3.1.63 \(\int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx\) [63]

3.1.63.1 Optimal result
3.1.63.2 Mathematica [A] (verified)
3.1.63.3 Rubi [A] (verified)
3.1.63.4 Maple [A] (verified)
3.1.63.5 Fricas [A] (verification not implemented)
3.1.63.6 Sympy [F]
3.1.63.7 Maxima [A] (verification not implemented)
3.1.63.8 Giac [A] (verification not implemented)
3.1.63.9 Mupad [B] (verification not implemented)

3.1.63.1 Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\csc ^4(c+d x)}{4 a d} \]

output
-1/8*arctanh(cos(d*x+c))/a/d-1/8*cot(d*x+c)*csc(d*x+c)/a/d+1/4*cot(d*x+c)* 
csc(d*x+c)^3/a/d-1/4*csc(d*x+c)^4/a/d
 
3.1.63.2 Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\left (2 \cot ^2\left (\frac {1}{2} (c+d x)\right )+4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{16 a d (1+\sec (c+d x))} \]

input
Integrate[Csc[c + d*x]^3/(a + a*Sec[c + d*x]),x]
 
output
-1/16*((2*Cot[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^2*(Log[Cos[(c + d*x)/2]] 
 - Log[Sin[(c + d*x)/2]]) + Sec[(c + d*x)/2]^2)*Sec[c + d*x])/(a*d*(1 + Se 
c[c + d*x]))
 
3.1.63.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.810, Rules used = {3042, 4360, 25, 25, 3042, 25, 3314, 25, 3042, 25, 3086, 15, 3091, 3042, 4255, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3314

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\)

\(\Big \downarrow \) 3086

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

\(\Big \downarrow \) 15

\(\displaystyle -\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

\(\Big \downarrow \) 3091

\(\displaystyle -\frac {-\frac {1}{4} \int \csc ^3(c+d x)dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {1}{4} \int \csc (c+d x)^3dx-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

\(\Big \downarrow \) 4255

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {\cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \csc (c+d x)dx\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {\cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} \int \csc (c+d x)dx\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {\frac {1}{4} \left (\frac {\text {arctanh}(\cos (c+d x))}{2 d}+\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 d}}{a}-\frac {\csc ^4(c+d x)}{4 a d}\)

input
Int[Csc[c + d*x]^3/(a + a*Sec[c + d*x]),x]
 
output
-1/4*Csc[c + d*x]^4/(a*d) - (-1/4*(Cot[c + d*x]*Csc[c + d*x]^3)/d + (ArcTa 
nh[Cos[c + d*x]]/(2*d) + (Cot[c + d*x]*Csc[c + d*x])/(2*d))/4)/a
 

3.1.63.3.1 Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3086
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[a/f   Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 
), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 
] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])
 

rule 3091
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3314
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( 
a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[Cos[e + f 
*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d)   Int[Cos[e + f*x]^(p 
 - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & 
& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p 
+ 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
 

rule 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

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]
 
3.1.63.4 Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67

method result size
derivativedivides \(\frac {-\frac {1}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}+\frac {1}{8 \cos \left (d x +c \right )-8}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}}{d a}\) \(55\)
default \(\frac {-\frac {1}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}+\frac {1}{8 \cos \left (d x +c \right )-8}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}}{d a}\) \(55\)
parallelrisch \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}\) \(61\)
norman \(\frac {-\frac {1}{16 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\) \(79\)
risch \(\frac {{\mathrm e}^{5 i \left (d x +c \right )}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{4 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\) \(128\)

input
int(csc(d*x+c)^3/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/d/a*(-1/8/(cos(d*x+c)+1)^2-1/16*ln(cos(d*x+c)+1)+1/8/(cos(d*x+c)-1)+1/16 
*ln(cos(d*x+c)-1))
 
3.1.63.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, \cos \left (d x + c\right ) + 4}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}} \]

input
integrate(csc(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="fricas")
 
output
1/16*(2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 
 1)*log(1/2*cos(d*x + c) + 1/2) + (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d 
*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*cos(d*x + c) + 4)/(a*d*cos(d 
*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)
 
3.1.63.6 Sympy [F]

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

input
integrate(csc(d*x+c)**3/(a+a*sec(d*x+c)),x)
 
output
Integral(csc(c + d*x)**3/(sec(c + d*x) + 1), x)/a
 
3.1.63.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 2\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]

input
integrate(csc(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="maxima")
 
output
1/16*(2*(cos(d*x + c)^2 + cos(d*x + c) + 2)/(a*cos(d*x + c)^3 + a*cos(d*x 
+ c)^2 - a*cos(d*x + c) - a) - log(cos(d*x + c) + 1)/a + log(cos(d*x + c) 
- 1)/a)/d
 
3.1.63.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.57 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {2 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac {\frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}}}{32 \, d} \]

input
integrate(csc(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="giac")
 
output
-1/32*(2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)/(a 
*(cos(d*x + c) - 1)) - 2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) 
/a - (2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/( 
cos(d*x + c) + 1)^2)/a^2)/d
 
3.1.63.9 Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a\,d}-\frac {\frac {{\cos \left (c+d\,x\right )}^2}{8}+\frac {\cos \left (c+d\,x\right )}{8}+\frac {1}{4}}{d\,\left (-a\,{\cos \left (c+d\,x\right )}^3-a\,{\cos \left (c+d\,x\right )}^2+a\,\cos \left (c+d\,x\right )+a\right )} \]

input
int(1/(sin(c + d*x)^3*(a + a/cos(c + d*x))),x)
 
output
- atanh(cos(c + d*x))/(8*a*d) - (cos(c + d*x)/8 + cos(c + d*x)^2/8 + 1/4)/ 
(d*(a + a*cos(c + d*x) - a*cos(c + d*x)^2 - a*cos(c + d*x)^3))