3.712 \(\int \frac {\cos ^5(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=150 \[ -\frac {5 \cos ^3(c+d x)}{6 a d}-\frac {5 \cos (c+d x)}{2 a d}+\frac {15 \cot (c+d x)}{8 a d}-\frac {\cos ^4(c+d x) \cot (c+d x)}{4 a d}-\frac {\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}-\frac {5 \cos ^2(c+d x) \cot (c+d x)}{8 a d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {15 x}{8 a} \]

[Out]

15/8*x/a+5/2*arctanh(cos(d*x+c))/a/d-5/2*cos(d*x+c)/a/d-5/6*cos(d*x+c)^3/a/d+15/8*cot(d*x+c)/a/d-5/8*cos(d*x+c
)^2*cot(d*x+c)/a/d-1/4*cos(d*x+c)^4*cot(d*x+c)/a/d-1/2*cos(d*x+c)^3*cot(d*x+c)^2/a/d

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Rubi [A]  time = 0.19, antiderivative size = 150, 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 = {2839, 2592, 288, 302, 206, 2591, 321, 203} \[ -\frac {5 \cos ^3(c+d x)}{6 a d}-\frac {5 \cos (c+d x)}{2 a d}+\frac {15 \cot (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \cot ^2(c+d x)}{2 a d}-\frac {\cos ^4(c+d x) \cot (c+d x)}{4 a d}-\frac {5 \cos ^2(c+d x) \cot (c+d x)}{8 a d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {15 x}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

(15*x)/(8*a) + (5*ArcTanh[Cos[c + d*x]])/(2*a*d) - (5*Cos[c + d*x])/(2*a*d) - (5*Cos[c + d*x]^3)/(6*a*d) + (15
*Cot[c + d*x])/(8*a*d) - (5*Cos[c + d*x]^2*Cot[c + d*x])/(8*a*d) - (Cos[c + d*x]^4*Cot[c + d*x])/(4*a*d) - (Co
s[c + d*x]^3*Cot[c + d*x]^2)/(2*a*d)

Rule 203

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

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 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 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 2591

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.53, size = 179, normalized size = 1.19 \[ -\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-285 \sin (2 (c+d x))+42 \sin (4 (c+d x))+3 \sin (6 (c+d x))+400 \cos (c+d x)-200 \cos (3 (c+d x))-8 \cos (5 (c+d x))+480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 \cos (2 (c+d x)) \left (-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 c+3 d x\right )-360 c-360 d x\right )}{1536 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]

[Out]

-1/1536*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(-360*c - 360*d*x + 400*Cos[c + d*x] - 200*Cos[3*(c + d*x)] -
8*Cos[5*(c + d*x)] - 480*Log[Cos[(c + d*x)/2]] + 120*Cos[2*(c + d*x)]*(3*c + 3*d*x + 4*Log[Cos[(c + d*x)/2]]
- 4*Log[Sin[(c + d*x)/2]]) + 480*Log[Sin[(c + d*x)/2]] - 285*Sin[2*(c + d*x)] + 42*Sin[4*(c + d*x)] + 3*Sin[6*
(c + d*x)]))/(a*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.49, size = 148, normalized size = 0.99 \[ -\frac {8 \, \cos \left (d x + c\right )^{5} - 45 \, d x \cos \left (d x + c\right )^{2} + 40 \, \cos \left (d x + c\right )^{3} + 45 \, d x - 30 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 30 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{3} - 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 60 \, \cos \left (d x + c\right )}{24 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

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)),x, algorithm="fricas")

[Out]

-1/24*(8*cos(d*x + c)^5 - 45*d*x*cos(d*x + c)^2 + 40*cos(d*x + c)^3 + 45*d*x - 30*(cos(d*x + c)^2 - 1)*log(1/2
*cos(d*x + c) + 1/2) + 30*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) - 3*(2*cos(d*x + c)^5 + 5*cos(d*x
+ c)^3 - 15*cos(d*x + c))*sin(d*x + c) - 60*cos(d*x + c))/(a*d*cos(d*x + c)^2 - a*d)

