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

Optimal. Leaf size=138 \[ -\frac {15 \cos (c+d x)}{8 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac {15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {x}{a} \]

[Out]

-x/a+15/8*arctanh(cos(d*x+c))/a/d-15/8*cos(d*x+c)/a/d-cot(d*x+c)/a/d-5/8*cos(d*x+c)*cot(d*x+c)^2/a/d+1/3*cot(d
*x+c)^3/a/d+1/4*cos(d*x+c)*cot(d*x+c)^4/a/d-1/5*cot(d*x+c)^5/a/d

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Rubi [A]  time = 0.15, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2839, 3473, 8, 2592, 288, 321, 206} \[ -\frac {15 \cos (c+d x)}{8 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac {15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) + (15*ArcTanh[Cos[c + d*x]])/(8*a*d) - (15*Cos[c + d*x])/(8*a*d) - Cot[c + d*x]/(a*d) - (5*Cos[c + d*x]
*Cot[c + d*x]^2)/(8*a*d) + Cot[c + d*x]^3/(3*a*d) + (Cos[c + d*x]*Cot[c + d*x]^4)/(4*a*d) - Cot[c + d*x]^5/(5*
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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \cos (c+d x) \cot ^5(c+d x) \, dx}{a}+\frac {\int \cot ^6(c+d x) \, dx}{a}\\ &=-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\int \cot ^4(c+d x) \, dx}{a}+\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {\int \cot ^2(c+d x) \, dx}{a}-\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{4 a d}\\ &=-\frac {\cot (c+d x)}{a d}-\frac {5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\int 1 \, dx}{a}+\frac {15 \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a d}\\ &=-\frac {x}{a}-\frac {15 \cos (c+d x)}{8 a d}-\frac {\cot (c+d x)}{a d}-\frac {5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{8 a d}\\ &=-\frac {x}{a}+\frac {15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {15 \cos (c+d x)}{8 a d}-\frac {\cot (c+d x)}{a d}-\frac {5 \cos (c+d x) \cot ^2(c+d x)}{8 a d}+\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cos (c+d x) \cot ^4(c+d x)}{4 a d}-\frac {\cot ^5(c+d x)}{5 a d}\\ \end {align*}

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Mathematica [A]  time = 1.00, size = 264, normalized size = 1.91 \[ -\frac {\csc ^5(c+d x) \left (1200 c \sin (c+d x)+1200 d x \sin (c+d x)+600 \sin (2 (c+d x))-600 c \sin (3 (c+d x))-600 d x \sin (3 (c+d x))-510 \sin (4 (c+d x))+120 c \sin (5 (c+d x))+120 d x \sin (5 (c+d x))+60 \sin (6 (c+d x))+400 \cos (c+d x)-200 \cos (3 (c+d x))+184 \cos (5 (c+d x))+2250 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1125 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+225 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2250 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1125 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-225 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1920 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1920*(Csc[c + d*x]^5*(400*Cos[c + d*x] - 200*Cos[3*(c + d*x)] + 184*Cos[5*(c + d*x)] + 1200*c*Sin[c + d*x]
+ 1200*d*x*Sin[c + d*x] - 2250*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 2250*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] +
600*Sin[2*(c + d*x)] - 600*c*Sin[3*(c + d*x)] - 600*d*x*Sin[3*(c + d*x)] + 1125*Log[Cos[(c + d*x)/2]]*Sin[3*(c
+ d*x)] - 1125*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 510*Sin[4*(c + d*x)] + 120*c*Sin[5*(c + d*x)] + 120*d
*x*Sin[5*(c + d*x)] - 225*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] + 225*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]
+ 60*Sin[6*(c + d*x)]))/(a*d)

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fricas [A]  time = 0.48, size = 211, normalized size = 1.53 \[ -\frac {368 \, \cos \left (d x + c\right )^{5} - 560 \, \cos \left (d x + c\right )^{3} - 225 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 225 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (8 \, d x \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{5} - 16 \, d x \cos \left (d x + c\right )^{2} - 25 \, \cos \left (d x + c\right )^{3} + 8 \, d x + 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 240 \, \cos \left (d x + c\right )}{240 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

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)),x, algorithm="fricas")

[Out]

-1/240*(368*cos(d*x + c)^5 - 560*cos(d*x + c)^3 - 225*(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) + 225*(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*(8*d*x*cos(d*x + c)^4 + 8*cos(d*x + c)^5 - 16*d*x*cos(d*x + c)^2 - 25*cos(d*x + c)^3 + 8*d*x + 15*cos(
d*x + c))*sin(d*x + c) + 240*cos(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))

