\(\int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [82]

Optimal result

Integrand size = 21, antiderivative size = 146 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\text {arctanh}(\cos (c+d x))}{64 a^2 d}-\frac {1}{64 d (a-a \cos (c+d x))^2}+\frac {a^2}{32 d (a+a \cos (c+d x))^4}-\frac {a}{48 d (a+a \cos (c+d x))^3}-\frac {1}{32 d (a+a \cos (c+d x))^2}-\frac {1}{64 d \left (a^2-a^2 \cos (c+d x)\right )}-\frac {1}{32 d \left (a^2+a^2 \cos (c+d x)\right )} \]

[Out]

1/64*arctanh(cos(d*x+c))/a^2/d-1/64/d/(a-a*cos(d*x+c))^2+1/32*a^2/d/(a+a*cos(d*x+c))^4-1/48*a/d/(a+a*cos(d*x+c
))^3-1/32/d/(a+a*cos(d*x+c))^2-1/64/d/(a^2-a^2*cos(d*x+c))-1/32/d/(a^2+a^2*cos(d*x+c))

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2915, 12, 90, 212} \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\text {arctanh}(\cos (c+d x))}{64 a^2 d}+\frac {a^2}{32 d (a \cos (c+d x)+a)^4}-\frac {1}{64 d \left (a^2-a^2 \cos (c+d x)\right )}-\frac {1}{32 d \left (a^2 \cos (c+d x)+a^2\right )}-\frac {a}{48 d (a \cos (c+d x)+a)^3}-\frac {1}{64 d (a-a \cos (c+d x))^2}-\frac {1}{32 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(64*a^2*d) - 1/(64*d*(a - a*Cos[c + d*x])^2) + a^2/(32*d*(a + a*Cos[c + d*x])^4) - a/(48
*d*(a + a*Cos[c + d*x])^3) - 1/(32*d*(a + a*Cos[c + d*x])^2) - 1/(64*d*(a^2 - a^2*Cos[c + d*x])) - 1/(32*d*(a^
2 + a^2*Cos[c + d*x]))

Rule 12

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

Rule 90

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 212

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

Rule 2915

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/b)*x
)^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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {a^5 \text {Subst}\left (\int \frac {x^2}{a^2 (-a-x)^3 (-a+x)^5} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \frac {x^2}{(-a-x)^3 (-a+x)^5} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \left (\frac {1}{8 a (a-x)^5}-\frac {1}{16 a^2 (a-x)^4}-\frac {1}{16 a^3 (a-x)^3}-\frac {1}{32 a^4 (a-x)^2}+\frac {1}{32 a^3 (a+x)^3}+\frac {1}{64 a^4 (a+x)^2}-\frac {1}{64 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {1}{64 d (a-a \cos (c+d x))^2}+\frac {a^2}{32 d (a+a \cos (c+d x))^4}-\frac {a}{48 d (a+a \cos (c+d x))^3}-\frac {1}{32 d (a+a \cos (c+d x))^2}-\frac {1}{64 d \left (a^2-a^2 \cos (c+d x)\right )}-\frac {1}{32 d \left (a^2+a^2 \cos (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,-a \cos (c+d x)\right )}{64 a d} \\ & = \frac {\text {arctanh}(\cos (c+d x))}{64 a^2 d}-\frac {1}{64 d (a-a \cos (c+d x))^2}+\frac {a^2}{32 d (a+a \cos (c+d x))^4}-\frac {a}{48 d (a+a \cos (c+d x))^3}-\frac {1}{32 d (a+a \cos (c+d x))^2}-\frac {1}{64 d \left (a^2-a^2 \cos (c+d x)\right )}-\frac {1}{32 d \left (a^2+a^2 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (12 \csc ^2\left (\frac {1}{2} (c+d x)\right )+6 \csc ^4\left (\frac {1}{2} (c+d x)\right )+24 \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+24 \sec ^2\left (\frac {1}{2} (c+d x)\right )+12 \sec ^4\left (\frac {1}{2} (c+d x)\right )+4 \sec ^6\left (\frac {1}{2} (c+d x)\right )-3 \sec ^8\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{384 a^2 d (1+\sec (c+d x))^2} \]

[In]

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

[Out]

