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

Optimal. Leaf size=116 \[ -\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {9 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {2 x}{a^2} \]

[Out]

-2*x/a^2+9/8*arctanh(cos(d*x+c))/a^2/d-cos(d*x+c)/a^2/d-2*cot(d*x+c)/a^2/d+2/3*cot(d*x+c)^3/a^2/d+1/8*cot(d*x+
c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.28, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2875, 2872, 3770, 3767, 8, 3768, 2638} \[ -\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {9 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/a^2 + (9*ArcTanh[Cos[c + d*x]])/(8*a^2*d) - Cos[c + d*x]/(a^2*d) - (2*Cot[c + d*x])/(a^2*d) + (2*Cot[c
+ d*x]^3)/(3*a^2*d) + (Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^2*d)

Rule 8

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

Rule 2638

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

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 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 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 {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^4(c+d x) \csc (c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (-2 a^6-a^6 \csc (c+d x)+4 a^6 \csc ^2(c+d x)-a^6 \csc ^3(c+d x)-2 a^6 \csc ^4(c+d x)+a^6 \csc ^5(c+d x)+a^6 \sin (c+d x)\right ) \, dx}{a^8}\\ &=-\frac {2 x}{a^2}-\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {\int \csc ^3(c+d x) \, dx}{a^2}+\frac {\int \csc ^5(c+d x) \, dx}{a^2}+\frac {\int \sin (c+d x) \, dx}{a^2}-\frac {2 \int \csc ^4(c+d x) \, dx}{a^2}+\frac {4 \int \csc ^2(c+d x) \, dx}{a^2}\\ &=-\frac {2 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 {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a^2}+\frac {2 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}-\frac {4 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}\\ &=-\frac {2 x}{a^2}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {3 \int \csc (c+d x) \, dx}{8 a^2}\\ &=-\frac {2 x}{a^2}+\frac {9 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.88, size = 219, normalized size = 1.89 \[ -\frac {\sin ^5(c+d x) \left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (192 \cot (c+d x)+(3 \csc (c+d x)-8) \csc ^4\left (\frac {1}{2} (c+d x)\right )+(128-6 \csc (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )+8 \left (-(8 \cos (c+d x)+7) \sec ^4\left (\frac {1}{2} (c+d x)\right )-6 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)+3 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+3 \csc (c+d x) \left (16 (c+d x)+9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )\right )}{3072 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3072*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^4*(192*Cot[c + d*x] + Csc[(c + d*x)/2]^2*(128 - 6*Csc[c + d*x])
+ Csc[(c + d*x)/2]^4*(-8 + 3*Csc[c + d*x]) + 8*(3*Csc[c + d*x]*(16*(c + d*x) - 9*Log[Cos[(c + d*x)/2]] + 9*Lo
g[Sin[(c + d*x)/2]]) - (7 + 8*Cos[c + d*x])*Sec[(c + d*x)/2]^4 + 3*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2 - 6*Csc[c
+ d*x]^5*Sin[(c + d*x)/2]^4))*Sin[c + d*x]^5)/(a^2*d*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.47, size = 187, normalized size = 1.61 \[ -\frac {96 \, d x \cos \left (d x + c\right )^{4} + 48 \, \cos \left (d x + c\right )^{5} - 192 \, d x \cos \left (d x + c\right )^{2} - 90 \, \cos \left (d x + c\right )^{3} + 96 \, d x - 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 54 \, \cos \left (d x + c\right )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/48*(96*d*x*cos(d*x + c)^4 + 48*cos(d*x + c)^5 - 192*d*x*cos(d*x + c)^2 - 90*cos(d*x + c)^3 + 96*d*x - 27*(c
os(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 27*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)
*log(-1/2*cos(d*x + c) + 1/2) - 32*(4*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x + c) + 54*cos(d*x + c))/(a^2*d*
cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)

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giac [A]  time = 0.25, size = 159, normalized size = 1.37 \[ -\frac {\frac {384 \, {\left (d x + c\right )}}{a^{2}} + \frac {216 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {384}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} - \frac {3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(384*(d*x + c)/a^2 + 216*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 384/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) -
(450*tan(1/2*d*x + 1/2*c)^4 - 240*tan(1/2*d*x + 1/2*c)^3 + 16*tan(1/2*d*x + 1/2*c) - 3)/(a^2*tan(1/2*d*x + 1/2
*c)^4) - (3*a^6*tan(1/2*d*x + 1/2*c)^4 - 16*a^6*tan(1/2*d*x + 1/2*c)^3 + 240*a^6*tan(1/2*d*x + 1/2*c))/a^8)/d

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maple [A]  time = 0.71, size = 173, normalized size = 1.49 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a^{2} d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d \,a^{2}}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2}}-\frac {2}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{64 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{12 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {5}{4 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {9 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x)

[Out]

1/64/d/a^2*tan(1/2*d*x+1/2*c)^4-1/12/d/a^2*tan(1/2*d*x+1/2*c)^3+5/4/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)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/64/a^2/d/tan(1/2*d*x+1/2*c)^4+1/12/a^2/d/tan(1/2*d*x+1/2*
c)^3-5/4/d/a^2/tan(1/2*d*x+1/2*c)-9/8/d/a^2*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 263, normalized size = 2.27 \[ \frac {\frac {\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {224 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {384 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {240 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 3}{\frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {240 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{2}} - \frac {768 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {216 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/192*((16*sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 224*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 384*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 240*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3)/(a^2*s
in(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (240*sin(d*x + c)/(cos(d*x + c
) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^2 - 768*arctan(sin(
d*x + c)/(cos(d*x + c) + 1))/a^2 - 216*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 9.08, size = 232, normalized size = 2.00 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^2\,d}+\frac {4\,\mathrm {atan}\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}\right )}{a^2\,d}-\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}-\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{4}}{d\,\left (16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^5*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^4/(64*a^2*d) - tan(c/2 + (d*x)/2)^3/(12*a^2*d) + (4*atan((9*tan(c/2 + (d*x)/2))/(16*tan(c/2
+ (d*x)/2) - 9) + 16/(16*tan(c/2 + (d*x)/2) - 9)))/(a^2*d) - (9*log(tan(c/2 + (d*x)/2)))/(8*a^2*d) - (tan(c/2
+ (d*x)/2)^2/4 - (4*tan(c/2 + (d*x)/2))/3 + (56*tan(c/2 + (d*x)/2)^3)/3 + 32*tan(c/2 + (d*x)/2)^4 + 20*tan(c/
2 + (d*x)/2)^5 + 1/4)/(d*(16*a^2*tan(c/2 + (d*x)/2)^4 + 16*a^2*tan(c/2 + (d*x)/2)^6)) + (5*tan(c/2 + (d*x)/2))
/(4*a^2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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