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giac [A]  time = 0.21, size = 216, normalized size = 1.44 \[ \frac {\frac {45 \, {\left (d x + c\right )}}{a} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{2}} + \frac {3 \, {\left (30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {2 \, {\left (27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 168 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 152 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 56\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

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)),x, algorithm="giac")

[Out]

1/24*(45*(d*x + c)/a - 60*log(abs(tan(1/2*d*x + 1/2*c)))/a + 3*(a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1
/2*c))/a^2 + 3*(30*tan(1/2*d*x + 1/2*c)^2 + 4*tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c)^2) - 2*(27*tan
(1/2*d*x + 1/2*c)^7 + 72*tan(1/2*d*x + 1/2*c)^6 + 3*tan(1/2*d*x + 1/2*c)^5 + 168*tan(1/2*d*x + 1/2*c)^4 - 3*ta
n(1/2*d*x + 1/2*c)^3 + 152*tan(1/2*d*x + 1/2*c)^2 - 27*tan(1/2*d*x + 1/2*c) + 56)/((tan(1/2*d*x + 1/2*c)^2 + 1
)^4*a))/d

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maple [B]  time = 0.56, size = 371, normalized size = 2.47 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {9 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {6 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {14 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {38 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {14}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d} \]

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)),x)

[Out]

1/8/a/d*tan(1/2*d*x+1/2*c)^2-1/2/a/d*tan(1/2*d*x+1/2*c)-9/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^
7-6/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6-1/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^
5-14/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4+1/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)
^3-38/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2+9/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)-14/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4+15/4/a/d*arctan(tan(1/2*d*x+1/2*c))-1/8/a/d/tan(1/2*d*x+1/2*c)^2+1/2/a
/d/tan(1/2*d*x+1/2*c)-5/2/a/d*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.42, size = 383, normalized size = 2.55 \[ \frac {\frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {124 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {102 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {322 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {78 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {348 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {42 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {147 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {42 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 3}{\frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {3 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a} + \frac {90 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{24 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/24*((12*sin(d*x + c)/(cos(d*x + c) + 1) - 124*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 102*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 322*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 78*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 348*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 42*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 147*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 - 42*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 3)/(a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a*sin(d*x + c)^4/
(cos(d*x + c) + 1)^4 + 6*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a*s
in(d*x + c)^10/(cos(d*x + c) + 1)^10) - 3*(4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) +
1)^2)/a + 90*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 9.02, size = 303, normalized size = 2.02 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {15\,\mathrm {atan}\left (\frac {225}{16\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {75}{4}\right )}-\frac {75\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {225\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}+\frac {75}{4}\right )}\right )}{4\,a\,d}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {49\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \]

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))),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - (15*atan(225/(16*((225*tan(c/2 + (d*x)/2))/16 + 75/4)) - (75*tan(c/2 + (d*x)/2)
)/(4*((225*tan(c/2 + (d*x)/2))/16 + 75/4))))/(4*a*d) - (5*log(tan(c/2 + (d*x)/2)))/(2*a*d) - ((62*tan(c/2 + (d
*x)/2)^2)/3 - 2*tan(c/2 + (d*x)/2) - 17*tan(c/2 + (d*x)/2)^3 + (161*tan(c/2 + (d*x)/2)^4)/3 - 13*tan(c/2 + (d*
x)/2)^5 + 58*tan(c/2 + (d*x)/2)^6 - 7*tan(c/2 + (d*x)/2)^7 + (49*tan(c/2 + (d*x)/2)^8)/2 + 7*tan(c/2 + (d*x)/2
)^9 + 1/2)/(d*(4*a*tan(c/2 + (d*x)/2)^2 + 16*a*tan(c/2 + (d*x)/2)^4 + 24*a*tan(c/2 + (d*x)/2)^6 + 16*a*tan(c/2
+ (d*x)/2)^8 + 4*a*tan(c/2 + (d*x)/2)^10)) - tan(c/2 + (d*x)/2)/(2*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)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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