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giac [A]  time = 0.22, size = 217, normalized size = 1.57 \[ -\frac {\frac {960 \, {\left (d x + c\right )}}{a} + \frac {1800 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {1920}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a} - \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 660 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {4110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 660 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

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)),x, algorithm="giac")

[Out]

-1/960*(960*(d*x + c)/a + 1800*log(abs(tan(1/2*d*x + 1/2*c)))/a + 1920/((tan(1/2*d*x + 1/2*c)^2 + 1)*a) - (6*a
^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*tan(1/2*d*x + 1/2*c)^4 - 70*a^4*tan(1/2*d*x + 1/2*c)^3 + 240*a^4*tan(1/2*d*
x + 1/2*c)^2 + 660*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (4110*tan(1/2*d*x + 1/2*c)^5 - 660*tan(1/2*d*x + 1/2*c)^4 -
240*tan(1/2*d*x + 1/2*c)^3 + 70*tan(1/2*d*x + 1/2*c)^2 + 15*tan(1/2*d*x + 1/2*c) - 6)/(a*tan(1/2*d*x + 1/2*c)
^5))/d

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maple [A]  time = 0.52, size = 249, normalized size = 1.80 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 a d}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}-\frac {2}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {1}{160 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{64 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {7}{96 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {1}{4 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {11}{16 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d} \]

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)),x)

[Out]

1/160/a/d*tan(1/2*d*x+1/2*c)^5-1/64/a/d*tan(1/2*d*x+1/2*c)^4-7/96/a/d*tan(1/2*d*x+1/2*c)^3+1/4/a/d*tan(1/2*d*x
+1/2*c)^2+11/16/a/d*tan(1/2*d*x+1/2*c)-2/a/d/(1+tan(1/2*d*x+1/2*c)^2)-2/a/d*arctan(tan(1/2*d*x+1/2*c))-1/160/a
/d/tan(1/2*d*x+1/2*c)^5+1/64/a/d/tan(1/2*d*x+1/2*c)^4+7/96/a/d/tan(1/2*d*x+1/2*c)^3-1/4/a/d/tan(1/2*d*x+1/2*c)
^2-11/16/a/d/tan(1/2*d*x+1/2*c)-15/8/a/d*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 319, normalized size = 2.31 \[ \frac {\frac {\frac {660 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {240 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} + \frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {64 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {225 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {590 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2160 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {660 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 6}{\frac {a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} - \frac {1920 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {1800 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{960 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/960*((660*sin(d*x + c)/(cos(d*x + c) + 1) + 240*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 70*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a + (15*sin
(d*x + c)/(cos(d*x + c) + 1) + 64*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 225*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 - 590*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2160*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 660*sin(d*x + c)^6/(c
os(d*x + c) + 1)^6 - 6)/(a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - 1920
*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 1800*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 9.03, size = 279, normalized size = 2.02 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a\,d}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}+\frac {2\,\mathrm {atan}\left (\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {15}{2}\right )}+\frac {4}{4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {15}{2}}\right )}{a\,d}-\frac {15\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+72\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {59\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}}{d\,\left (32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d} \]

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))),x)

[Out]

tan(c/2 + (d*x)/2)^2/(4*a*d) - (7*tan(c/2 + (d*x)/2)^3)/(96*a*d) - tan(c/2 + (d*x)/2)^4/(64*a*d) + tan(c/2 + (
d*x)/2)^5/(160*a*d) + (2*atan((15*tan(c/2 + (d*x)/2))/(2*(4*tan(c/2 + (d*x)/2) - 15/2)) + 4/(4*tan(c/2 + (d*x)
/2) - 15/2)))/(a*d) - (15*log(tan(c/2 + (d*x)/2)))/(8*a*d) - ((15*tan(c/2 + (d*x)/2)^3)/2 - (32*tan(c/2 + (d*x
)/2)^2)/15 - tan(c/2 + (d*x)/2)/2 + (59*tan(c/2 + (d*x)/2)^4)/3 + 72*tan(c/2 + (d*x)/2)^5 + 22*tan(c/2 + (d*x)
/2)^6 + 1/5)/(d*(32*a*tan(c/2 + (d*x)/2)^5 + 32*a*tan(c/2 + (d*x)/2)^7)) + (11*tan(c/2 + (d*x)/2))/(16*a*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)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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