3.639 \(\int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=73 \[ \frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2875, 2872, 3767, 8, 3768, 3770, 2638} \[ \frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*x)/a^2 - ArcTanh[Cos[c + d*x]]/(2*a^2*d) + Cos[c + d*x]/(a^2*d) + (2*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*
Csc[c + d*x])/(2*a^2*d)

Rule 8

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

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Mathematica [A]  time = 0.58, size = 134, normalized size = 1.84 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (16 (c+d x)+8 \cos (c+d x)-8 \tan \left (\frac {1}{2} (c+d x)\right )+8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(16*(c + d*x) + 8*Cos[c + d*x] + 8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2
]^2 - 4*Log[Cos[(c + d*x)/2]] + 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 8*Tan[(c + d*x)/2]))/(8*d*(a +
a*Sin[c + d*x])^2)

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fricas [A]  time = 0.76, size = 118, normalized size = 1.62 \[ \frac {8 \, d x \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right )^{3} - 8 \, d x - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 8 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 128, normalized size = 1.75 \[ \frac {\frac {16 \, {\left (d x + c\right )}}{a^{2}} + \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {16}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(16*(d*x + c)/a^2 + 4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x
+ 1/2*c))/a^4 + 16/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) - (6*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) +
1)/(a^2*tan(1/2*d*x + 1/2*c)^2))/d

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maple [A]  time = 0.56, size = 134, normalized size = 1.84 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2}}+\frac {2}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)+4/d/a^2*arctan(tan(
1/2*d*x+1/2*c))-1/8/a^2/d/tan(1/2*d*x+1/2*c)^2+1/d/a^2/tan(1/2*d*x+1/2*c)+1/2/d/a^2*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.41, size = 204, normalized size = 2.79 \[ \frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {32 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((8*sin(d*x + c)/(cos(d*x + c) + 1) + 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 1)/(a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - (8*sin(d
*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 32*arctan(sin(d*x + c)/(cos(d*x + c) +
1))/a^2 + 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 9.10, size = 186, normalized size = 2.55 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {4\,\mathrm {atan}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (4*atan((4*tan(c/2 + (d*x)/2))/(16*tan(c/2 + (d*x)/2) - 4) + 16/(16*tan(c/2 +
(d*x)/2) - 4)))/(a^2*d) + log(tan(c/2 + (d*x)/2))/(2*a^2*d) + (4*tan(c/2 + (d*x)/2) + (15*tan(c/2 + (d*x)/2)^
2)/2 + 4*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^4)) - tan(c/2 +
(d*x)/2)/(a^2*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)**6*csc(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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