3.439 \(\int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=117 \[ \frac {\cot ^3(c+d x)}{a^3 d}+\frac {7 \cot (c+d x)}{a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {51 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac {19 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

[Out]

-51/8*arctanh(cos(d*x+c))/a^3/d+7*cot(d*x+c)/a^3/d+cot(d*x+c)^3/a^3/d-19/8*cot(d*x+c)*csc(d*x+c)/a^3/d-1/4*cot
(d*x+c)*csc(d*x+c)^3/a^3/d+4*cos(d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.30, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2875, 2872, 3770, 3767, 8, 3768, 2648} \[ \frac {\cot ^3(c+d x)}{a^3 d}+\frac {7 \cot (c+d x)}{a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {51 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac {19 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^4*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-51*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (7*Cot[c + d*x])/(a^3*d) + Cot[c + d*x]^3/(a^3*d) - (19*Cot[c + d*x]*C
sc[c + d*x])/(8*a^3*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d) + (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

Rule 8

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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 {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc ^5(c+d x) \sec ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (4 a \csc (c+d x)-4 a \csc ^2(c+d x)+4 a \csc ^3(c+d x)-3 a \csc ^4(c+d x)+a \csc ^5(c+d x)-\frac {4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac {\int \csc ^5(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^4(c+d x) \, dx}{a^3}+\frac {4 \int \csc (c+d x) \, dx}{a^3}-\frac {4 \int \csc ^2(c+d x) \, dx}{a^3}+\frac {4 \int \csc ^3(c+d x) \, dx}{a^3}-\frac {4 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {2 \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a^3}+\frac {2 \int \csc (c+d x) \, dx}{a^3}+\frac {3 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}+\frac {4 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=-\frac {6 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {7 \cot (c+d x)}{a^3 d}+\frac {\cot ^3(c+d x)}{a^3 d}-\frac {19 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {3 \int \csc (c+d x) \, dx}{8 a^3}\\ &=-\frac {51 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac {7 \cot (c+d x)}{a^3 d}+\frac {\cot ^3(c+d x)}{a^3 d}-\frac {19 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 6.16, size = 601, normalized size = 5.14 \[ -\frac {8 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}{d (a \sin (c+d x)+a)^3}-\frac {51 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a \sin (c+d x)+a)^3}+\frac {51 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a \sin (c+d x)+a)^3}-\frac {3 \tan \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{d (a \sin (c+d x)+a)^3}+\frac {3 \cot \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{d (a \sin (c+d x)+a)^3}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{64 d (a \sin (c+d x)+a)^3}-\frac {19 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{32 d (a \sin (c+d x)+a)^3}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{64 d (a \sin (c+d x)+a)^3}+\frac {19 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{32 d (a \sin (c+d x)+a)^3}+\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a \sin (c+d x)+a)^3}-\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{8 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^4*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-8*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)/(d*(a + a*Sin[c + d*x])^3) + (3*Cot[(c + d*x)/2]
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(d*(a + a*Sin[c + d*x])^3) - (19*Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^6)/(32*d*(a + a*Sin[c + d*x])^3) + (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^6)/(8*d*(a + a*Sin[c + d*x])^3) - (Csc[(c + d*x)/2]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^6)/(64*d*(a + a*Sin[c + d*x])^3) - (51*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(8*d
*(a + a*Sin[c + d*x])^3) + (51*Log[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(8*d*(a + a*Sin[
c + d*x])^3) + (19*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(32*d*(a + a*Sin[c + d*x])^3) +
(Sec[(c + d*x)/2]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(64*d*(a + a*Sin[c + d*x])^3) - (3*(Cos[(c + d*x
)/2] + Sin[(c + d*x)/2])^6*Tan[(c + d*x)/2])/(d*(a + a*Sin[c + d*x])^3) - (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^6*Tan[(c + d*x)/2])/(8*d*(a + a*Sin[c + d*x])^3)

