3.28 \(\int \cot ^9(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=169 \[ \frac{51 a^2}{32 d (1-\cos (c+d x))}+\frac{9 a^2}{64 d (\cos (c+d x)+1)}-\frac{3 a^2}{4 d (1-\cos (c+d x))^2}-\frac{a^2}{64 d (\cos (c+d x)+1)^2}+\frac{11 a^2}{48 d (1-\cos (c+d x))^3}-\frac{a^2}{32 d (1-\cos (c+d x))^4}+\frac{99 a^2 \log (1-\cos (c+d x))}{128 d}+\frac{29 a^2 \log (\cos (c+d x)+1)}{128 d} \]

[Out]

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

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Rubi [A]  time = 0.111059, antiderivative size = 169, normalized size of antiderivative = 1., 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} \[ \frac{51 a^2}{32 d (1-\cos (c+d x))}+\frac{9 a^2}{64 d (\cos (c+d x)+1)}-\frac{3 a^2}{4 d (1-\cos (c+d x))^2}-\frac{a^2}{64 d (\cos (c+d x)+1)^2}+\frac{11 a^2}{48 d (1-\cos (c+d x))^3}-\frac{a^2}{32 d (1-\cos (c+d x))^4}+\frac{99 a^2 \log (1-\cos (c+d x))}{128 d}+\frac{29 a^2 \log (\cos (c+d x)+1)}{128 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^9*(a + a*Sec[c + d*x])^2,x]

[Out]

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

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]

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]))

Rubi steps

\begin{align*} \int \cot ^9(c+d x) (a+a \sec (c+d x))^2 \, dx &=-\frac{a^{10} \operatorname{Subst}\left (\int \frac{x^7}{(a-a x)^5 (a+a x)^3} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^{10} \operatorname{Subst}\left (\int \left (-\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)}-\frac{1}{32 a^8 (1+x)^3}+\frac{9}{64 a^8 (1+x)^2}-\frac{29}{128 a^8 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^2}{32 d (1-\cos (c+d x))^4}+\frac{11 a^2}{48 d (1-\cos (c+d x))^3}-\frac{3 a^2}{4 d (1-\cos (c+d x))^2}+\frac{51 a^2}{32 d (1-\cos (c+d x))}-\frac{a^2}{64 d (1+\cos (c+d x))^2}+\frac{9 a^2}{64 d (1+\cos (c+d x))}+\frac{99 a^2 \log (1-\cos (c+d x))}{128 d}+\frac{29 a^2 \log (1+\cos (c+d x))}{128 d}\\ \end{align*}

Mathematica [A]  time = 0.331827, size = 146, normalized size = 0.86 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (3 \csc ^8\left (\frac{1}{2} (c+d x)\right )-44 \csc ^6\left (\frac{1}{2} (c+d x)\right )+288 \csc ^4\left (\frac{1}{2} (c+d x)\right )-1224 \csc ^2\left (\frac{1}{2} (c+d x)\right )-6 \left (-\sec ^4\left (\frac{1}{2} (c+d x)\right )+18 \sec ^2\left (\frac{1}{2} (c+d x)\right )+396 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+116 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{6144 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^9*(a + a*Sec[c + d*x])^2,x]

[Out]

-(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-1224*Csc[(c + d*x)/2]^2 + 288*Csc[(c + d*x)/2]^4 - 44*Csc[(c +
d*x)/2]^6 + 3*Csc[(c + d*x)/2]^8 - 6*(116*Log[Cos[(c + d*x)/2]] + 396*Log[Sin[(c + d*x)/2]] + 18*Sec[(c + d*x
)/2]^2 - Sec[(c + d*x)/2]^4)))/(6144*d)

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Maple [A]  time = 0.095, size = 159, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2}}{64\,d \left ( 1+\sec \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,{a}^{2}}{64\,d \left ( 1+\sec \left ( dx+c \right ) \right ) }}+{\frac{29\,{a}^{2}\ln \left ( 1+\sec \left ( dx+c \right ) \right ) }{128\,d}}-{\frac{{a}^{2}}{32\,d \left ( -1+\sec \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,{a}^{2}}{48\,d \left ( -1+\sec \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{a}^{2}}{4\,d \left ( -1+\sec \left ( dx+c \right ) \right ) ^{2}}}+{\frac{21\,{a}^{2}}{32\,d \left ( -1+\sec \left ( dx+c \right ) \right ) }}+{\frac{99\,{a}^{2}\ln \left ( -1+\sec \left ( dx+c \right ) \right ) }{128\,d}}-{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^9*(a+a*sec(d*x+c))^2,x)

[Out]

-1/64/d*a^2/(1+sec(d*x+c))^2-7/64/d*a^2/(1+sec(d*x+c))+29/128/d*a^2*ln(1+sec(d*x+c))-1/32/d*a^2/(-1+sec(d*x+c)
)^4+5/48/d*a^2/(-1+sec(d*x+c))^3-1/4/d*a^2/(-1+sec(d*x+c))^2+21/32/d*a^2/(-1+sec(d*x+c))+99/128/d*a^2*ln(-1+se
c(d*x+c))-1/d*a^2*ln(sec(d*x+c))

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Maxima [A]  time = 1.19769, size = 223, normalized size = 1.32 \begin{align*} \frac{87 \, a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) + 297 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (279 \, a^{2} \cos \left (d x + c\right )^{5} - 78 \, a^{2} \cos \left (d x + c\right )^{4} - 634 \, a^{2} \cos \left (d x + c\right )^{3} + 338 \, a^{2} \cos \left (d x + c\right )^{2} + 343 \, a^{2} \cos \left (d x + c\right ) - 224 \, a^{2}\right )}}{\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}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^9*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.969955, size = 815, normalized size = 4.82 \begin{align*} -\frac{558 \, a^{2} \cos \left (d x + c\right )^{5} - 156 \, a^{2} \cos \left (d x + c\right )^{4} - 1268 \, a^{2} \cos \left (d x + c\right )^{3} + 676 \, a^{2} \cos \left (d x + c\right )^{2} + 686 \, a^{2} \cos \left (d x + c\right ) - 448 \, a^{2} - 87 \,{\left (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}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 297 \,{\left (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}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{384 \,{\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{5} - d \cos \left (d x + c\right )^{4} + 4 \, d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^9*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**9*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.71284, size = 321, normalized size = 1.9 \begin{align*} \frac{1188 \, a^{2} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 1536 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac{96 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{6 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{{\left (3 \, a^{2} + \frac{32 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{174 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{768 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{2475 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}}{1536 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^9*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/1536*(1188*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 1536*a^2*log(abs(-(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 1)) - 96*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c)
+ 1)^2 - (3*a^2 + 32*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 174*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 + 768*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 2475*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4
)*(cos(d*x + c) + 1)^4/(cos(d*x + c) - 1)^4)/d