3.99 \(\int \frac{\csc ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=128 \[ -\frac{a^2}{40 d (a \cos (c+d x)+a)^5}-\frac{3}{128 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{128 a^3 d}+\frac{3 a}{64 d (a \cos (c+d x)+a)^4}-\frac{1}{128 a d (a-a \cos (c+d x))^2}-\frac{1}{64 a d (a \cos (c+d x)+a)^2} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(128*a^3*d) - 1/(128*a*d*(a - a*Cos[c + d*x])^2) - a^2/(40*d*(a + a*Cos[c + d*x])^5)
+ (3*a)/(64*d*(a + a*Cos[c + d*x])^4) - 1/(64*a*d*(a + a*Cos[c + d*x])^2) - 3/(128*d*(a^3 + a^3*Cos[c + d*x])
)

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Rubi [A]  time = 0.205667, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2836, 12, 88, 206} \[ -\frac{a^2}{40 d (a \cos (c+d x)+a)^5}-\frac{3}{128 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{128 a^3 d}+\frac{3 a}{64 d (a \cos (c+d x)+a)^4}-\frac{1}{128 a d (a-a \cos (c+d x))^2}-\frac{1}{64 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(128*a^3*d) - 1/(128*a*d*(a - a*Cos[c + d*x])^2) - a^2/(40*d*(a + a*Cos[c + d*x])^5)
+ (3*a)/(64*d*(a + a*Cos[c + d*x])^4) - 1/(64*a*d*(a + a*Cos[c + d*x])^2) - 3/(128*d*(a^3 + a^3*Cos[c + d*x])
)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2836

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cot ^3(c+d x) \csc ^2(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=\frac{a^5 \operatorname{Subst}\left (\int \frac{x^3}{a^3 (-a-x)^3 (-a+x)^6} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \frac{x^3}{(-a-x)^3 (-a+x)^6} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \left (-\frac{1}{8 (a-x)^6}+\frac{3}{16 a (a-x)^5}-\frac{1}{32 a^3 (a-x)^3}-\frac{3}{128 a^4 (a-x)^2}+\frac{1}{64 a^3 (a+x)^3}-\frac{3}{128 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac{1}{128 a d (a-a \cos (c+d x))^2}-\frac{a^2}{40 d (a+a \cos (c+d x))^5}+\frac{3 a}{64 d (a+a \cos (c+d x))^4}-\frac{1}{64 a d (a+a \cos (c+d x))^2}-\frac{3}{128 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,-a \cos (c+d x)\right )}{128 a^2 d}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{128 a^3 d}-\frac{1}{128 a d (a-a \cos (c+d x))^2}-\frac{a^2}{40 d (a+a \cos (c+d x))^5}+\frac{3 a}{64 d (a+a \cos (c+d x))^4}-\frac{1}{64 a d (a+a \cos (c+d x))^2}-\frac{3}{128 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 5.16965, size = 137, normalized size = 1.07 \[ -\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (60 \cos ^8\left (\frac{1}{2} (c+d x)\right )-15 \cos ^2\left (\frac{1}{2} (c+d x)\right )+10 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\cot ^4\left (\frac{1}{2} (c+d x)\right )+2\right )-120 \cos ^{10}\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+4\right )}{640 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

-((4 - 15*Cos[(c + d*x)/2]^2 + 60*Cos[(c + d*x)/2]^8 + 10*Cos[(c + d*x)/2]^6*(2 + Cot[(c + d*x)/2]^4) - 120*Co
s[(c + d*x)/2]^10*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]))*Sec[(c + d*x)/2]^4*Sec[c + d*x]^3)/(640*a^3
*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.081, size = 126, normalized size = 1. \begin{align*} -{\frac{1}{40\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{5}}}+{\frac{3}{64\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{4}}}-{\frac{1}{64\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{3}{128\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}+{\frac{3\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{256\,d{a}^{3}}}-{\frac{1}{128\,d{a}^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{256\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a+a*sec(d*x+c))^3,x)

