3.710 \(\int \frac {\cos ^7(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=143 \[ \frac {\cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos (c+d x)}{a d}-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 a d}-\frac {5 \sin (c+d x) \cos ^3(c+d x)}{24 a d}-\frac {5 \sin (c+d x) \cos (c+d x)}{16 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {5 x}{16 a} \]

[Out]

-5/16*x/a-arctanh(cos(d*x+c))/a/d+cos(d*x+c)/a/d+1/3*cos(d*x+c)^3/a/d+1/5*cos(d*x+c)^5/a/d-5/16*cos(d*x+c)*sin
(d*x+c)/a/d-5/24*cos(d*x+c)^3*sin(d*x+c)/a/d-1/6*cos(d*x+c)^5*sin(d*x+c)/a/d

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Rubi [A]  time = 0.15, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2839, 2592, 302, 206, 2635, 8} \[ \frac {\cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos (c+d x)}{a d}-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 a d}-\frac {5 \sin (c+d x) \cos ^3(c+d x)}{24 a d}-\frac {5 \sin (c+d x) \cos (c+d x)}{16 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {5 x}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(-5*x)/(16*a) - ArcTanh[Cos[c + d*x]]/(a*d) + Cos[c + d*x]/(a*d) + Cos[c + d*x]^3/(3*a*d) + Cos[c + d*x]^5/(5*
a*d) - (5*Cos[c + d*x]*Sin[c + d*x])/(16*a*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])/(24*a*d) - (Cos[c + d*x]^5*Sin
[c + d*x])/(6*a*d)

Rule 8

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^7(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \cos ^6(c+d x) \, dx}{a}+\frac {\int \cos ^5(c+d x) \cot (c+d x) \, dx}{a}\\ &=-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}-\frac {5 \int \cos ^4(c+d x) \, dx}{6 a}-\frac {\operatorname {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}-\frac {5 \int \cos ^2(c+d x) \, dx}{8 a}-\frac {\operatorname {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {5 \cos (c+d x) \sin (c+d x)}{16 a d}-\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}-\frac {5 \int 1 \, dx}{16 a}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac {5 x}{16 a}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {5 \cos (c+d x) \sin (c+d x)}{16 a d}-\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 102, normalized size = 0.71 \[ -\frac {225 \sin (2 (c+d x))+45 \sin (4 (c+d x))+5 \sin (6 (c+d x))-1320 \cos (c+d x)-140 \cos (3 (c+d x))-12 \cos (5 (c+d x))-960 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+960 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+300 c+300 d x}{960 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

-1/960*(300*c + 300*d*x - 1320*Cos[c + d*x] - 140*Cos[3*(c + d*x)] - 12*Cos[5*(c + d*x)] + 960*Log[Cos[(c + d*
x)/2]] - 960*Log[Sin[(c + d*x)/2]] + 225*Sin[2*(c + d*x)] + 45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)])/(a*d)

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fricas [A]  time = 0.50, size = 104, normalized size = 0.73 \[ \frac {48 \, \cos \left (d x + c\right )^{5} + 80 \, \cos \left (d x + c\right )^{3} - 75 \, d x - 5 \, {\left (8 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 240 \, \cos \left (d x + c\right ) - 120 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 120 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{240 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(48*cos(d*x + c)^5 + 80*cos(d*x + c)^3 - 75*d*x - 5*(8*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 15*cos(d*x +
c))*sin(d*x + c) + 240*cos(d*x + c) - 120*log(1/2*cos(d*x + c) + 1/2) + 120*log(-1/2*cos(d*x + c) + 1/2))/(a*
d)

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giac [A]  time = 0.17, size = 195, normalized size = 1.36 \[ -\frac {\frac {75 \, {\left (d x + c\right )}}{a} - \frac {240 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, {\left (165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 720 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1488 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 368\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(75*(d*x + c)/a - 240*log(abs(tan(1/2*d*x + 1/2*c)))/a - 2*(165*tan(1/2*d*x + 1/2*c)^11 + 720*tan(1/2*d
*x + 1/2*c)^10 - 25*tan(1/2*d*x + 1/2*c)^9 + 2160*tan(1/2*d*x + 1/2*c)^8 + 450*tan(1/2*d*x + 1/2*c)^7 + 3680*t
an(1/2*d*x + 1/2*c)^6 - 450*tan(1/2*d*x + 1/2*c)^5 + 3360*tan(1/2*d*x + 1/2*c)^4 + 25*tan(1/2*d*x + 1/2*c)^3 +
1488*tan(1/2*d*x + 1/2*c)^2 - 165*tan(1/2*d*x + 1/2*c) + 368)/((tan(1/2*d*x + 1/2*c)^2 + 1)^6*a))/d

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maple [B]  time = 0.44, size = 432, normalized size = 3.02 \[ \frac {11 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {6 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {18 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {15 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {92 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {15 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {28 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {62 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {46}{15 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11+6/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^
10-5/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9+18/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*
c)^8+15/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7+92/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+
1/2*c)^6-15/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5+28/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*
x+1/2*c)^4+5/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3+62/5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/
2*d*x+1/2*c)^2-11/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)+46/15/a/d/(1+tan(1/2*d*x+1/2*c)^2)^6-5/8
/a/d*arctan(tan(1/2*d*x+1/2*c))+1/a/d*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 402, normalized size = 2.81 \[ -\frac {\frac {\frac {165 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {1488 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {25 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3360 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {450 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3680 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {450 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {2160 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {25 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {720 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {165 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - 368}{a + \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} + \frac {75 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*((165*sin(d*x + c)/(cos(d*x + c) + 1) - 1488*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 25*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 - 3360*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 450*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3680
*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 450*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 2160*sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8 + 25*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 720*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 165*sin(d*x
+ c)^11/(cos(d*x + c) + 1)^11 - 368)/(a + 6*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 + 20*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6*a*si
n(d*x + c)^10/(cos(d*x + c) + 1)^10 + a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 75*arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/a - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 11.83, size = 305, normalized size = 2.13 \[ \frac {5\,\mathrm {atan}\left (\frac {25}{64\,\left (\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {5}{4}\right )}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {5}{4}\right )}\right )}{8\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {92\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {46}{15}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atan(25/(64*((25*tan(c/2 + (d*x)/2))/64 + 5/4)) - (5*tan(c/2 + (d*x)/2))/(4*((25*tan(c/2 + (d*x)/2))/64 + 5
/4))))/(8*a*d) + log(tan(c/2 + (d*x)/2))/(a*d) + ((62*tan(c/2 + (d*x)/2)^2)/5 - (11*tan(c/2 + (d*x)/2))/8 + (5
*tan(c/2 + (d*x)/2)^3)/24 + 28*tan(c/2 + (d*x)/2)^4 - (15*tan(c/2 + (d*x)/2)^5)/4 + (92*tan(c/2 + (d*x)/2)^6)/
3 + (15*tan(c/2 + (d*x)/2)^7)/4 + 18*tan(c/2 + (d*x)/2)^8 - (5*tan(c/2 + (d*x)/2)^9)/24 + 6*tan(c/2 + (d*x)/2)
^10 + (11*tan(c/2 + (d*x)/2)^11)/8 + 46/15)/(d*(a + 6*a*tan(c/2 + (d*x)/2)^2 + 15*a*tan(c/2 + (d*x)/2)^4 + 20*
a*tan(c/2 + (d*x)/2)^6 + 15*a*tan(c/2 + (d*x)/2)^8 + 6*a*tan(c/2 + (d*x)/2)^10 + a*tan(c/2 + (d*x)/2)^12))

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

[Out]

Timed out

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