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

Optimal. Leaf size=78 \[ \frac {3 \cot (c+d x)}{a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {9 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

[Out]

-9/2*arctanh(cos(d*x+c))/a^3/d+3*cot(d*x+c)/a^3/d-1/2*cot(d*x+c)*csc(d*x+c)/a^3/d+4*cos(d*x+c)/a^3/d/(1+sin(d*
x+c))

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Rubi [A]  time = 0.25, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {3 \cot (c+d x)}{a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {9 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (3*Cot[c + d*x])/(a^3*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^3*d) + (4*C
os[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 {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc ^3(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)-3 a \csc ^2(c+d x)+a \csc ^3(c+d x)-\frac {4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac {\int \csc ^3(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^2(c+d x) \, dx}{a^3}+\frac {4 \int \csc (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 {\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{2 a^3}+\frac {3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=-\frac {9 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 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 = 5.90, size = 213, normalized size = 2.73 \[ -\frac {\sin ^8\left (\frac {1}{2} (c+d x)\right ) \sin ^7(c+d x) \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )+2 \csc (c+d x)\right )^5 \left ((\csc (c+d x)-6) \csc ^6\left (\frac {1}{2} (c+d x)\right )-8 (\csc (c+d x)-6) \csc ^3(c+d x)+2 \csc (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (\csc (c+d x)-18 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6\right )-4 \csc ^2(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (\csc (c+d x)+18 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-38\right )\right )}{512 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/512*((Csc[(c + d*x)/2]^2 + 2*Csc[c + d*x])^5*(Csc[(c + d*x)/2]^6*(-6 + Csc[c + d*x]) - 8*(-6 + Csc[c + d*x]
)*Csc[c + d*x]^3 + 2*Csc[(c + d*x)/2]^4*Csc[c + d*x]*(-6 + Csc[c + d*x] + 18*Log[Cos[(c + d*x)/2]] - 18*Log[Si
n[(c + d*x)/2]]) - 4*Csc[(c + d*x)/2]^2*Csc[c + d*x]^2*(-38 + Csc[c + d*x] - 18*Log[Cos[(c + d*x)/2]] + 18*Log
[Sin[(c + d*x)/2]]))*Sin[(c + d*x)/2]^8*Sin[c + d*x]^7)/(a^3*d*(1 + Sin[c + d*x])^3)

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fricas [B]  time = 0.45, size = 246, normalized size = 3.15 \[ \frac {28 \, \cos \left (d x + c\right )^{3} + 18 \, \cos \left (d x + c\right )^{2} - 9 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (14 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) - 26 \, \cos \left (d x + c\right ) - 16}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(28*cos(d*x + c)^3 + 18*cos(d*x + c)^2 - 9*(cos(d*x + c)^3 + cos(d*x + c)^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) + 9*(cos(d*x + c)^3 + cos(d*x + c)^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*(14*cos(d*x + c)^2 + 5*cos(d*x + c) - 8)
*sin(d*x + c) - 26*cos(d*x + c) - 16)/(a^3*d*cos(d*x + c)^3 + 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)^2 - a^3*d)*sin(d*x + c))

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giac [A]  time = 0.25, size = 116, normalized size = 1.49 \[ \frac {\frac {36 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {64}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {54 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(36*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 64/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) - (54*tan(1/2*d*x + 1/2*c)^2
- 12*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^2) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x
+ 1/2*c))/a^6)/d

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maple [A]  time = 0.68, size = 115, normalized size = 1.47 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {9 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}+\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*tan(1/2*d*x+1/2*c)-1/8/d/a^3/tan(1/2*d*x+1/2*c)^2+3/2/d/a^3/tan(1/2*d
*x+1/2*c)+9/2/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 = 0.33, size = 161, normalized size = 2.06 \[ \frac {\frac {\frac {11 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {76 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{3}} + \frac {36 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((11*sin(d*x + c)/(cos(d*x + c) + 1) + 76*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a^3*sin(d*x + c)^2/(co
s(d*x + c) + 1)^2 + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) - (12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x +
c)^2/(cos(d*x + c) + 1)^2)/a^3 + 36*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 8.69, size = 120, normalized size = 1.54 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (9*log(tan(c/2 + (d*x)/2)))/(2*a^3*d) + ((11*tan(c/2 + (d*x)/2))/2 + 38*tan(c
/2 + (d*x)/2)^2 - 1/2)/(d*(4*a^3*tan(c/2 + (d*x)/2)^2 + 4*a^3*tan(c/2 + (d*x)/2)^3)) - (3*tan(c/2 + (d*x)/2))/
(2*a^3*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)**4*csc(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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