3.78 \(\int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=165 \[ \frac {9}{64 a^2 d (1-\cos (c+d x))}+\frac {51}{32 a^2 d (\cos (c+d x)+1)}-\frac {1}{64 a^2 d (1-\cos (c+d x))^2}-\frac {3}{4 a^2 d (\cos (c+d x)+1)^2}+\frac {11}{48 a^2 d (\cos (c+d x)+1)^3}-\frac {1}{32 a^2 d (\cos (c+d x)+1)^4}+\frac {29 \log (1-\cos (c+d x))}{128 a^2 d}+\frac {99 \log (\cos (c+d x)+1)}{128 a^2 d} \]

[Out]

-1/64/a^2/d/(1-cos(d*x+c))^2+9/64/a^2/d/(1-cos(d*x+c))-1/32/a^2/d/(1+cos(d*x+c))^4+11/48/a^2/d/(1+cos(d*x+c))^
3-3/4/a^2/d/(1+cos(d*x+c))^2+51/32/a^2/d/(1+cos(d*x+c))+29/128*ln(1-cos(d*x+c))/a^2/d+99/128*ln(1+cos(d*x+c))/
a^2/d

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Rubi [A]  time = 0.11, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 88} \[ \frac {9}{64 a^2 d (1-\cos (c+d x))}+\frac {51}{32 a^2 d (\cos (c+d x)+1)}-\frac {1}{64 a^2 d (1-\cos (c+d x))^2}-\frac {3}{4 a^2 d (\cos (c+d x)+1)^2}+\frac {11}{48 a^2 d (\cos (c+d x)+1)^3}-\frac {1}{32 a^2 d (\cos (c+d x)+1)^4}+\frac {29 \log (1-\cos (c+d x))}{128 a^2 d}+\frac {99 \log (\cos (c+d x)+1)}{128 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(64*a^2*d*(1 - Cos[c + d*x])^2) + 9/(64*a^2*d*(1 - Cos[c + d*x])) - 1/(32*a^2*d*(1 + Cos[c + d*x])^4) + 11/
(48*a^2*d*(1 + Cos[c + d*x])^3) - 3/(4*a^2*d*(1 + Cos[c + d*x])^2) + 51/(32*a^2*d*(1 + Cos[c + d*x])) + (29*Lo
g[1 - Cos[c + d*x]])/(128*a^2*d) + (99*Log[1 + Cos[c + d*x]])/(128*a^2*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {a^6 \operatorname {Subst}\left (\int \frac {x^7}{(a-a x)^3 (a+a x)^5} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^6 \operatorname {Subst}\left (\int \left (-\frac {1}{32 a^8 (-1+x)^3}-\frac {9}{64 a^8 (-1+x)^2}-\frac {29}{128 a^8 (-1+x)}-\frac {1}{8 a^8 (1+x)^5}+\frac {11}{16 a^8 (1+x)^4}-\frac {3}{2 a^8 (1+x)^3}+\frac {51}{32 a^8 (1+x)^2}-\frac {99}{128 a^8 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {1}{64 a^2 d (1-\cos (c+d x))^2}+\frac {9}{64 a^2 d (1-\cos (c+d x))}-\frac {1}{32 a^2 d (1+\cos (c+d x))^4}+\frac {11}{48 a^2 d (1+\cos (c+d x))^3}-\frac {3}{4 a^2 d (1+\cos (c+d x))^2}+\frac {51}{32 a^2 d (1+\cos (c+d x))}+\frac {29 \log (1-\cos (c+d x))}{128 a^2 d}+\frac {99 \log (1+\cos (c+d x))}{128 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 154, normalized size = 0.93 \[ -\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (6 \csc ^4\left (\frac {1}{2} (c+d x)\right )-108 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \sec ^8\left (\frac {1}{2} (c+d x)\right )-44 \sec ^6\left (\frac {1}{2} (c+d x)\right )+288 \sec ^4\left (\frac {1}{2} (c+d x)\right )-1224 \sec ^2\left (\frac {1}{2} (c+d x)\right )-24 \left (29 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+99 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{384 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(Cos[(c + d*x)/2]^4*(-108*Csc[(c + d*x)/2]^2 + 6*Csc[(c + d*x)/2]^4 - 24*(99*Log[Cos[(c + d*x)/2]] + 29
*Log[Sin[(c + d*x)/2]]) - 1224*Sec[(c + d*x)/2]^2 + 288*Sec[(c + d*x)/2]^4 - 44*Sec[(c + d*x)/2]^6 + 3*Sec[(c
+ d*x)/2]^8)*Sec[c + d*x]^2)/(a^2*d*(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.60, size = 283, normalized size = 1.72 \[ \frac {558 \, \cos \left (d x + c\right )^{5} + 156 \, \cos \left (d x + c\right )^{4} - 1268 \, \cos \left (d x + c\right )^{3} - 676 \, \cos \left (d x + c\right )^{2} + 297 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 87 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 686 \, \cos \left (d x + c\right ) + 448}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} + 2 \, a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/384*(558*cos(d*x + c)^5 + 156*cos(d*x + c)^4 - 1268*cos(d*x + c)^3 - 676*cos(d*x + c)^2 + 297*(cos(d*x + c)^
6 + 2*cos(d*x + c)^5 - cos(d*x + c)^4 - 4*cos(d*x + c)^3 - cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*
x + c) + 1/2) + 87*(cos(d*x + c)^6 + 2*cos(d*x + c)^5 - cos(d*x + c)^4 - 4*cos(d*x + c)^3 - cos(d*x + c)^2 + 2
*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 686*cos(d*x + c) + 448)/(a^2*d*cos(d*x + c)^6 + 2*a^2*d*cos(
d*x + c)^5 - a^2*d*cos(d*x + c)^4 - 4*a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2
*d)