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fricas [B]  time = 0.48, size = 381, normalized size = 3.26 \[ \frac {160 \, \cos \left (d x + c\right )^{5} + 102 \, \cos \left (d x + c\right )^{4} - 298 \, \cos \left (d x + c\right )^{3} - 170 \, \cos \left (d x + c\right )^{2} - 51 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 51 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (80 \, \cos \left (d x + c\right )^{4} + 29 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} - 35 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right ) + 134 \, \cos \left (d x + c\right ) + 64}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right ) + a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(160*cos(d*x + c)^5 + 102*cos(d*x + c)^4 - 298*cos(d*x + c)^3 - 170*cos(d*x + c)^2 - 51*(cos(d*x + c)^5 +
cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c) +
cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 51*(cos(d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2*co
s(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c) + cos(d*x + c) + 1)*log(-1/2*cos(d*x + c)
+ 1/2) - 2*(80*cos(d*x + c)^4 + 29*cos(d*x + c)^3 - 120*cos(d*x + c)^2 - 35*cos(d*x + c) + 32)*sin(d*x + c) +
134*cos(d*x + c) + 64)/(a^3*d*cos(d*x + c)^5 + a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x
+ c)^2 + a^3*d*cos(d*x + c) + a^3*d + (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 [A]  time = 0.28, size = 174, normalized size = 1.49 \[ \frac {\frac {408 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {512}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {850 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 200 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} + \frac {a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 200 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{12}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/64*(408*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 512/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) - (850*tan(1/2*d*x + 1/2*c
)^4 - 200*tan(1/2*d*x + 1/2*c)^3 + 40*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x +
1/2*c)^4) + (a^9*tan(1/2*d*x + 1/2*c)^4 - 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 40*a^9*tan(1/2*d*x + 1/2*c)^2 - 200*a
^9*tan(1/2*d*x + 1/2*c))/a^12)/d

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maple [A]  time = 0.68, size = 191, normalized size = 1.63 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \,a^{3}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {1}{64 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {5}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {25}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {51 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}+\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x)

[Out]

1/64/d/a^3*tan(1/2*d*x+1/2*c)^4-1/8/d/a^3*tan(1/2*d*x+1/2*c)^3+5/8/d/a^3*tan(1/2*d*x+1/2*c)^2-25/8/d/a^3*tan(1
/2*d*x+1/2*c)-1/64/d/a^3/tan(1/2*d*x+1/2*c)^4+1/8/d/a^3/tan(1/2*d*x+1/2*c)^3-5/8/d/a^3/tan(1/2*d*x+1/2*c)^2+25
/8/d/a^3/tan(1/2*d*x+1/2*c)+51/8/d/a^3*ln(tan(1/2*d*x+1/2*c))+8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.38, size = 241, normalized size = 2.06 \[ \frac {\frac {\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {32 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {160 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {712 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} - \frac {\frac {200 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{3}} + \frac {408 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/64*((7*sin(d*x + c)/(cos(d*x + c) + 1) - 32*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 160*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 + 712*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)/(a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*si
n(d*x + c)^5/(cos(d*x + c) + 1)^5) - (200*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^3 + 408*log(sin(d*x + c)
/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 9.13, size = 176, normalized size = 1.50 \[ \frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}+\frac {51\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^3\,d}-\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}-\frac {1}{64}\right )}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^5*(a + a*sin(c + d*x))^3),x)

[Out]

(5*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - tan(c/2 + (d*x)/2)^3/(8*a^3*d) + tan(c/2 + (d*x)/2)^4/(64*a^3*d) + (51*lo
g(tan(c/2 + (d*x)/2)))/(8*a^3*d) - (25*tan(c/2 + (d*x)/2))/(8*a^3*d) + (cot(c/2 + (d*x)/2)^4*((7*tan(c/2 + (d*
x)/2))/64 - tan(c/2 + (d*x)/2)^2/2 + (5*tan(c/2 + (d*x)/2)^3)/2 + (89*tan(c/2 + (d*x)/2)^4)/8 - 1/64))/(a^3*d*
(tan(c/2 + (d*x)/2) + 1))

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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)**4*csc(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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