3.418 \(\int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2839, 2607, 14, 2611, 3768, 3770} \[ -\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^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 {\cot ^4(c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a}+\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a}\\ &=\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \csc ^3(c+d x) \, dx}{4 a}+\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \csc (c+d x) \, dx}{8 a}+\frac {\operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 189, normalized size = 1.89 \[ -\frac {\csc ^5(c+d x) \left (-180 \sin (2 (c+d x))-30 \sin (4 (c+d x))+320 \cos (c+d x)+80 \cos (3 (c+d x))-16 \cos (5 (c+d x))-150 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+75 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+150 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-75 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1920 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1920*(Csc[c + d*x]^5*(320*Cos[c + d*x] + 80*Cos[3*(c + d*x)] - 16*Cos[5*(c + d*x)] + 150*Log[Cos[(c + d*x)/
2]]*Sin[c + d*x] - 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 180*Sin[2*(c + d*x)] - 75*Log[Cos[(c + d*x)/2]]*Si
n[3*(c + d*x)] + 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 30*Sin[4*(c + d*x)] + 15*Log[Cos[(c + d*x)/2]]*Si
n[5*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(a*d)

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fricas [A]  time = 0.46, size = 161, normalized size = 1.61 \[ \frac {32 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(32*cos(d*x + c)^5 - 80*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)*sin(d*x + c) + 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) +
30*(cos(d*x + c)^3 + cos(d*x + c))*sin(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x +
c))

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giac [A]  time = 0.20, size = 157, normalized size = 1.57 \[ \frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a + (6*a^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*tan(1/2*d*x + 1/2*c)^4 +
10*a^4*tan(1/2*d*x + 1/2*c)^3 - 60*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (274*tan(1/2*d*x + 1/2*c)^5 - 60*tan(1/2*d*
x + 1/2*c)^4 + 10*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 6)/(a*tan(1/2*d*x + 1/2*c)^5))/d

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maple [A]  time = 0.46, size = 170, normalized size = 1.70 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {1}{16 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {1}{160 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{64 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{96 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160/a/d*tan(1/2*d*x+1/2*c)^5-1/64/a/d*tan(1/2*d*x+1/2*c)^4+1/96/a/d*tan(1/2*d*x+1/2*c)^3-1/16/a/d*tan(1/2*d*
x+1/2*c)+1/16/a/d/tan(1/2*d*x+1/2*c)+1/8/a/d*ln(tan(1/2*d*x+1/2*c))-1/160/a/d/tan(1/2*d*x+1/2*c)^5+1/64/a/d/ta
n(1/2*d*x+1/2*c)^4-1/96/a/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.32, size = 195, normalized size = 1.95 \[ -\frac {\frac {\frac {60 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*((60*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 - 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (15*
sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 6)*(cos(d*x + c) + 1)^5/(a*sin(d*x + c)^5))/d

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mupad [B]  time = 8.75, size = 151, normalized size = 1.51 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{5}\right )}{32\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^3/(96*a*d) - tan(c/2 + (d*x)/2)^4/(64*a*d) + tan(c/2 + (d*x)/2)^5/(160*a*d) + log(tan(c/2 +
(d*x)/2))/(8*a*d) - tan(c/2 + (d*x)/2)/(16*a*d) + (cot(c/2 + (d*x)/2)^5*(tan(c/2 + (d*x)/2)/2 - tan(c/2 + (d*
x)/2)^2/3 + 2*tan(c/2 + (d*x)/2)^4 - 1/5))/(32*a*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)**4*csc(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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