\(\int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx\) [26]

Optimal result

Integrand size = 21, antiderivative size = 119 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^4}{4 d (a-a \cos (c+d x))^2}-\frac {5 a^5}{4 d \left (a^3-a^3 \cos (c+d x)\right )}+\frac {17 a^2 \log (1-\cos (c+d x))}{8 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (1+\cos (c+d x))}{8 d}+\frac {a^2 \sec (c+d x)}{d} \] Output:

-1/4*a^4/d/(a-a*cos(d*x+c))^2-5/4*a^5/d/(a^3-a^3*cos(d*x+c))+17/8*a^2*ln(1 
-cos(d*x+c))/d-2*a^2*ln(cos(d*x+c))/d-1/8*a^2*ln(1+cos(d*x+c))/d+a^2*sec(d 
*x+c)/d
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (10 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\csc ^4\left (\frac {1}{2} (c+d x)\right )+4 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 \log (\cos (c+d x))-17 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 \sec (c+d x)\right )\right )}{64 d} \] Input:

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

Output:

-1/64*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(10*Csc[(c + d*x)/2]^2 
+ Csc[(c + d*x)/2]^4 + 4*(Log[Cos[(c + d*x)/2]] + 8*Log[Cos[c + d*x]] - 17 
*Log[Sin[(c + d*x)/2]] - 4*Sec[c + d*x])))/d
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4360, 3042, 3315, 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.

\(\displaystyle \int \csc ^5(c+d x) (a \sec (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int \csc ^5(c+d x) \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {a^5 \int \frac {\sec ^2(c+d x)}{(a-a \cos (c+d x))^3 (\cos (c+d x) a+a)}d(-a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^7 \int \frac {\sec ^2(c+d x)}{a^2 (a-a \cos (c+d x))^3 (\cos (c+d x) a+a)}d(-a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {a^7 \int \left (\frac {\sec ^2(c+d x)}{a^6}+\frac {2 \sec (c+d x)}{a^6}+\frac {17}{8 a^5 (a-a \cos (c+d x))}+\frac {1}{8 a^5 (\cos (c+d x) a+a)}+\frac {5}{4 a^4 (a-a \cos (c+d x))^2}+\frac {1}{2 a^3 (a-a \cos (c+d x))^3}\right )d(-a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^7 \left (\frac {\sec (c+d x)}{a^5}-\frac {2 \log (-a \cos (c+d x))}{a^5}+\frac {17 \log (a-a \cos (c+d x))}{8 a^5}-\frac {\log (a \cos (c+d x)+a)}{8 a^5}-\frac {5}{4 a^4 (a-a \cos (c+d x))}-\frac {1}{4 a^3 (a-a \cos (c+d x))^2}\right )}{d}\)

Input:

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

Output:

(a^7*(-1/4*1/(a^3*(a - a*Cos[c + d*x])^2) - 5/(4*a^4*(a - a*Cos[c + d*x])) 
 - (2*Log[-(a*Cos[c + d*x])])/a^5 + (17*Log[a - a*Cos[c + d*x]])/(8*a^5) - 
 Log[a + a*Cos[c + d*x]]/(8*a^5) + Sec[c + d*x]/a^5))/d
 

Defintions of rubi rules used

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 = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {17 a^{2} \left (-\frac {8 \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{17}-\frac {8 \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{17}+\cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {45 \left (\cos \left (d x +c \right )-\frac {3 \cos \left (2 d x +2 c \right )}{10}-\frac {59}{90}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{136}\right )}{4 d \cos \left (d x +c \right )}\)	\(118\)
norman	\(\frac {\frac {a^{2}}{16 d}+\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 d}-\frac {11 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {17 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(134\)
risch	\(\frac {a^{2} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}-28 \,{\mathrm e}^{4 i \left (d x +c \right )}+34 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}+\frac {17 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(152\)
derivativedivides	\(\frac {a^{2} \left (-\frac {1}{4 \sin \left (d x +c \right )^{4} \cos \left (d x +c \right )}-\frac {5}{8 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {15}{8 \cos \left (d x +c \right )}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(157\)
default	\(\frac {a^{2} \left (-\frac {1}{4 \sin \left (d x +c \right )^{4} \cos \left (d x +c \right )}-\frac {5}{8 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {15}{8 \cos \left (d x +c \right )}+\frac {15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\)	\(157\)