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giac [A]  time = 0.41, size = 236, normalized size = 1.43 \[ -\frac {\frac {6 \, {\left (\frac {16 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {87 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {348 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {1536 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} + \frac {\frac {768 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {174 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {32 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{8}}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/1536*(6*(16*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 87*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1)*(cos(
d*x + c) + 1)^2/(a^2*(cos(d*x + c) - 1)^2) - 348*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + 1536*
log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 + (768*a^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 17
4*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 32*a^6*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3*a^6*(co
s(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/a^8)/d

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maple [A]  time = 1.02, size = 144, normalized size = 0.87 \[ -\frac {1}{64 a^{2} d \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {9}{64 a^{2} d \left (-1+\cos \left (d x +c \right )\right )}+\frac {29 \ln \left (-1+\cos \left (d x +c \right )\right )}{128 a^{2} d}-\frac {1}{32 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{4}}+\frac {11}{48 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {3}{4 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {51}{32 a^{2} d \left (1+\cos \left (d x +c \right )\right )}+\frac {99 \ln \left (1+\cos \left (d x +c \right )\right )}{128 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x)

[Out]

-1/64/a^2/d/(-1+cos(d*x+c))^2-9/64/a^2/d/(-1+cos(d*x+c))+29/128/a^2/d*ln(-1+cos(d*x+c))-1/32/a^2/d/(1+cos(d*x+
c))^4+11/48/a^2/d/(1+cos(d*x+c))^3-3/4/a^2/d/(1+cos(d*x+c))^2+51/32/a^2/d/(1+cos(d*x+c))+99/128*ln(1+cos(d*x+c
))/a^2/d

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maxima [A]  time = 0.45, size = 167, normalized size = 1.01 \[ \frac {\frac {2 \, {\left (279 \, \cos \left (d x + c\right )^{5} + 78 \, \cos \left (d x + c\right )^{4} - 634 \, \cos \left (d x + c\right )^{3} - 338 \, \cos \left (d x + c\right )^{2} + 343 \, \cos \left (d x + c\right ) + 224\right )}}{a^{2} \cos \left (d x + c\right )^{6} + 2 \, a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} + \frac {297 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {87 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/384*(2*(279*cos(d*x + c)^5 + 78*cos(d*x + c)^4 - 634*cos(d*x + c)^3 - 338*cos(d*x + c)^2 + 343*cos(d*x + c)
+ 224)/(a^2*cos(d*x + c)^6 + 2*a^2*cos(d*x + c)^5 - a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^3 - a^2*cos(d*x +
c)^2 + 2*a^2*cos(d*x + c) + a^2) + 297*log(cos(d*x + c) + 1)/a^2 + 87*log(cos(d*x + c) - 1)/a^2)/d

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mupad [B]  time = 1.24, size = 151, normalized size = 0.92 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,a^2\,d}-\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{256\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{512\,a^2\,d}+\frac {29\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {1}{4}\right )}{64\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(2*a^2*d) - (29*tan(c/2 + (d*x)/2)^4)/(256*a^2*d) + tan(c/2 + (d*x)/2)^6/(48*a^2*d) - tan
(c/2 + (d*x)/2)^8/(512*a^2*d) + (29*log(tan(c/2 + (d*x)/2)))/(64*a^2*d) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d
) + (cot(c/2 + (d*x)/2)^4*(4*tan(c/2 + (d*x)/2)^2 - 1/4))/(64*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**5/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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