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

Optimal. Leaf size=60 \[ \frac {3 \cot (c+d x)}{a^3 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {x}{a^3} \]

[Out]

-x/a^3-7/2*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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {3 \cot (c+d x)}{a^3 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^3) - (7*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)

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

________________________________________________________________________________________

Mathematica [B]  time = 0.47, size = 126, normalized size = 2.10 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (-8 (c+d x)-12 \tan \left (\frac {1}{2} (c+d x)\right )+12 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+28 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-28 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-8*(c + d*x) + 12*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 28*Log[Cos
[(c + d*x)/2]] + 28*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 12*Tan[(c + d*x)/2]))/(8*d*(a + a*Sin[c + d*x
])^3)

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 109, normalized size = 1.82 \[ -\frac {4 \, d x \cos \left (d x + c\right )^{2} - 4 \, d x + 7 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 7 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

-1/4*(4*d*x*cos(d*x + c)^2 - 4*d*x + 7*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - 7*(cos(d*x + c)^2 -
1)*log(-1/2*cos(d*x + c) + 1/2) + 12*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))/(a^3*d*cos(d*x + c)^2 - a^3*d
)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 108, normalized size = 1.80 \[ -\frac {\frac {8 \, {\left (d x + c\right )}}{a^{3}} - \frac {28 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*(d*x + c)/a^3 - 28*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (42*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

________________________________________________________________________________________

maple [A]  time = 0.68, size = 112, normalized size = 1.87 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3} d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}-\frac {1}{8 a^{3} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{3} d} \]

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))^3,x)

[Out]

1/8/a^3/d*tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*tan(1/2*d*x+1/2*c)-2/a^3/d*arctan(tan(1/2*d*x+1/2*c))-1/8/a^3/d/tan(1
/2*d*x+1/2*c)^2+3/2/d/a^3/tan(1/2*d*x+1/2*c)+7/2/a^3/d*ln(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

maxima [B]  time = 0.41, size = 138, normalized size = 2.30 \[ -\frac {\frac {\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{3}} + \frac {16 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {28 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (\frac {12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{8 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/8*((12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 16*arctan(sin(d*x + c)/
(cos(d*x + c) + 1))/a^3 - 28*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (12*sin(d*x + c)/(cos(d*x + c) + 1) -
1)*(cos(d*x + c) + 1)^2/(a^3*sin(d*x + c)^2))/d

________________________________________________________________________________________

mupad [B]  time = 9.35, size = 161, normalized size = 2.68 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {2\,\mathrm {atan}\left (\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{7\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d}+\frac {7\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,a^3\,d}+\frac {3\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(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) - cot(c/2 + (d*x)/2)^2/(8*a^3*d) + (2*atan((2*cos(c/2 + (d*x)/2) - 7*sin(c/2 +
(d*x)/2))/(7*cos(c/2 + (d*x)/2) + 2*sin(c/2 + (d*x)/2))))/(a^3*d) + (7*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2)))/(2*a^3*d) + (3*cot(c/2 + (d*x)/2))/(2*a^3*d) - (3*tan(c/2 + (d*x)/2))/(2*a^3*d)

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________