\(\int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [97]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {1}{6 d (a+a \cos (c+d x))^3}+\frac {5 a}{8 d \left (a^2+a^2 \cos (c+d x)\right )^2}-\frac {7}{8 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

-1/8*arctanh(cos(d*x+c))/a^3/d-1/6/d/(a+a*cos(d*x+c))^3+5/8*a/d/(a^2+a^2*c 
os(d*x+c))^2-7/8/d/(a^3+a^3*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.15 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\left (2-15 \cos ^2\left (\frac {1}{2} (c+d x)\right )+42 \cos ^4\left (\frac {1}{2} (c+d x)\right )+12 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sec ^3(c+d x)}{12 a^3 d (1+\sec (c+d x))^3} \] Input:

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

Output:

-1/12*((2 - 15*Cos[(c + d*x)/2]^2 + 42*Cos[(c + d*x)/2]^4 + 12*Cos[(c + d* 
x)/2]^6*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]))*Sec[c + d*x]^3)/( 
a^3*d*(1 + Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 4360, 25, 25, 3042, 25, 3315, 25, 27, 99, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int -\frac {\cos ^2(c+d x) \cot (c+d x)}{(a (-\cos (c+d x))-a)^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\frac {\cos ^2(c+d x) \cot (c+d x)}{(\cos (c+d x) a+a)^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {\cos ^2(c+d x) \cot (c+d x)}{(a \cos (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right ) \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a \int -\frac {\cos ^3(c+d x)}{(a-a \cos (c+d x)) (\cos (c+d x) a+a)^4}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a \int \frac {\cos ^3(c+d x)}{(a-a \cos (c+d x)) (\cos (c+d x) a+a)^4}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {a^3 \cos ^3(c+d x)}{(a-a \cos (c+d x)) (\cos (c+d x) a+a)^4}d(a \cos (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 99

\(\displaystyle -\frac {\int \left (-\frac {a^2}{2 (\cos (c+d x) a+a)^4}+\frac {5 a}{4 (\cos (c+d x) a+a)^3}+\frac {1}{8 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {7}{8 (\cos (c+d x) a+a)^2}\right )d(a \cos (c+d x))}{a^2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {a^2}{6 (a \cos (c+d x)+a)^3}+\frac {\text {arctanh}(\cos (c+d x))}{8 a}-\frac {5 a}{8 (a \cos (c+d x)+a)^2}+\frac {7}{8 (a \cos (c+d x)+a)}}{a^2 d}\)

Input:

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

Output:

-((ArcTanh[Cos[c + d*x]]/(8*a) + a^2/(6*(a + a*Cos[c + d*x])^3) - (5*a)/(8 
*(a + a*Cos[c + d*x])^2) + 7/(8*(a + a*Cos[c + d*x])))/(a^2*d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3315

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 

rule 4360

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d \,a^{3}}\)	\(61\)
derivativedivides	\(\frac {-\frac {1}{6 \left (1+\cos \left (d x +c \right )\right )^{3}}+\frac {5}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {7}{8 \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{16}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{16}}{d \,a^{3}}\)	\(67\)
default	\(\frac {-\frac {1}{6 \left (1+\cos \left (d x +c \right )\right )^{3}}+\frac {5}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {7}{8 \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{16}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{16}}{d \,a^{3}}\)	\(67\)
norman	\(\frac {-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{32 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{48 d a}}{a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}\)	\(82\)
risch	\(-\frac {21 \,{\mathrm e}^{5 i \left (d x +c \right )}+54 \,{\mathrm e}^{4 i \left (d x +c \right )}+82 \,{\mathrm e}^{3 i \left (d x +c \right )}+54 \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{i \left (d x +c \right )}}{12 a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}\)	\(119\)

Input:

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

Output:

1/96*(-2*tan(1/2*d*x+1/2*c)^6+9*tan(1/2*d*x+1/2*c)^4-18*tan(1/2*d*x+1/2*c) 
^2+12*ln(tan(1/2*d*x+1/2*c)))/d/a^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {42 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 54 \, \cos \left (d x + c\right ) + 20}{48 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

-1/48*(42*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 + 3*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)^3 + 3*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) 
+ 20)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c 
) + a^3*d)
 

Sympy [F]

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

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

Output:

Integral(csc(c + d*x)/(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)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.17 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (21 \, \cos \left (d x + c\right )^{2} + 27 \, \cos \left (d x + c\right ) + 10\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \] Input:

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

Output:

-1/48*(2*(21*cos(d*x + c)^2 + 27*cos(d*x + c) + 10)/(a^3*cos(d*x + c)^3 + 
3*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)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{16 \, a^{3} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{16 \, a^{3} d} - \frac {21 \, \cos \left (d x + c\right )^{2} + 27 \, \cos \left (d x + c\right ) + 10}{24 \, a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}^{3}} \] Input:

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

Output:

-1/16*log(abs(cos(d*x + c) + 1))/(a^3*d) + 1/16*log(abs(cos(d*x + c) - 1)) 
/(a^3*d) - 1/24*(21*cos(d*x + c)^2 + 27*cos(d*x + c) + 10)/(a^3*d*(cos(d*x 
 + c) + 1)^3)
 

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {7\,{\cos \left (c+d\,x\right )}^2}{8}+\frac {9\,\cos \left (c+d\,x\right )}{8}+\frac {5}{12}}{d\,\left (a^3\,{\cos \left (c+d\,x\right )}^3+3\,a^3\,{\cos \left (c+d\,x\right )}^2+3\,a^3\,\cos \left (c+d\,x\right )+a^3\right )}-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a^3\,d} \] Input:

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

Output:

- ((9*cos(c + d*x))/8 + (7*cos(c + d*x)^2)/8 + 5/12)/(d*(3*a^3*cos(c + d*x 
) + a^3 + 3*a^3*cos(c + d*x)^2 + a^3*cos(c + d*x)^3)) - atanh(cos(c + d*x) 
)/(8*a^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{96 a^{3} d} \] Input:

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

Output:

(12*log(tan((c + d*x)/2)) - 2*tan((c + d*x)/2)**6 + 9*tan((c + d*x)/2)**4 
- 18*tan((c + d*x)/2)**2)/(96*a**3*d)