3.27 \(\int \cot ^7(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=127 \[ -\frac {23 a^2}{16 d (1-\cos (c+d x))}-\frac {a^2}{16 d (\cos (c+d x)+1)}+\frac {a^2}{2 d (1-\cos (c+d x))^2}-\frac {a^2}{12 d (1-\cos (c+d x))^3}-\frac {13 a^2 \log (1-\cos (c+d x))}{16 d}-\frac {3 a^2 \log (\cos (c+d x)+1)}{16 d} \]

[Out]

-1/12*a^2/d/(1-cos(d*x+c))^3+1/2*a^2/d/(1-cos(d*x+c))^2-23/16*a^2/d/(1-cos(d*x+c))-1/16*a^2/d/(1+cos(d*x+c))-1
3/16*a^2*ln(1-cos(d*x+c))/d-3/16*a^2*ln(1+cos(d*x+c))/d

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Rubi [A]  time = 0.09, antiderivative size = 127, 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} \[ -\frac {23 a^2}{16 d (1-\cos (c+d x))}-\frac {a^2}{16 d (\cos (c+d x)+1)}+\frac {a^2}{2 d (1-\cos (c+d x))^2}-\frac {a^2}{12 d (1-\cos (c+d x))^3}-\frac {13 a^2 \log (1-\cos (c+d x))}{16 d}-\frac {3 a^2 \log (\cos (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^7*(a + a*Sec[c + d*x])^2,x]

[Out]

-a^2/(12*d*(1 - Cos[c + d*x])^3) + a^2/(2*d*(1 - Cos[c + d*x])^2) - (23*a^2)/(16*d*(1 - Cos[c + d*x])) - a^2/(
16*d*(1 + Cos[c + d*x])) - (13*a^2*Log[1 - Cos[c + d*x]])/(16*d) - (3*a^2*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 ^7(c+d x) (a+a \sec (c+d x))^2 \, dx &=-\frac {a^8 \operatorname {Subst}\left (\int \frac {x^5}{(a-a x)^4 (a+a x)^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^8 \operatorname {Subst}\left (\int \left (\frac {1}{4 a^6 (-1+x)^4}+\frac {1}{a^6 (-1+x)^3}+\frac {23}{16 a^6 (-1+x)^2}+\frac {13}{16 a^6 (-1+x)}-\frac {1}{16 a^6 (1+x)^2}+\frac {3}{16 a^6 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2}{12 d (1-\cos (c+d x))^3}+\frac {a^2}{2 d (1-\cos (c+d x))^2}-\frac {23 a^2}{16 d (1-\cos (c+d x))}-\frac {a^2}{16 d (1+\cos (c+d x))}-\frac {13 a^2 \log (1-\cos (c+d x))}{16 d}-\frac {3 a^2 \log (1+\cos (c+d x))}{16 d}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 114, normalized size = 0.90 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\csc ^6\left (\frac {1}{2} (c+d x)\right )-12 \csc ^4\left (\frac {1}{2} (c+d x)\right )+69 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \left (\sec ^2\left (\frac {1}{2} (c+d x)\right )+52 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^7*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/384*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(69*Csc[(c + d*x)/2]^2 - 12*Csc[(c + d*x)/2]^4 + Csc[(c +
d*x)/2]^6 + 3*(12*Log[Cos[(c + d*x)/2]] + 52*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2)))/d

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fricas [A]  time = 0.70, size = 191, normalized size = 1.50 \[ \frac {66 \, a^{2} \cos \left (d x + c\right )^{3} - 36 \, a^{2} \cos \left (d x + c\right )^{2} - 74 \, a^{2} \cos \left (d x + c\right ) + 52 \, a^{2} - 9 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 39 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(66*a^2*cos(d*x + c)^3 - 36*a^2*cos(d*x + c)^2 - 74*a^2*cos(d*x + c) + 52*a^2 - 9*(a^2*cos(d*x + c)^4 - 2
*a^2*cos(d*x + c)^3 + 2*a^2*cos(d*x + c) - a^2)*log(1/2*cos(d*x + c) + 1/2) - 39*(a^2*cos(d*x + c)^4 - 2*a^2*c
os(d*x + c)^3 + 2*a^2*cos(d*x + c) - a^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3
+ 2*d*cos(d*x + c) - d)

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giac [A]  time = 0.42, size = 186, normalized size = 1.46 \[ -\frac {78 \, a^{2} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 96 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {3 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{2} + \frac {9 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {48 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {143 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/96*(78*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 96*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) + 1)) - 3*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - (a^2 + 9*a^2*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 48*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 143*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)*(
cos(d*x + c) + 1)^3/(cos(d*x + c) - 1)^3)/d

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maple [A]  time = 0.63, size = 122, normalized size = 0.96 \[ \frac {a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}-\frac {a^{2}}{12 d \left (-1+\sec \left (d x +c \right )\right )^{3}}+\frac {a^{2}}{4 d \left (-1+\sec \left (d x +c \right )\right )^{2}}-\frac {11 a^{2}}{16 d \left (-1+\sec \left (d x +c \right )\right )}-\frac {13 a^{2} \ln \left (-1+\sec \left (d x +c \right )\right )}{16 d}+\frac {a^{2}}{16 d \left (1+\sec \left (d x +c \right )\right )}-\frac {3 a^{2} \ln \left (1+\sec \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^7*(a+a*sec(d*x+c))^2,x)

[Out]

a^2/d*ln(sec(d*x+c))-1/12*a^2/d/(-1+sec(d*x+c))^3+1/4*a^2/d/(-1+sec(d*x+c))^2-11/16*a^2/d/(-1+sec(d*x+c))-13/1
6*a^2/d*ln(-1+sec(d*x+c))+1/16*a^2/d/(1+sec(d*x+c))-3/16*a^2/d*ln(1+sec(d*x+c))

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maxima [A]  time = 0.56, size = 109, normalized size = 0.86 \[ -\frac {9 \, a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) + 39 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (33 \, a^{2} \cos \left (d x + c\right )^{3} - 18 \, a^{2} \cos \left (d x + c\right )^{2} - 37 \, a^{2} \cos \left (d x + c\right ) + 26 \, a^{2}\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right ) - 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/48*(9*a^2*log(cos(d*x + c) + 1) + 39*a^2*log(cos(d*x + c) - 1) - 2*(33*a^2*cos(d*x + c)^3 - 18*a^2*cos(d*x
+ c)^2 - 37*a^2*cos(d*x + c) + 26*a^2)/(cos(d*x + c)^4 - 2*cos(d*x + c)^3 + 2*cos(d*x + c) - 1))/d

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mupad [B]  time = 1.27, size = 113, normalized size = 0.89 \[ \frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {13\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {a^2}{6}\right )}{16\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^7*(a + a/cos(c + d*x))^2,x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (13*a^2*log(tan(c/2 + (d*x)/2)))/(8*d) - (cot(c/2 + (d*x)/2)^6*(8*a^2*
tan(c/2 + (d*x)/2)^4 - (3*a^2*tan(c/2 + (d*x)/2)^2)/2 + a^2/6))/(16*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(32*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \cot ^{7}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \cot ^{7}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cot ^{7}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**7*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*cot(c + d*x)**7*sec(c + d*x), x) + Integral(cot(c + d*x)**7*sec(c + d*x)**2, x) + Integral(co
t(c + d*x)**7, x))

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