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]

5/2*x/a^3-5/2*arctanh(cos(d*x+c))/a^3/d+3*cos(d*x+c)/a^3/d+3*cot(d*x+c)/a^3/d-1/2*cot(d*x+c)*csc(d*x+c)/a^3/d-
1/2*cos(d*x+c)*sin(d*x+c)/a^3/d

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Rubi [A]  time = 0.25, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 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 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 ^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*}

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Mathematica [A]  time = 0.89, 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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fricas [A]  time = 0.48, size = 130, normalized size = 1.33 \[ \frac {10 \, d x \cos \left (d x + c\right )^{2} + 12 \, \cos \left (d x + c\right )^{3} - 10 \, d x - 5 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (\cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 10 \, \cos \left (d x + c\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]

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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giac [A]  time = 0.26, size = 172, normalized size = 1.76 \[ \frac {\frac {20 \, {\left (d x + c\right )}}{a^{3}} + \frac {20 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} a^{3}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{8 \, d} \]

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

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maple [B]  time = 0.68, size = 234, normalized size = 2.39 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3} d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {6}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}-\frac {1}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}} \]

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/a^3/d*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 = 0.57, size = 267, normalized size = 2.72 \[ \frac {\frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {46 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {47 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{3}} + \frac {40 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {20 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{8 \, d} \]

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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mupad [B]  time = 9.24, size = 228, normalized size = 2.33 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {5\,\mathrm {atan}\left (\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-25}+\frac {25}{25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-25}\right )}{a^3\,d}+\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {47\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^3*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^3*d) - (5*atan((25*tan(c/2 + (d*x)/2))/(25*tan(c/2 + (d*x)/2) - 25) + 25/(25*tan(c/2
+ (d*x)/2) - 25)))/(a^3*d) + (5*log(tan(c/2 + (d*x)/2)))/(2*a^3*d) + (6*tan(c/2 + (d*x)/2) + 23*tan(c/2 + (d*
x)/2)^2 + 8*tan(c/2 + (d*x)/2)^3 + (47*tan(c/2 + (d*x)/2)^4)/2 + 10*tan(c/2 + (d*x)/2)^5 - 1/2)/(d*(4*a^3*tan(
c/2 + (d*x)/2)^2 + 8*a^3*tan(c/2 + (d*x)/2)^4 + 4*a^3*tan(c/2 + (d*x)/2)^6)) - (3*tan(c/2 + (d*x)/2))/(2*a^3*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)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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