Input:

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

Output:

17/4*a^2*(-8/17*cos(d*x+c)*ln(tan(1/2*d*x+1/2*c)-1)-8/17*cos(d*x+c)*ln(tan 
(1/2*d*x+1/2*c)+1)+cos(d*x+c)*ln(tan(1/2*d*x+1/2*c))-45/136*(cos(d*x+c)-3/ 
10*cos(2*d*x+2*c)-59/90)*cot(1/2*d*x+1/2*c)^2*csc(1/2*d*x+1/2*c)^2)/d/cos( 
d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.76 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {18 \, a^{2} \cos \left (d x + c\right )^{2} - 28 \, a^{2} \cos \left (d x + c\right ) + 8 \, a^{2} - 16 \, {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 17 \, {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

1/8*(18*a^2*cos(d*x + c)^2 - 28*a^2*cos(d*x + c) + 8*a^2 - 16*(a^2*cos(d*x 
 + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(-cos(d*x + c)) - (a 
^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(1/2*cos(d 
*x + c) + 1/2) + 17*(a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d 
*x + c))*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^3 - 2*d*cos(d*x + c 
)^2 + d*cos(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) - 17 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) + 16 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (9 \, a^{2} \cos \left (d x + c\right )^{2} - 14 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )}}{\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )}}{8 \, d} \] Input:

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

Output:

-1/8*(a^2*log(cos(d*x + c) + 1) - 17*a^2*log(cos(d*x + c) - 1) + 16*a^2*lo 
g(cos(d*x + c)) - 2*(9*a^2*cos(d*x + c)^2 - 14*a^2*cos(d*x + c) + 4*a^2)/( 
cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {1}{8} \, a^{2} {\left (\frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {17 \, \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {16 \, \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{d} - \frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 14 \, \cos \left (d x + c\right ) + 4\right )}}{d {\left (\cos \left (d x + c\right ) - 1\right )}^{2} \cos \left (d x + c\right )}\right )} \] Input:

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

Output:

-1/8*a^2*(log(abs(cos(d*x + c) + 1))/d - 17*log(abs(cos(d*x + c) - 1))/d + 
 16*log(abs(cos(d*x + c)))/d - 2*(9*cos(d*x + c)^2 - 14*cos(d*x + c) + 4)/ 
(d*(cos(d*x + c) - 1)^2*cos(d*x + c)))
 

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {17\,a^2\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{8\,d}-\frac {a^2\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{8\,d}+\frac {\frac {9\,a^2\,{\cos \left (c+d\,x\right )}^2}{4}-\frac {7\,a^2\,\cos \left (c+d\,x\right )}{2}+a^2}{d\,\left ({\cos \left (c+d\,x\right )}^3-2\,{\cos \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\right )}-\frac {2\,a^2\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \] Input:

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

Output:

(17*a^2*log(cos(c + d*x) - 1))/(8*d) - (a^2*log(cos(c + d*x) + 1))/(8*d) + 
 (a^2 - (7*a^2*cos(c + d*x))/2 + (9*a^2*cos(c + d*x)^2)/4)/(d*(cos(c + d*x 
) - 2*cos(c + d*x)^2 + cos(c + d*x)^3)) - (2*a^2*log(cos(c + d*x)))/d
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.75 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^{2} \left (-32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+68 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-68 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-44 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:

int(csc(d*x+c)^5*(a+a*sec(d*x+c))^2,x)
 

Output:

(a**2*( - 32*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**6 + 32*log(tan((c 
 + d*x)/2) - 1)*tan((c + d*x)/2)**4 - 32*log(tan((c + d*x)/2) + 1)*tan((c 
+ d*x)/2)**6 + 32*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4 + 68*log(t 
an((c + d*x)/2))*tan((c + d*x)/2)**6 - 68*log(tan((c + d*x)/2))*tan((c + d 
*x)/2)**4 - 44*tan((c + d*x)/2)**6 + 11*tan((c + d*x)/2)**2 + 1))/(16*tan( 
(c + d*x)/2)**4*d*(tan((c + d*x)/2)**2 - 1))