\(\int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [764]

Optimal result

Integrand size = 29, antiderivative size = 82 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {5 a^2 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {2 a^2 \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a^2 \sec (c+d x)}{d}+\frac {2 a^2 \tan (c+d x)}{d} \] Output:

-5/2*a^2*arctanh(cos(d*x+c))/d-2*a^2*cot(d*x+c)/d-1/2*a^2*cot(d*x+c)*csc(d 
*x+c)/d+2*a^2*sec(d*x+c)/d+2*a^2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.51 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )-20 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+20 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {32 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+8 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \] Input:

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
 

Output:

(a^2*(-8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 20*Log[Cos[(c + d*x)/2]] 
+ 20*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + (32*Sin[(c + d*x)/2])/(C 
os[(c + d*x)/2] - Sin[(c + d*x)/2]) + 8*Tan[(c + d*x)/2]))/(8*d)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 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.

\(\displaystyle \int \csc ^3(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (a^2 \csc ^3(c+d x) \sec ^2(c+d x)+2 a^2 \csc ^2(c+d x) \sec ^2(c+d x)+a^2 \csc (c+d x) \sec ^2(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {5 a^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {5 a^2 \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}\)

Input:

Int[Csc[c + d*x]^3*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
 

Output:

(-5*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (2*a^2*Cot[c + d*x])/d + (5*a^2*Sec 
[c + d*x])/(2*d) - (a^2*Csc[c + d*x]^2*Sec[c + d*x])/(2*d) + (2*a^2*Tan[c 
+ d*x])/d
 

Defintions of rubi rules used

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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.13

method	result	size
parallelrisch	\(\frac {\left (-48+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{8 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(93\)
derivativedivides	\(\frac {a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(117\)
default	\(\frac {a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+a^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(117\)
risch	\(\frac {a^{2} \left (-5 i {\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i {\mathrm e}^{i \left (d x +c \right )}-11 \,{\mathrm e}^{2 i \left (d x +c \right )}+8\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d}-\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\)	\(124\)
norman	\(\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {a^{2}}{8 d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {10 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8 d}-\frac {15 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 d}-\frac {35 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}-\frac {65 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {5 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(238\)

Input:

int(csc(d*x+c)^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/8*(-48+20*ln(tan(1/2*d*x+1/2*c))*(tan(1/2*d*x+1/2*c)-1)+tan(1/2*d*x+1/2* 
c)^3+cot(1/2*d*x+1/2*c)^2+7*tan(1/2*d*x+1/2*c)^2+7*cot(1/2*d*x+1/2*c))*a^2 
/d/(tan(1/2*d*x+1/2*c)-1)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (78) = 156\).

Time = 0.08 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.66 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {16 \, a^{2} \cos \left (d x + c\right )^{3} + 10 \, a^{2} \cos \left (d x + c\right )^{2} - 14 \, a^{2} \cos \left (d x + c\right ) - 8 \, a^{2} - 5 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

1/4*(16*a^2*cos(d*x + c)^3 + 10*a^2*cos(d*x + c)^2 - 14*a^2*cos(d*x + c) - 
 8*a^2 - 5*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - a 
^2 - (a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) 
+ 5*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - a^2 - (a 
^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(8 
*a^2*cos(d*x + c)^2 + 3*a^2*cos(d*x + c) - 4*a^2)*sin(d*x + c))/(d*cos(d*x 
 + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) - (d*cos(d*x + c)^2 - d)*sin(d 
*x + c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:

integrate(csc(d*x+c)**3*sec(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.51 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{4 \, d} \] Input:

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.41 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {32 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} - \frac {30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 20*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 
8*a^2*tan(1/2*d*x + 1/2*c) - 32*a^2/(tan(1/2*d*x + 1/2*c) - 1) - (30*a^2*t 
an(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/ 
2*c)^2)/d
 

Mupad [B] (verification not implemented)

Time = 31.32 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.51 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \] Input:

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^2*sin(c + d*x)^3),x)
 

Output:

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (5*a^2*log(tan(c/2 + (d*x)/2)))/(2*d) + 
 (a^2*tan(c/2 + (d*x)/2))/d - (a^2/2 - 20*a^2*tan(c/2 + (d*x)/2)^2 + (7*a^ 
2*tan(c/2 + (d*x)/2))/2)/(d*(4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) 
^3))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.56 \[ \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (20 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-20 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:

int(csc(d*x+c)^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
 

Output:

(a**2*(20*log(tan((c + d*x)/2))*tan((c + d*x)/2)**3 - 20*log(tan((c + d*x) 
/2))*tan((c + d*x)/2)**2 + tan((c + d*x)/2)**5 + 7*tan((c + d*x)/2)**4 - 4 
8*tan((c + d*x)/2)**3 + 7*tan((c + d*x)/2) + 1))/(8*tan((c + d*x)/2)**2*d* 
(tan((c + d*x)/2) - 1))