\(\int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\text {arctanh}(\cos (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \cos (c+d x))^4}+\frac {1}{6 d (a+a \cos (c+d x))^3}-\frac {3}{32 a d (a+a \cos (c+d x))^2}-\frac {1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac {1}{16 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/32*arctanh(cos(d*x+c))/a^3/d-1/16*a/d/(a+a*cos(d*x+c))^4+1/6/d/(a+a*cos(d*x+c))^3-3/32/a/d/(a+a*cos(d*x+c))^
2-1/32/d/(a^3-a^3*cos(d*x+c))-1/16/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2786, 90, 212} \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\text {arctanh}(\cos (c+d x))}{32 a^3 d}-\frac {1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac {1}{16 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {a}{16 d (a \cos (c+d x)+a)^4}+\frac {1}{6 d (a \cos (c+d x)+a)^3}-\frac {3}{32 a d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(32*a^3*d) - a/(16*d*(a + a*Cos[c + d*x])^4) + 1/(6*d*(a + a*Cos[c + d*x])^3) - 3/(32*a*
d*(a + a*Cos[c + d*x])^2) - 1/(32*d*(a^3 - a^3*Cos[c + d*x])) - 1/(16*d*(a^3 + a^3*Cos[c + d*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 2786

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

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 ^3(c+d x)}{(-a-a \cos (c+d x))^3} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{(-a-x)^2 (-a+x)^5} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a}{4 (a-x)^5}+\frac {1}{2 (a-x)^4}-\frac {3}{16 a (a-x)^3}-\frac {1}{16 a^2 (a-x)^2}+\frac {1}{32 a^2 (a+x)^2}-\frac {1}{32 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {a}{16 d (a+a \cos (c+d x))^4}+\frac {1}{6 d (a+a \cos (c+d x))^3}-\frac {3}{32 a d (a+a \cos (c+d x))^2}-\frac {1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac {1}{16 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,-a \cos (c+d x)\right )}{32 a^2 d} \\ & = \frac {\text {arctanh}(\cos (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \cos (c+d x))^4}+\frac {1}{6 d (a+a \cos (c+d x))^3}-\frac {3}{32 a d (a+a \cos (c+d x))^2}-\frac {1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac {1}{16 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (12 \csc ^2\left (\frac {1}{2} (c+d x)\right )+24 \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+24 \sec ^2\left (\frac {1}{2} (c+d x)\right )+18 \sec ^4\left (\frac {1}{2} (c+d x)\right )-16 \sec ^6\left (\frac {1}{2} (c+d x)\right )+3 \sec ^8\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^3(c+d x)}{96 a^3 d (1+\sec (c+d x))^3} \]

[In]

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

[Out]

-1/96*(Cos[(c + d*x)/2]^6*(12*Csc[(c + d*x)/2]^2 + 24*(-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]]) + 24*Se
c[(c + d*x)/2]^2 + 18*Sec[(c + d*x)/2]^4 - 16*Sec[(c + d*x)/2]^6 + 3*Sec[(c + d*x)/2]^8)*Sec[c + d*x]^3)/(a^3*
d*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768 a^{3} d}\)	\(87\)
derivativedivides	\(\frac {\frac {1}{32 \cos \left (d x +c \right )-32}-\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )^{4}}+\frac {1}{6 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {3}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{64}}{d \,a^{3}}\)	\(91\)
default	\(\frac {\frac {1}{32 \cos \left (d x +c \right )-32}-\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )^{4}}+\frac {1}{6 \left (\cos \left (d x +c \right )+1\right )^{3}}-\frac {3}{32 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{64}}{d \,a^{3}}\)	\(91\)
norman	\(\frac {-\frac {1}{64 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{32 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{64 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{192 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{256 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a^{3} d}\)	\(120\)
risch	\(-\frac {3 \,{\mathrm e}^{9 i \left (d x +c \right )}+18 \,{\mathrm e}^{8 i \left (d x +c \right )}-88 \,{\mathrm e}^{7 i \left (d x +c \right )}-162 \,{\mathrm e}^{6 i \left (d x +c \right )}-310 \,{\mathrm e}^{5 i \left (d x +c \right )}-162 \,{\mathrm e}^{4 i \left (d x +c \right )}-88 \,{\mathrm e}^{3 i \left (d x +c \right )}+18 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{48 a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{32 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{32 a^{3} d}\)	\(176\)

[In]

int(csc(d*x+c)^3/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/768*(-3*tan(1/2*d*x+1/2*c)^8+4*tan(1/2*d*x+1/2*c)^6+12*tan(1/2*d*x+1/2*c)^4-24*tan(1/2*d*x+1/2*c)^2-12*cot(1
/2*d*x+1/2*c)^2-24*ln(tan(1/2*d*x+1/2*c)))/a^3/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (115) = 230\).

Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.90 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{4} + 18 \, \cos \left (d x + c\right )^{3} - 50 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 54 \, \cos \left (d x + c\right ) - 16}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} - 3 \, a^{3} d \cos \left (d x + c\right ) - a^{3} d\right )}} \]

[In]

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

[Out]

-1/192*(6*cos(d*x + c)^4 + 18*cos(d*x + c)^3 - 50*cos(d*x + c)^2 - 3*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*co
s(d*x + c)^3 - 2*cos(d*x + c)^2 - 3*cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^5 + 3*cos(
d*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 - 3*cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 54*cos(d
*x + c) - 16)/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x + c)^2
- 3*a^3*d*cos(d*x + c) - a^3*d)

Sympy [F]

\[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\csc ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(csc(d*x+c)**3/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 9 \, \cos \left (d x + c\right )^{3} - 25 \, \cos \left (d x + c\right )^{2} - 27 \, \cos \left (d x + c\right ) - 8\right )}}{a^{3} \cos \left (d x + c\right )^{5} + 3 \, a^{3} \cos \left (d x + c\right )^{4} + 2 \, a^{3} \cos \left (d x + c\right )^{3} - 2 \, a^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} \cos \left (d x + c\right ) - a^{3}} - \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, d} \]

[In]

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

[Out]

-1/192*(2*(3*cos(d*x + c)^4 + 9*cos(d*x + c)^3 - 25*cos(d*x + c)^2 - 27*cos(d*x + c) - 8)/(a^3*cos(d*x + c)^5
+ 3*a^3*cos(d*x + c)^4 + 2*a^3*cos(d*x + c)^3 - 2*a^3*cos(d*x + c)^2 - 3*a^3*cos(d*x + c) - a^3) - 3*log(cos(d
*x + c) + 1)/a^3 + 3*log(cos(d*x + c) - 1)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {12 \, {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {12 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac {\frac {24 \, a^{9} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{9} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, a^{9} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, a^{9} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{12}}}{768 \, d} \]

[In]

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

[Out]

1/768*(12*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a^3*(cos(d*x + c) - 1)) - 12*log(abs
(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (24*a^9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a^9*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 4*a^9*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3*a^9*(cos(d*x + c) -
1)^4/(cos(d*x + c) + 1)^4)/a^12)/d

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.03 \[ \int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{32\,a^3\,d}-\frac {-\frac {{\cos \left (c+d\,x\right )}^4}{32}-\frac {3\,{\cos \left (c+d\,x\right )}^3}{32}+\frac {25\,{\cos \left (c+d\,x\right )}^2}{96}+\frac {9\,\cos \left (c+d\,x\right )}{32}+\frac {1}{12}}{d\,\left (-a^3\,{\cos \left (c+d\,x\right )}^5-3\,a^3\,{\cos \left (c+d\,x\right )}^4-2\,a^3\,{\cos \left (c+d\,x\right )}^3+2\,a^3\,{\cos \left (c+d\,x\right )}^2+3\,a^3\,\cos \left (c+d\,x\right )+a^3\right )} \]

[In]

int(1/(sin(c + d*x)^3*(a + a/cos(c + d*x))^3),x)

[Out]

atanh(cos(c + d*x))/(32*a^3*d) - ((9*cos(c + d*x))/32 + (25*cos(c + d*x)^2)/96 - (3*cos(c + d*x)^3)/32 - cos(c
+ d*x)^4/32 + 1/12)/(d*(3*a^3*cos(c + d*x) + a^3 + 2*a^3*cos(c + d*x)^2 - 2*a^3*cos(c + d*x)^3 - 3*a^3*cos(c
+ d*x)^4 - a^3*cos(c + d*x)^5))