-1/384*(Cos[(c + d*x)/2]^4*(12*Csc[(c + d*x)/2]^2 + 6*Csc[(c + d*x)/2]^4 + 24*(-Log[Cos[(c + d*x)/2]] + Log[Si
n[(c + d*x)/2]]) + 24*Sec[(c + d*x)/2]^2 + 12*Sec[(c + d*x)/2]^4 + 4*Sec[(c + d*x)/2]^6 - 3*Sec[(c + d*x)/2]^8
)*Sec[c + d*x]^2)/(a^2*d*(1 + Sec[c + d*x])^2)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-6 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1536 a^{2} d}\)	\(100\)
derivativedivides	\(\frac {-\frac {1}{64 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {1}{64 \cos \left (d x +c \right )-64}-\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{4}}-\frac {1}{48 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{128}}{d \,a^{2}}\)	\(103\)
default	\(\frac {-\frac {1}{64 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {1}{64 \cos \left (d x +c \right )-64}-\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{4}}-\frac {1}{48 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {1}{32 \left (\cos \left (d x +c \right )+1\right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{128}}{d \,a^{2}}\)	\(103\)
norman	\(\frac {-\frac {1}{256 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{512 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{32 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{256 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{192 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a^{2} d}\)	\(139\)
risch	\(-\frac {3 \,{\mathrm e}^{11 i \left (d x +c \right )}+12 \,{\mathrm e}^{10 i \left (d x +c \right )}+7 \,{\mathrm e}^{9 i \left (d x +c \right )}-32 \,{\mathrm e}^{8 i \left (d x +c \right )}+566 \,{\mathrm e}^{7 i \left (d x +c \right )}+424 \,{\mathrm e}^{6 i \left (d x +c \right )}+566 \,{\mathrm e}^{5 i \left (d x +c \right )}-32 \,{\mathrm e}^{4 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}+12 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{96 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{64 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{64 a^{2} d}\)	\(198\)

[In]

int(csc(d*x+c)^5/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/1536*(3*tan(1/2*d*x+1/2*c)^8+8*tan(1/2*d*x+1/2*c)^6-6*cot(1/2*d*x+1/2*c)^4-6*tan(1/2*d*x+1/2*c)^4-24*cot(1/2
*d*x+1/2*c)^2-48*tan(1/2*d*x+1/2*c)^2-24*ln(tan(1/2*d*x+1/2*c)))/a^2/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (134) = 268\).

Time = 0.28 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.94 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{5} + 12 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - 20 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 70 \, \cos \left (d x + c\right ) + 32}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} + 2 \, a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-1/384*(6*cos(d*x + c)^5 + 12*cos(d*x + c)^4 - 4*cos(d*x + c)^3 - 20*cos(d*x + c)^2 - 3*(cos(d*x + c)^6 + 2*co
s(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)*log(1/2*cos(d*x + c) +
1/2) + 3*(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)*log(-1/2*cos(d*x + c) + 1/2) + 70*cos(d*x + c) + 32)/(a^2*d*cos(d*x + c)^6 + 2*a^2*d*cos(d*x + c)^5
- a^2*d*cos(d*x + c)^4 - 4*a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F]

\[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

Integral(csc(c + d*x)**5/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 10 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right ) + 16\right )}}{a^{2} \cos \left (d x + c\right )^{6} + 2 \, a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} - \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \]

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/384*(2*(3*cos(d*x + c)^5 + 6*cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 10*cos(d*x + c)^2 + 35*cos(d*x + c) + 16)/
(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) - 3*log(cos(d*x + c) + 1)/a^2 + 3*log(cos(d*x + c) - 1)/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.42 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (\frac {4 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {12 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {\frac {48 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {6 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{8}}}{1536 \, d} \]

[In]

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

[Out]

1/1536*(6*(4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x
+ c) + 1)^2/(a^2*(cos(d*x + c) - 1)^2) - 12*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + (48*a^6*(
cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 8*a^6*(cos(d*x + c) -
1)^3/(cos(d*x + c) + 1)^3 + 3*a^6*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/a^8)/d

Mupad [B] (verification not implemented)

Time = 13.64 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{64\,a^2\,d}-\frac {\frac {{\cos \left (c+d\,x\right )}^5}{64}+\frac {{\cos \left (c+d\,x\right )}^4}{32}-\frac {{\cos \left (c+d\,x\right )}^3}{96}-\frac {5\,{\cos \left (c+d\,x\right )}^2}{96}+\frac {35\,\cos \left (c+d\,x\right )}{192}+\frac {1}{12}}{d\,\left (a^2\,{\cos \left (c+d\,x\right )}^6+2\,a^2\,{\cos \left (c+d\,x\right )}^5-a^2\,{\cos \left (c+d\,x\right )}^4-4\,a^2\,{\cos \left (c+d\,x\right )}^3-a^2\,{\cos \left (c+d\,x\right )}^2+2\,a^2\,\cos \left (c+d\,x\right )+a^2\right )} \]

[In]

int(1/(sin(c + d*x)^5*(a + a/cos(c + d*x))^2),x)

[Out]

atanh(cos(c + d*x))/(64*a^2*d) - ((35*cos(c + d*x))/192 - (5*cos(c + d*x)^2)/96 - cos(c + d*x)^3/96 + cos(c +
d*x)^4/32 + cos(c + d*x)^5/64 + 1/12)/(d*(2*a^2*cos(c + d*x) + a^2 - a^2*cos(c + d*x)^2 - 4*a^2*cos(c + d*x)^3
- a^2*cos(c + d*x)^4 + 2*a^2*cos(c + d*x)^5 + a^2*cos(c + d*x)^6))