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

Optimal. Leaf size=92 \[ -\frac {\cos (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {3 \cot (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {3 x}{a^3} \]

[Out]

-3*x/a^3-1/2*arctanh(cos(d*x+c))/a^3/d-cos(d*x+c)/a^3/d-3*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 = 0.27, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2875, 2872, 3770, 3767, 8, 3768, 2638} \[ -\frac {\cos (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {3 \cot (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {3 x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/a^3 - ArcTanh[Cos[c + d*x]]/(2*a^3*d) - Cos[c + d*x]/(a^3*d) - (3*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^
3/(3*a^3*d) + (3*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 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 ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \cot ^2(c+d x) \csc ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (-3 a^5+2 a^5 \csc (c+d x)+2 a^5 \csc ^2(c+d x)-3 a^5 \csc ^3(c+d x)+a^5 \csc ^4(c+d x)+a^5 \sin (c+d x)\right ) \, dx}{a^8}\\ &=-\frac {3 x}{a^3}+\frac {\int \csc ^4(c+d x) \, dx}{a^3}+\frac {\int \sin (c+d x) \, dx}{a^3}+\frac {2 \int \csc (c+d x) \, dx}{a^3}+\frac {2 \int \csc ^2(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^3(c+d x) \, dx}{a^3}\\ &=-\frac {3 x}{a^3}-\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {\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 {2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=-\frac {3 x}{a^3}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cos (c+d x)}{a^3 d}-\frac {3 \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 = 2.65, size = 132, normalized size = 1.43 \[ \frac {\csc ^3(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (2 (3 \sin (c+d x)+8) \cos (3 (c+d x))+6 (5 \sin (c+d x)-4) \cos (c+d x)-12 \sin ^3(c+d x) \left (6 (c+d x)-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{24 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-12*(6*(c + d*x) + Log[Cos[(c + d*x)/2]] - Log[Sin[(c
+ d*x)/2]])*Sin[c + d*x]^3 + 2*Cos[3*(c + d*x)]*(8 + 3*Sin[c + d*x]) + 6*Cos[c + d*x]*(-4 + 5*Sin[c + d*x])))
/(24*a^3*d*(1 + Sin[c + d*x])^3)

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fricas [A]  time = 0.48, size = 150, normalized size = 1.63 \[ -\frac {32 \, \cos \left (d x + c\right )^{3} + 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (6 \, d x \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right )^{3} - 6 \, d x + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 36 \, \cos \left (d x + c\right )}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(32*cos(d*x + c)^3 + 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(cos(d*x + c)^2
- 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(6*d*x*cos(d*x + c)^2 + 2*cos(d*x + c)^3 - 6*d*x + cos(d*x
+ c))*sin(d*x + c) - 36*cos(d*x + c))/((a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))

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giac [A]  time = 0.27, size = 157, normalized size = 1.71 \[ -\frac {\frac {72 \, {\left (d x + c\right )}}{a^{3}} - \frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {48}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {22 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 33 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/24*(72*(d*x + c)/a^3 - 12*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 48/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + (22*
tan(1/2*d*x + 1/2*c)^3 + 33*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 + 33*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d

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maple [A]  time = 0.70, size = 173, normalized size = 1.88 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{3}}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{3} d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {2}{a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}-\frac {1}{24 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {11}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*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/d/a^3*tan(1/2*d*x+1/2*c)^2+11/8/d/a^3*tan(1/2*d*x+1/2*c)-2/d/a^3/(1+tan(1/
2*d*x+1/2*c)^2)-6/a^3/d*arctan(tan(1/2*d*x+1/2*c))-1/24/d/a^3/tan(1/2*d*x+1/2*c)^3+3/8/d/a^3/tan(1/2*d*x+1/2*c
)^2-11/8/d/a^3/tan(1/2*d*x+1/2*c)+1/2/d/a^3*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 242, normalized size = 2.63 \[ \frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {34 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {39 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {33 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac {144 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 34*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 39*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 33*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)/(a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(
d*x + c)^5/(cos(d*x + c) + 1)^5) + (33*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 - 144*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 12*log(sin(d*x
+ c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 9.26, size = 219, normalized size = 2.38 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {6\,\mathrm {atan}\left (\frac {36}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{36\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}}{d\,\left (8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(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) + (6*atan(36/(36*tan(c/2 + (d*x)/2) + 6)
- (6*tan(c/2 + (d*x)/2))/(36*tan(c/2 + (d*x)/2) + 6)))/(a^3*d) + log(tan(c/2 + (d*x)/2))/(2*a^3*d) + (11*tan(c
/2 + (d*x)/2))/(8*a^3*d) - ((34*tan(c/2 + (d*x)/2)^2)/3 - 3*tan(c/2 + (d*x)/2) + 13*tan(c/2 + (d*x)/2)^3 + 11*
tan(c/2 + (d*x)/2)^4 + 1/3)/(d*(8*a^3*tan(c/2 + (d*x)/2)^3 + 8*a^3*tan(c/2 + (d*x)/2)^5))

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

[Out]

Timed out

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