3.747 \(\int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=100 \[ -\frac {\cot ^5(c+d x)}{5 a^3 d}-\frac {4 \cot ^3(c+d x)}{3 a^3 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

[Out]

-7/8*arctanh(cos(d*x+c))/a^3/d-4/3*cot(d*x+c)^3/a^3/d-1/5*cot(d*x+c)^5/a^3/d+1/8*cot(d*x+c)*csc(d*x+c)/a^3/d+3
/4*cot(d*x+c)*csc(d*x+c)^3/a^3/d

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Rubi [A]  time = 0.34, antiderivative size = 100, 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, 2873, 2611, 3770, 2607, 30, 3768, 14} \[ -\frac {\cot ^5(c+d x)}{5 a^3 d}-\frac {4 \cot ^3(c+d x)}{3 a^3 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-7*ArcTanh[Cos[c + d*x]])/(8*a^3*d) - (4*Cot[c + d*x]^3)/(3*a^3*d) - Cot[c + d*x]^5/(5*a^3*d) + (Cot[c + d*x]
*Csc[c + d*x])/(8*a^3*d) + (3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^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] && IGtQ[m, 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 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 ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (-a^3 \cot ^2(c+d x) \csc (c+d x)+3 a^3 \cot ^2(c+d x) \csc ^2(c+d x)-3 a^3 \cot ^2(c+d x) \csc ^3(c+d x)+a^3 \cot ^2(c+d x) \csc ^4(c+d x)\right ) \, dx}{a^6}\\ &=-\frac {\int \cot ^2(c+d x) \csc (c+d x) \, dx}{a^3}+\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a^3}+\frac {3 \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx}{a^3}-\frac {3 \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a^3}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {\int \csc (c+d x) \, dx}{2 a^3}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a^3}+\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^3 d}+\frac {3 \operatorname {Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{a^3 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot ^3(c+d x)}{a^3 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {3 \int \csc (c+d x) \, dx}{8 a^3}+\frac {\operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a^3 d}\\ &=-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac {4 \cot ^3(c+d x)}{3 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^3 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.76, size = 189, normalized size = 1.89 \[ -\frac {\csc ^5(c+d x) \left (-780 \sin (2 (c+d x))+30 \sin (4 (c+d x))+560 \cos (c+d x)-40 \cos (3 (c+d x))-136 \cos (5 (c+d x))-1050 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+525 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-105 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+1050 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-525 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+105 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1920 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^6)/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/1920*(Csc[c + d*x]^5*(560*Cos[c + d*x] - 40*Cos[3*(c + d*x)] - 136*Cos[5*(c + d*x)] + 1050*Log[Cos[(c + d*x
)/2]]*Sin[c + d*x] - 1050*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 780*Sin[2*(c + d*x)] - 525*Log[Cos[(c + d*x)/2]
]*Sin[3*(c + d*x)] + 525*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 30*Sin[4*(c + d*x)] + 105*Log[Cos[(c + d*x)/
2]]*Sin[5*(c + d*x)] - 105*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(a^3*d)

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fricas [A]  time = 0.46, size = 169, normalized size = 1.69 \[ \frac {272 \, \cos \left (d x + c\right )^{5} - 320 \, \cos \left (d x + c\right )^{3} - 105 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 105 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (\cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(272*cos(d*x + c)^5 - 320*cos(d*x + c)^3 - 105*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x +
c) + 1/2)*sin(d*x + c) + 105*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c
) - 30*(cos(d*x + c)^3 - 7*cos(d*x + c))*sin(d*x + c))/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d
)*sin(d*x + c))

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giac [B]  time = 0.31, size = 186, normalized size = 1.86 \[ \frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {1918 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 130 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {6 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 130 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 420 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (1918*tan(1/2*d*x + 1/2*c)^5 - 420*tan(1/2*d*x + 1/2*c)^4 - 12
0*tan(1/2*d*x + 1/2*c)^3 + 130*tan(1/2*d*x + 1/2*c)^2 - 45*tan(1/2*d*x + 1/2*c) + 6)/(a^3*tan(1/2*d*x + 1/2*c)
^5) + (6*a^12*tan(1/2*d*x + 1/2*c)^5 - 45*a^12*tan(1/2*d*x + 1/2*c)^4 + 130*a^12*tan(1/2*d*x + 1/2*c)^3 - 120*
a^12*tan(1/2*d*x + 1/2*c)^2 - 420*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [B]  time = 0.71, size = 208, normalized size = 2.08 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d \,a^{3}}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d \,a^{3}}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d \,a^{3}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3} d}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{3}}+\frac {7}{16 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}-\frac {1}{160 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3}{64 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {13}{96 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x)

[Out]

1/160/d/a^3*tan(1/2*d*x+1/2*c)^5-3/64/d/a^3*tan(1/2*d*x+1/2*c)^4+13/96/d/a^3*tan(1/2*d*x+1/2*c)^3-1/8/d/a^3*ta
n(1/2*d*x+1/2*c)^2-7/16/d/a^3*tan(1/2*d*x+1/2*c)+7/16/d/a^3/tan(1/2*d*x+1/2*c)+7/8/d/a^3*ln(tan(1/2*d*x+1/2*c)
)-1/160/d/a^3/tan(1/2*d*x+1/2*c)^5+1/8/d/a^3/tan(1/2*d*x+1/2*c)^2+3/64/d/a^3/tan(1/2*d*x+1/2*c)^4-13/96/d/a^3/
tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.40, size = 235, normalized size = 2.35 \[ -\frac {\frac {\frac {420 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {130 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {840 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {130 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {420 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{3} \sin \left (d x + c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/960*((420*sin(d*x + c)/(cos(d*x + c) + 1) + 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 130*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 + 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 840
*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (45*sin(d*x + c)/(cos(d*x + c) + 1) - 130*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 420*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6)*(cos(d*x
+ c) + 1)^5/(a^3*sin(d*x + c)^5))/d

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mupad [B]  time = 9.60, size = 291, normalized size = 2.91 \[ \frac {6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-45\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+45\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+130\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-130\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+840\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^6*(a + a*sin(c + d*x))^3),x)

[Out]

(6*sin(c/2 + (d*x)/2)^10 - 6*cos(c/2 + (d*x)/2)^10 - 45*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 + 45*cos(c/2 +
(d*x)/2)^9*sin(c/2 + (d*x)/2) + 130*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 120*cos(c/2 + (d*x)/2)^3*sin(
c/2 + (d*x)/2)^7 - 420*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 + 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)
^4 + 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3 - 130*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 840*log(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)/(960*a^3*d*cos(c/2 + (d*x)/2)
^5*sin(c/2 + (d*x)/2)^5)

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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)**6/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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