3.26 \(\int \cot ^5(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=85 \[ \frac {5 a^2}{4 d (1-\cos (c+d x))}-\frac {a^2}{4 d (1-\cos (c+d x))^2}+\frac {7 a^2 \log (1-\cos (c+d x))}{8 d}+\frac {a^2 \log (\cos (c+d x)+1)}{8 d} \]

[Out]

-1/4*a^2/d/(1-cos(d*x+c))^2+5/4*a^2/d/(1-cos(d*x+c))+7/8*a^2*ln(1-cos(d*x+c))/d+1/8*a^2*ln(1+cos(d*x+c))/d

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Rubi [A]  time = 0.07, antiderivative size = 85, 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} \[ \frac {5 a^2}{4 d (1-\cos (c+d x))}-\frac {a^2}{4 d (1-\cos (c+d x))^2}+\frac {7 a^2 \log (1-\cos (c+d x))}{8 d}+\frac {a^2 \log (\cos (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a^2/(4*d*(1 - Cos[c + d*x])^2) + (5*a^2)/(4*d*(1 - Cos[c + d*x])) + (7*a^2*Log[1 - Cos[c + d*x]])/(8*d) + (a^
2*Log[1 + Cos[c + d*x]])/(8*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))^2 \, dx &=-\frac {a^6 \operatorname {Subst}\left (\int \frac {x^3}{(a-a x)^3 (a+a x)} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^6 \operatorname {Subst}\left (\int \left (-\frac {1}{2 a^4 (-1+x)^3}-\frac {5}{4 a^4 (-1+x)^2}-\frac {7}{8 a^4 (-1+x)}-\frac {1}{8 a^4 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2}{4 d (1-\cos (c+d x))^2}+\frac {5 a^2}{4 d (1-\cos (c+d x))}+\frac {7 a^2 \log (1-\cos (c+d x))}{8 d}+\frac {a^2 \log (1+\cos (c+d x))}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 86, normalized size = 1.01 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\csc ^4\left (\frac {1}{2} (c+d x)\right )-10 \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (7 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(a^2*(1 + Cos[c + d*x])^2*(-10*Csc[(c + d*x)/2]^2 + Csc[(c + d*x)/2]^4 - 4*(Log[Cos[(c + d*x)/2]] + 7*Lo
g[Sin[(c + d*x)/2]]))*Sec[(c + d*x)/2]^4)/d

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fricas [A]  time = 0.58, size = 122, normalized size = 1.44 \[ -\frac {10 \, a^{2} \cos \left (d x + c\right ) - 8 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(10*a^2*cos(d*x + c) - 8*a^2 - (a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*log(1/2*cos(d*x + c) + 1/2
) - 7*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^2 - 2*d*co
s(d*x + c) + d)

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giac [A]  time = 0.56, size = 138, normalized size = 1.62 \[ \frac {14 \, a^{2} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 16 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a^{2} + \frac {8 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {21 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(14*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 16*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 1)) - (a^2 + 8*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 21*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c
) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2)/d

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maple [A]  time = 0.58, size = 87, normalized size = 1.02 \[ -\frac {a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}-\frac {a^{2}}{4 d \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {3 a^{2}}{4 d \left (-1+\sec \left (d x +c \right )\right )}+\frac {7 a^{2} \ln \left (-1+\sec \left (d x +c \right )\right )}{8 d}+\frac {a^{2} \ln \left (1+\sec \left (d x +c \right )\right )}{8 d} \]

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]

-a^2/d*ln(sec(d*x+c))-1/4*a^2/d/(-1+sec(d*x+c))^2+3/4*a^2/d/(-1+sec(d*x+c))+7/8*a^2/d*ln(-1+sec(d*x+c))+1/8*a^
2/d*ln(1+sec(d*x+c))

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maxima [A]  time = 0.64, size = 72, normalized size = 0.85 \[ \frac {a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a^{2} \cos \left (d x + c\right ) - 4 \, a^{2}\right )}}{\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}}{8 \, d} \]

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]

1/8*(a^2*log(cos(d*x + c) + 1) + 7*a^2*log(cos(d*x + c) - 1) - 2*(5*a^2*cos(d*x + c) - 4*a^2)/(cos(d*x + c)^2
- 2*cos(d*x + c) + 1))/d

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mupad [B]  time = 1.26, size = 62, normalized size = 0.73 \[ \frac {a^2\,\left (-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {7\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )}{d} \]

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]

(a^2*((7*log(tan(c/2 + (d*x)/2)))/4 - log(tan(c/2 + (d*x)/2)^2 + 1) + cot(c/2 + (d*x)/2)^2/2 - cot(c/2 + (d*x)
/2)^4/16))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*cot(c + d*x)**5*sec(c + d*x), x) + Integral(cot(c + d*x)**5*sec(c + d*x)**2, x) + Integral(co
t(c + d*x)**5, x))

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