3.730 \(\int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2709, 3770, 3767, 8, 3768, 2638, 2635} \[ \frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2638

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

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/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 ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^4(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (-a^6+4 a^6 \csc (c+d x)-a^6 \csc ^2(c+d x)-2 a^6 \csc ^3(c+d x)+a^6 \csc ^4(c+d x)-2 a^6 \sin (c+d x)+a^6 \sin ^2(c+d x)\right ) \, dx}{a^8}\\ &=-\frac {x}{a^2}-\frac {\int \csc ^2(c+d x) \, dx}{a^2}+\frac {\int \csc ^4(c+d x) \, dx}{a^2}+\frac {\int \sin ^2(c+d x) \, dx}{a^2}-\frac {2 \int \csc ^3(c+d x) \, dx}{a^2}-\frac {2 \int \sin (c+d x) \, dx}{a^2}+\frac {4 \int \csc (c+d x) \, dx}{a^2}\\ &=-\frac {x}{a^2}-\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\int 1 \, dx}{2 a^2}-\frac {\int \csc (c+d x) \, dx}{a^2}+\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}-\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}\\ &=-\frac {x}{2 a^2}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 2.45, size = 184, normalized size = 1.90 \[ -\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right )^4 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (30 \cos (c+d x)-\cos (3 (c+d x))+3 \left (\cos (5 (c+d x))+8 \sin (c+d x) \left (-6 \cos (c+d x)+2 \cos (3 (c+d x))-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\cos (2 (c+d x)) \left (-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )+c+d x\right )\right )\right )}{768 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/768*((1 + Cot[(c + d*x)/2])^4*Sec[(c + d*x)/2]^2*(30*Cos[c + d*x] - Cos[3*(c + d*x)] + 3*(Cos[5*(c + d*x)]
+ 8*(c + d*x - 6*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + 6*Log[Cos[(c + d*x)/2]] - Cos[2*(c + d*x)]*(c + d*x + 6*L
og[Cos[(c + d*x)/2]] - 6*Log[Sin[(c + d*x)/2]]) - 6*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))*Tan[(c + d*x)/2])/(a
^2*d*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.48, size = 161, normalized size = 1.66 \[ \frac {3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (d x \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right )^{3} - d x + 6 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^3 - 9*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 9
*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(d*x*cos(d*x + c)^2 - 4*cos(d*x + c)^3 - d
*x + 6*cos(d*x + c))*sin(d*x + c) + 3*cos(d*x + c))/((a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

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giac [B]  time = 0.23, size = 194, normalized size = 2.00 \[ -\frac {\frac {12 \, {\left (d x + c\right )}}{a^{2}} - \frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {24 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {132 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(12*(d*x + c)/a^2 - 72*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 24*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x +
1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (132*tan(1/2*d*x + 1/2*c)^3 - 3*t
an(1/2*d*x + 1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) - (a^4*tan(1/2*d*x + 1/2*c)^3
- 6*a^4*tan(1/2*d*x + 1/2*c)^2 - 3*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.65, size = 272, normalized size = 2.80 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{24 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/24/d/a^2*tan(1/2*d*x+1/2*c)^3-1/4/d/a^2*tan(1/2*d*x+1/2*c)^2-1/8/d/a^2*tan(1/2*d*x+1/2*c)+1/d/a^2/(1+tan(1/2
*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2/(1+tan(1
/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2-1/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/
24/a^2/d/tan(1/2*d*x+1/2*c)^3+1/4/a^2/d/tan(1/2*d*x+1/2*c)^2+1/8/d/a^2/tan(1/2*d*x+1/2*c)+3/d/a^2*ln(tan(1/2*d
*x+1/2*c))

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maxima [B]  time = 0.45, size = 306, normalized size = 3.15 \[ \frac {\frac {\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {108 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {19 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {102 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {27 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} - \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {72 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*((6*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 108*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 19*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 102*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 27*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 - 1)/(a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*a^2*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 + a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - (3*sin(d*x + c)/(cos(d*x + c) + 1) + 6*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 7
2*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 9.16, size = 253, normalized size = 2.61 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d}+\frac {\mathrm {atan}\left (\frac {1}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^2*d) - tan(c/2 + (d*x)/2)^2/(4*a^2*d) + (3*log(tan(c/2 + (d*x)/2)))/(a^2*d) + (2*ta
n(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2/3 + 36*tan(c/2 + (d*x)/2)^3 - (19*tan(c/2 + (d*x)/2)^4)/3 + 34*tan(c/2
+ (d*x)/2)^5 + 9*tan(c/2 + (d*x)/2)^6 - 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 16*a^2*tan(c/2 + (d*x)/2)^5 + 8
*a^2*tan(c/2 + (d*x)/2)^7)) - tan(c/2 + (d*x)/2)/(8*a^2*d) + atan(1/(tan(c/2 + (d*x)/2) + 6) - (6*tan(c/2 + (d
*x)/2))/(tan(c/2 + (d*x)/2) + 6))/(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)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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