3.650 \(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=72 \[ -\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2869, 2757, 3770, 3767, 8, 3768} \[ -\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(5*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (4*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a^3*d) + (3*Cot[c + d*x]*Cs
c[c + d*x])/(2*a^3*d)

Rule 8

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

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 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 ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (-a^3 \csc (c+d x)+3 a^3 \csc ^2(c+d x)-3 a^3 \csc ^3(c+d x)+a^3 \csc ^4(c+d x)\right ) \, dx}{a^6}\\ &=-\frac {\int \csc (c+d x) \, dx}{a^3}+\frac {\int \csc ^4(c+d x) \, dx}{a^3}+\frac {3 \int \csc ^2(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^3(c+d x) \, dx}{a^3}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {3 \int \csc (c+d x) \, dx}{2 a^3}-\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 115, normalized size = 1.60 \[ -\frac {\csc ^3(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (-18 \sin (2 (c+d x))+30 \cos (c+d x)-22 \cos (3 (c+d x))-60 \sin ^3(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{24 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/24*(Csc[c + d*x]^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(30*Cos[c + d*x] - 22*Cos[3*(c + d*x)] - 60*(Log
[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3 - 18*Sin[2*(c + d*x)]))/(a^3*d*(1 + Sin[c + d*x])^3
)

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fricas [A]  time = 0.77, size = 123, normalized size = 1.71 \[ -\frac {44 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 18 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 48 \, \cos \left (d x + c\right )}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(44*cos(d*x + c)^3 - 15*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(cos(d*x + c)
^2 - 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 18*cos(d*x + c)*sin(d*x + c) - 48*cos(d*x + c))/((a^3*d*co
s(d*x + c)^2 - a^3*d)*sin(d*x + c))

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giac [A]  time = 0.24, size = 128, normalized size = 1.78 \[ -\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/24*(60*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (110*tan(1/2*d*x + 1/2*c)^3 - 45*tan(1/2*d*x + 1/2*c)^2 + 9*tan
(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)^3) - (a^6*tan(1/2*d*x + 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)
^2 + 45*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d

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maple [A]  time = 0.68, size = 132, normalized size = 1.83 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{3}}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{3} d}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {15}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{3} d}+\frac {3}{8 a^{3} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{24 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

1/24/d/a^3*tan(1/2*d*x+1/2*c)^3-3/8/a^3/d*tan(1/2*d*x+1/2*c)^2+15/8/d/a^3*tan(1/2*d*x+1/2*c)-15/8/d/a^3/tan(1/
2*d*x+1/2*c)-5/2/a^3/d*ln(tan(1/2*d*x+1/2*c))+3/8/a^3/d/tan(1/2*d*x+1/2*c)^2-1/24/d/a^3/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.32, size = 153, normalized size = 2.12 \[ \frac {\frac {\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {{\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {45 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{3} \sin \left (d x + c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*((45*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + (9*sin(d*x + c)/(cos(d*x + c) + 1) - 45*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a^3*sin(d*x + c)^3))/d

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mupad [B]  time = 9.25, size = 119, normalized size = 1.65 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^4*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - (5*log(tan(c/2 + (d*x)/2)))/(2*a^3*d) +
(15*tan(c/2 + (d*x)/2))/(8*a^3*d) - (cot(c/2 + (d*x)/2)^3*(15*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2) + 1
/3))/(8*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)**6*csc(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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