3.8 \(\int \cot ^5(c+d x) (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=95 \[ \frac {3 a}{4 d (1-\cos (c+d x))}+\frac {a}{8 d (\cos (c+d x)+1)}-\frac {a}{8 d (1-\cos (c+d x))^2}+\frac {11 a \log (1-\cos (c+d x))}{16 d}+\frac {5 a \log (\cos (c+d x)+1)}{16 d} \]

[Out]

-1/8*a/d/(1-cos(d*x+c))^2+3/4*a/d/(1-cos(d*x+c))+1/8*a/d/(1+cos(d*x+c))+11/16*a*ln(1-cos(d*x+c))/d+5/16*a*ln(1
+cos(d*x+c))/d

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3879, 88} \[ \frac {3 a}{4 d (1-\cos (c+d x))}+\frac {a}{8 d (\cos (c+d x)+1)}-\frac {a}{8 d (1-\cos (c+d x))^2}+\frac {11 a \log (1-\cos (c+d x))}{16 d}+\frac {5 a \log (\cos (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(8*d*(1 - Cos[c + d*x])^2) + (3*a)/(4*d*(1 - Cos[c + d*x])) + a/(8*d*(1 + Cos[c + d*x])) + (11*a*Log[1 - Co
s[c + d*x]])/(16*d) + (5*a*Log[1 + Cos[c + d*x]])/(16*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 \cot ^5(c+d x) (a+a \sec (c+d x)) \, dx &=-\frac {a^6 \operatorname {Subst}\left (\int \frac {x^4}{(a-a x)^3 (a+a x)^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^6 \operatorname {Subst}\left (\int \left (-\frac {1}{4 a^5 (-1+x)^3}-\frac {3}{4 a^5 (-1+x)^2}-\frac {11}{16 a^5 (-1+x)}+\frac {1}{8 a^5 (1+x)^2}-\frac {5}{16 a^5 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a}{8 d (1-\cos (c+d x))^2}+\frac {3 a}{4 d (1-\cos (c+d x))}+\frac {a}{8 d (1+\cos (c+d x))}+\frac {11 a \log (1-\cos (c+d x))}{16 d}+\frac {5 a \log (1+\cos (c+d x))}{16 d}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 127, normalized size = 1.34 \[ \frac {a \left (-16 \cot ^4(c+d x)+32 \cot ^2(c+d x)-\csc ^4\left (\frac {1}{2} (c+d x)\right )+10 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right )-10 \sec ^2\left (\frac {1}{2} (c+d x)\right )+24 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+64 \log (\tan (c+d x))-24 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+64 \log (\cos (c+d x))\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(32*Cot[c + d*x]^2 - 16*Cot[c + d*x]^4 + 10*Csc[(c + d*x)/2]^2 - Csc[(c + d*x)/2]^4 - 24*Log[Cos[(c + d*x)/
2]] + 64*Log[Cos[c + d*x]] + 24*Log[Sin[(c + d*x)/2]] + 64*Log[Tan[c + d*x]] - 10*Sec[(c + d*x)/2]^2 + Sec[(c
+ d*x)/2]^4))/(64*d)

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fricas [A]  time = 0.61, size = 150, normalized size = 1.58 \[ -\frac {10 \, a \cos \left (d x + c\right )^{2} + 6 \, a \cos \left (d x + c\right ) - 5 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 11 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(10*a*cos(d*x + c)^2 + 6*a*cos(d*x + c) - 5*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*l
og(1/2*cos(d*x + c) + 1/2) - 11*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(-1/2*cos(d*x +
c) + 1/2) - 12*a)/(d*cos(d*x + c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d)

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giac [A]  time = 0.30, size = 149, normalized size = 1.57 \[ \frac {22 \, a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 32 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + \frac {10 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {33 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(22*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 32*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 1)) - (a + 10*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 33*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)
*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/d

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maple [A]  time = 0.55, size = 93, normalized size = 0.98 \[ -\frac {a \ln \left (\sec \left (d x +c \right )\right )}{d}-\frac {a}{8 d \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {a}{2 d \left (-1+\sec \left (d x +c \right )\right )}+\frac {11 a \ln \left (-1+\sec \left (d x +c \right )\right )}{16 d}-\frac {a}{8 d \left (1+\sec \left (d x +c \right )\right )}+\frac {5 a \ln \left (1+\sec \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d*ln(sec(d*x+c))-1/8*a/d/(-1+sec(d*x+c))^2+1/2*a/d/(-1+sec(d*x+c))+11/16*a/d*ln(-1+sec(d*x+c))-1/8*a/d/(1+s
ec(d*x+c))+5/16*a/d*ln(1+sec(d*x+c))

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maxima [A]  time = 0.59, size = 86, normalized size = 0.91 \[ \frac {5 \, a \log \left (\cos \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) - 6 \, a\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*a*log(cos(d*x + c) + 1) + 11*a*log(cos(d*x + c) - 1) - 2*(5*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) - 6*a)
/(cos(d*x + c)^3 - cos(d*x + c)^2 - cos(d*x + c) + 1))/d

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mupad [B]  time = 1.18, size = 88, normalized size = 0.93 \[ \frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a}{4}-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}\right )}{8\,d}+\frac {11\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (cot(c/2 + (d*x)/2)^4*(a/4 - (5*a*tan(
c/2 + (d*x)/2)^2)/2))/(8*d) + (11*a*log(tan(c/2 + (d*x)/2)))/(8*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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