + 4*Sin[
2*(c + d*x)] - 8*Tan[(c + d*x)/2])))/(24*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [B]  time = 0.152, size = 234, normalized size = 2.4 \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 ) }-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{2}{3\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-{\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{1}{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)^8*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)-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^5+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2
*c)+2/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3+6/d/a^2*arctan(tan(1/2*d*x+1/2*c))-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)-1/2/d/a^2*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.62956, size = 446, normalized size = 4.6 \begin{align*} \frac{\frac{\frac{24 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{7 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{72 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{45 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{24 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 3}{\frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{3 \,{\left (\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}}\right )}}{a^{2}} + \frac{144 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{24 \, 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))^2,x, algorithm="maxima")

[Out]

1/24*((24*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 120*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 72*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 45*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 - 24*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3)/(a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)
^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^
8/(cos(d*x + c) + 1)^8) - 3*(8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 14
4*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.22781, size = 383, normalized size = 3.95 \begin{align*} \frac{4 \, \cos \left (d x + c\right )^{5} + 36 \, d x \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right )^{3} - 36 \, d x + 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 ) - 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 ) + 12 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/12*(4*cos(d*x + c)^5 + 36*d*x*cos(d*x + c)^2 - 4*cos(d*x + c)^3 - 36*d*x + 3*(cos(d*x + c)^2 - 1)*log(1/2*co
s(d*x + c) + 1/2) - 3*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) + 12*(cos(d*x + c)^3 - 3*cos(d*x + c))
*sin(d*x + c) + 6*cos(d*x + c))/(a^2*d*cos(d*x + c)^2 - a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33612, size = 227, normalized size = 2.34 \begin{align*} \frac{\frac{72 \,{\left (d x + c\right )}}{a^{2}} - \frac{12 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{3 \,{\left (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 )\right )}}{a^{4}} + \frac{3 \,{\left (6 \, \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\right )}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} - \frac{16 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}}}{24 \, 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))^2,x, algorithm="giac")

[Out]

1/24*(72*(d*x + c)/a^2 - 12*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*(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*(6*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) -
16*(3*tan(1/2*d*x + 1/2*c)^5 - 3*tan(1/2*d*x + 1/2*c)^4 - 3*tan(1/2*d*x + 1/2*c) - 1)/((tan(1/2*d*x + 1/2*c)^2
+ 1)^3*a^2))/d