[Out]

-1/40/d/a^3/(cos(d*x+c)+1)^5+3/64/d/a^3/(cos(d*x+c)+1)^4-1/64/d/a^3/(cos(d*x+c)+1)^2-3/128/d/a^3/(cos(d*x+c)+1
)+3/256*ln(cos(d*x+c)+1)/a^3/d-1/128/d/a^3/(-1+cos(d*x+c))^2-3/256/d/a^3*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.01282, size = 254, normalized size = 1.98 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{6} + 45 \, \cos \left (d x + c\right )^{5} + 20 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} + 61 \, \cos \left (d x + c\right )^{2} + 63 \, \cos \left (d x + c\right ) + 16\right )}}{a^{3} \cos \left (d x + c\right )^{7} + 3 \, a^{3} \cos \left (d x + c\right )^{6} + a^{3} \cos \left (d x + c\right )^{5} - 5 \, a^{3} \cos \left (d x + c\right )^{4} - 5 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} - \frac{15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{1280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/1280*(2*(15*cos(d*x + c)^6 + 45*cos(d*x + c)^5 + 20*cos(d*x + c)^4 - 60*cos(d*x + c)^3 + 61*cos(d*x + c)^2
+ 63*cos(d*x + c) + 16)/(a^3*cos(d*x + c)^7 + 3*a^3*cos(d*x + c)^6 + a^3*cos(d*x + c)^5 - 5*a^3*cos(d*x + c)^4
- 5*a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3) - 15*log(cos(d*x + c) + 1)/a^3 + 15*l
og(cos(d*x + c) - 1)/a^3)/d

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Fricas [B]  time = 1.8624, size = 857, normalized size = 6.7 \begin{align*} -\frac{30 \, \cos \left (d x + c\right )^{6} + 90 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - 120 \, \cos \left (d x + c\right )^{3} + 122 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (\cos \left (d x + c\right )^{7} + 3 \, \cos \left (d x + c\right )^{6} + \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 15 \,{\left (\cos \left (d x + c\right )^{7} + 3 \, \cos \left (d x + c\right )^{6} + \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 126 \, \cos \left (d x + c\right ) + 32}{1280 \,{\left (a^{3} d \cos \left (d x + c\right )^{7} + 3 \, a^{3} d \cos \left (d x + c\right )^{6} + a^{3} d \cos \left (d x + c\right )^{5} - 5 \, a^{3} d \cos \left (d x + c\right )^{4} - 5 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/1280*(30*cos(d*x + c)^6 + 90*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - 120*cos(d*x + c)^3 + 122*cos(d*x + c)^2 -
15*(cos(d*x + c)^7 + 3*cos(d*x + c)^6 + cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 5*cos(d*x + c)^3 + cos(d*x + c)^2
+ 3*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x + c)^7 + 3*cos(d*x + c)^6 + cos(d*x + c)^5 -
5*cos(d*x + c)^4 - 5*cos(d*x + c)^3 + cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 126*
cos(d*x + c) + 32)/(a^3*d*cos(d*x + c)^7 + 3*a^3*d*cos(d*x + c)^6 + a^3*d*cos(d*x + c)^5 - 5*a^3*d*cos(d*x + c
)^4 - 5*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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(csc(d*x+c)**5/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.40154, size = 313, normalized size = 2.45 \begin{align*} \frac{\frac{10 \,{\left (\frac{2 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \,{\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^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac{60 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac{\frac{60 \, a^{12}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{30 \, a^{12}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{20 \, a^{12}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{5 \, a^{12}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{12}{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{15}}}{5120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/5120*(10*(2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*
x + c) + 1)^2/(a^3*(cos(d*x + c) - 1)^2) - 60*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (60*a^12
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 30*a^12*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 20*a^12*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 - 5*a^12*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 4*a^12*(cos(d*x + c) - 1
)^5/(cos(d*x + c) + 1)^5)/a^15)/d