3.641 \(\int \frac {\cos (c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=82 \[ \frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

[Out]

5/8*arctanh(cos(d*x+c))/a^2/d+2/3*cot(d*x+c)^3/a^2/d-3/8*cot(d*x+c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*csc(d*x+c)
^3/a^2/d

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Rubi [A]  time = 0.30, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2875, 2873, 2611, 3770, 2607, 30, 3768} \[ \frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {5 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]

[Out]

(5*ArcTanh[Cos[c + d*x]])/(8*a^2*d) + (2*Cot[c + d*x]^3)/(3*a^2*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) -
(Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^2*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^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] && IGtQ[m, 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 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 (c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^2(c+d x) \csc (c+d x)-2 a^2 \cot ^2(c+d x) \csc ^2(c+d x)+a^2 \cot ^2(c+d x) \csc ^3(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^2(c+d x) \csc (c+d x) \, dx}{a^2}+\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx}{a^2}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\int \csc ^3(c+d x) \, dx}{4 a^2}-\frac {\int \csc (c+d x) \, dx}{2 a^2}-\frac {2 \operatorname {Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\int \csc (c+d x) \, dx}{8 a^2}\\ &=\frac {5 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 116, normalized size = 1.41 \[ \frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (24 \sin (2 (c+d x))-33 \cos (c+d x)+(16 \sin (c+d x)+9) \cos (3 (c+d x))+60 \sin ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{1536 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]

[Out]

((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^4*(-33*Cos[c + d*x] + 60*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]
)*Sin[c + d*x]^4 + Cos[3*(c + d*x)]*(9 + 16*Sin[c + d*x]) + 24*Sin[2*(c + d*x)]))/(1536*a^2*d*(1 + Sin[c + d*x
])^2)

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fricas [A]  time = 0.72, size = 138, normalized size = 1.68 \[ \frac {32 \, \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + 18 \, \cos \left (d x + c\right )^{3} + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, \cos \left (d x + c\right )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(32*cos(d*x + c)^3*sin(d*x + c) + 18*cos(d*x + c)^3 + 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*
cos(d*x + c) + 1/2) - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 30*cos(d*x + c
))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)

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giac [B]  time = 0.25, size = 158, normalized size = 1.93 \[ -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {250 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} - \frac {3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (250*tan(1/2*d*x + 1/2*c)^4 - 48*tan(1/2*d*x + 1/2*c)^3 - 24*
tan(1/2*d*x + 1/2*c)^2 + 16*tan(1/2*d*x + 1/2*c) - 3)/(a^2*tan(1/2*d*x + 1/2*c)^4) - (3*a^6*tan(1/2*d*x + 1/2*
c)^4 - 16*a^6*tan(1/2*d*x + 1/2*c)^3 + 24*a^6*tan(1/2*d*x + 1/2*c)^2 + 48*a^6*tan(1/2*d*x + 1/2*c))/a^8)/d

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maple [B]  time = 0.63, size = 170, normalized size = 2.07 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a^{2} d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d \,a^{2}}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2}}-\frac {1}{4 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{64 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{12 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x)

[Out]

1/64/d/a^2*tan(1/2*d*x+1/2*c)^4-1/12/d/a^2*tan(1/2*d*x+1/2*c)^3+1/8/d/a^2*tan(1/2*d*x+1/2*c)^2+1/4/d/a^2*tan(1
/2*d*x+1/2*c)-1/4/d/a^2/tan(1/2*d*x+1/2*c)-5/8/d/a^2*ln(tan(1/2*d*x+1/2*c))-1/8/a^2/d/tan(1/2*d*x+1/2*c)^2-1/6
4/a^2/d/tan(1/2*d*x+1/2*c)^4+1/12/a^2/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.33, size = 194, normalized size = 2.37 \[ \frac {\frac {\frac {48 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{2}} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {48 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a^{2} \sin \left (d x + c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/192*((48*sin(d*x + c)/(cos(d*x + c) + 1) + 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 16*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^2 - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + (
16*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 48*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 - 3)*(cos(d*x + c) + 1)^4/(a^2*sin(d*x + c)^4))/d

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mupad [B]  time = 9.03, size = 151, normalized size = 1.84 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{4}\right )}{16\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^5*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - tan(c/2 + (d*x)/2)^3/(12*a^2*d) + tan(c/2 + (d*x)/2)^4/(64*a^2*d) - (5*log(ta
n(c/2 + (d*x)/2)))/(8*a^2*d) + tan(c/2 + (d*x)/2)/(4*a^2*d) - (cot(c/2 + (d*x)/2)^4*(2*tan(c/2 + (d*x)/2)^2 -
(4*tan(c/2 + (d*x)/2))/3 + 4*tan(c/2 + (d*x)/2)^3 + 1/4))/(16*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)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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