3.1.59 \(\int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx\) [59]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2785, 2687, 30, 2691, 3855} \begin {gather*} -\frac {\cot ^5(c+d x)}{5 a d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 30

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

Rule 2687

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 2691

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 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3855

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 ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \cot ^4(c+d x) \csc (c+d x) \, dx}{a}+\frac {\int \cot ^4(c+d x) \csc ^2(c+d x) \, dx}{a}\\ &=\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d}+\frac {3 \int \cot ^2(c+d x) \csc (c+d x) \, dx}{4 a}+\frac {\text {Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=-\frac {\cot ^5(c+d x)}{5 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d}-\frac {3 \int \csc (c+d x) \, dx}{8 a}\\ &=\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(189\) vs. \(2(82)=164\).
time = 0.52, size = 189, normalized size = 2.30 \begin {gather*} -\frac {\csc ^5(c+d x) \left (80 \cos (c+d x)+40 \cos (3 (c+d x))+8 \cos (5 (c+d x))-150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+20 \sin (2 (c+d x))+75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-50 \sin (4 (c+d x))-15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{640 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

-1/640*(Csc[c + d*x]^5*(80*Cos[c + d*x] + 40*Cos[3*(c + d*x)] + 8*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] + 20*Sin[2*(c + d*x)] + 75*Log[Cos[(c + d*x)/2]]*Sin[3*
(c + d*x)] - 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 50*Sin[4*(c + d*x)] - 15*Log[Cos[(c + d*x)/2]]*Sin[5*
(c + d*x)] + 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(a*d)

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Maple [A]
time = 0.23, size = 148, normalized size = 1.80

method	result	size
risch	\(\frac {-40 i {\mathrm e}^{8 i \left (d x +c \right )}+25 \,{\mathrm e}^{9 i \left (d x +c \right )}-10 \,{\mathrm e}^{7 i \left (d x +c \right )}-80 i {\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i-25 \,{\mathrm e}^{i \left (d x +c \right )}}{20 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\)	\(134\)
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d a}\)	\(148\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d a}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/32/d/a*(1/5*tan(1/2*d*x+1/2*c)^5-1/2*tan(1/2*d*x+1/2*c)^4-tan(1/2*d*x+1/2*c)^3+4*tan(1/2*d*x+1/2*c)^2+2*tan(
1/2*d*x+1/2*c)-4/tan(1/2*d*x+1/2*c)^2-1/5/tan(1/2*d*x+1/2*c)^5+1/tan(1/2*d*x+1/2*c)^3-12*ln(tan(1/2*d*x+1/2*c)
)+1/2/tan(1/2*d*x+1/2*c)^4-2/tan(1/2*d*x+1/2*c))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (74) = 148\).
time = 0.30, size = 234, normalized size = 2.85 \begin {gather*} \frac {\frac {\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2 \, \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 {5 \, \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 {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {20 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 2\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{320 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*((20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 120*log(si
n(d*x + c)/(cos(d*x + c) + 1))/a + (5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 20*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2)*(cos(d*x + c) + 1)^5/(
a*sin(d*x + c)^5))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (74) = 148\).
time = 0.35, size = 155, normalized size = 1.89 \begin {gather*} -\frac {16 \, \cos \left (d x + c\right )^{5} - 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 ) - 10 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \, {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*(16*cos(d*x + c)^5 - 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) - 10*(5*cos(d*x + c)^3
- 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\cot ^{6}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**6/(sin(c + d*x) + 1), x)/a

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (74) = 148\).
time = 8.24, size = 187, normalized size = 2.28 \begin {gather*} -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, 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} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, 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} - 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{320 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/320*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - (2*a^4*tan(1/2*d*x + 1/2*c)^5 - 5*a^4*tan(1/2*d*x + 1/2*c)^4 -
10*a^4*tan(1/2*d*x + 1/2*c)^3 + 40*a^4*tan(1/2*d*x + 1/2*c)^2 + 20*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (274*tan(1/
2*d*x + 1/2*c)^5 - 20*tan(1/2*d*x + 1/2*c)^4 - 40*tan(1/2*d*x + 1/2*c)^3 + 10*tan(1/2*d*x + 1/2*c)^2 + 5*tan(1
/2*d*x + 1/2*c) - 2)/(a*tan(1/2*d*x + 1/2*c)^5))/d

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Mupad [B]
time = 6.66, size = 183, normalized size = 2.23 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,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 {3\,\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+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(32*a*d) - tan(c/2 + (d*x)/2)^4/(64*a*d) + tan(c/2 + (d*x)
/2)^5/(160*a*d) - (3*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*(4
*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)/2 + 2*tan(c/2 + (d*x)/2)^4 + 1/5))/(32*a*d)

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