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

Optimal. Leaf size=45 \[ \frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {x}{a^3} \]

[Out]

x/a^3-arctanh(cos(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 = 0.18, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2875, 2872, 3770, 2648} \[ \frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^3 - ArcTanh[Cos[c + d*x]]/(a^3*d) + (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*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 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 ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc (c+d x) \sec ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (a+a \csc (c+d x)-\frac {4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac {x}{a^3}+\frac {\int \csc (c+d x) \, dx}{a^3}-\frac {4 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac {x}{a^3}-\frac {\tanh ^{-1}(\cos (c+d x))}{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 = 0.28, size = 122, normalized size = 2.71 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x-8\right )\right )}{a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2]*(c + d*x - Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x
)/2]]) + (-8 + c + d*x - Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]])*Sin[(c + d*x)/2]))/(a^3*d*(1 + Sin[c +
d*x])^3)

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fricas [B]  time = 0.46, size = 117, normalized size = 2.60 \[ \frac {2 \, d x + 2 \, {\left (d x + 4\right )} \cos \left (d x + c\right ) - {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (d x - 4\right )} \sin \left (d x + c\right ) + 8}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d*x + 2*(d*x + 4)*cos(d*x + c) - (cos(d*x + c) + sin(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + (cos(d
*x + c) + sin(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*(d*x - 4)*sin(d*x + c) + 8)/(a^3*d*cos(d*x + c) +
a^3*d*sin(d*x + c) + a^3*d)

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giac [A]  time = 0.21, size = 47, normalized size = 1.04 \[ \frac {\frac {d x + c}{a^{3}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {8}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)/a^3 + log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 8/(a^3*(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.58, size = 58, normalized size = 1.29 \[ \frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3*arctan(tan(1/2*d*x+1/2*c))+1/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 [A]  time = 0.42, size = 78, normalized size = 1.73 \[ \frac {\frac {8}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1)) + 2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + log(sin(d*x +
c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 8.73, size = 115, normalized size = 2.56 \[ \frac {8}{d\,\left (a^3+a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {2\,\mathrm {atan}\left (\frac {4\,a^3}{4\,a^3-4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3-4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/(d*(a^3 + a^3*tan(c/2 + (d*x)/2))) + log(tan(c/2 + (d*x)/2))/(a^3*d) + (2*atan((4*a^3)/(4*a^3 - 4*a^3*tan(c/
2 + (d*x)/2)) + (4*a^3*tan(c/2 + (d*x)/2))/(4*a^3 - 4*a^3*tan(c/2 + (d*x)/2))))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)**4*csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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