Integrand size = 21, antiderivative size = 201 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {6 a^2 \cot (c+d x)}{d}-\frac {14 a^2 \cot ^3(c+d x)}{3 d}-\frac {16 a^2 \cot ^5(c+d x)}{5 d}-\frac {9 a^2 \cot ^7(c+d x)}{7 d}-\frac {2 a^2 \cot ^9(c+d x)}{9 d}-\frac {2 a^2 \csc (c+d x)}{d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}-\frac {2 a^2 \csc ^9(c+d x)}{9 d}+\frac {a^2 \tan (c+d x)}{d} \] Output:
2*a^2*arctanh(sin(d*x+c))/d-6*a^2*cot(d*x+c)/d-14/3*a^2*cot(d*x+c)^3/d-16/ 5*a^2*cot(d*x+c)^5/d-9/7*a^2*cot(d*x+c)^7/d-2/9*a^2*cot(d*x+c)^9/d-2*a^2*c sc(d*x+c)/d-2/3*a^2*csc(d*x+c)^3/d-2/5*a^2*csc(d*x+c)^5/d-2/7*a^2*csc(d*x+ c)^7/d-2/9*a^2*csc(d*x+c)^9/d+a^2*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(1050\) vs. \(2(201)=402\).
Time = 10.18 (sec) , antiderivative size = 1050, normalized size of antiderivative = 5.22 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx =\text {Too large to display} \] Input:
Integrate[Csc[c + d*x]^10*(a + a*Sec[c + d*x])^2,x]
Output:
(-6899*Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^2*Sec[c/2 + (d*x)/2]^4*( a + a*Sec[c + d*x])^2)/(80640*d) - (193*Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(13440*d) - (71*Co s[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^6*Sec[c/2 + (d*x)/2]^4*(a + a*Sec [c + d*x])^2)/(32256*d) - (Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^8*Se c[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(4608*d) - (Cos[c + d*x]^2*Log[ Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(2*d) + (Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d* x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(2*d) + (123041*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d *x])^2*Sin[(d*x)/2])/(161280*d) + (6899*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^3*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(8064 0*d) + (193*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^5*Sec[c/2 + (d*x)/2 ]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(13440*d) + (71*Cos[c + d*x]^2*Cs c[c/2]*Csc[c/2 + (d*x)/2]^7*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Si n[(d*x)/2])/(32256*d) + (Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^9*Sec[ c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(4608*d) + (803*Cos[ c + d*x]^2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(a + a*Sec[c + d*x])^2*Sin[(d*x)/ 2])/(7680*d) + (49*Cos[c + d*x]^2*Sec[c/2]*Sec[c/2 + (d*x)/2]^7*(a + a*Sec [c + d*x])^2*Sin[(d*x)/2])/(7680*d) + (Cos[c + d*x]^2*Sec[c/2]*Sec[c/2 ...
Time = 0.58 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4360, 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.
\(\displaystyle \int \csc ^{10}(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{10}}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \csc ^{10}(c+d x) \sec ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right )^{10}}dx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (a^2 \csc ^{10}(c+d x)+a^2 \csc ^{10}(c+d x) \sec ^2(c+d x)+2 a^2 \csc ^{10}(c+d x) \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x)}{d}-\frac {2 a^2 \cot ^9(c+d x)}{9 d}-\frac {9 a^2 \cot ^7(c+d x)}{7 d}-\frac {16 a^2 \cot ^5(c+d x)}{5 d}-\frac {14 a^2 \cot ^3(c+d x)}{3 d}-\frac {6 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \csc ^9(c+d x)}{9 d}-\frac {2 a^2 \csc ^7(c+d x)}{7 d}-\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}-\frac {2 a^2 \csc (c+d x)}{d}\) |
Input:
Int[Csc[c + d*x]^10*(a + a*Sec[c + d*x])^2,x]
Output:
(2*a^2*ArcTanh[Sin[c + d*x]])/d - (6*a^2*Cot[c + d*x])/d - (14*a^2*Cot[c + d*x]^3)/(3*d) - (16*a^2*Cot[c + d*x]^5)/(5*d) - (9*a^2*Cot[c + d*x]^7)/(7 *d) - (2*a^2*Cot[c + d*x]^9)/(9*d) - (2*a^2*Csc[c + d*x])/d - (2*a^2*Csc[c + d*x]^3)/(3*d) - (2*a^2*Csc[c + d*x]^5)/(5*d) - (2*a^2*Csc[c + d*x]^7)/( 7*d) - (2*a^2*Csc[c + d*x]^9)/(9*d) + (a^2*Tan[c + d*x])/d
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]
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]
Time = 0.64 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {a^{2} \left (35 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+460 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+63 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+3096 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+1092 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-80640 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+80640 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+16044 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16800 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+80640 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-80640 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+119910 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-238140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40320 d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-40320 d}\) | \(217\) |
derivativedivides | \(\frac {a^{2} \left (-\frac {1}{9 \sin \left (d x +c \right )^{9} \cos \left (d x +c \right )}-\frac {10}{63 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )}-\frac {16}{63 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {32}{63 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {128}{63 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {256 \cot \left (d x +c \right )}{63}\right )+2 a^{2} \left (-\frac {1}{9 \sin \left (d x +c \right )^{9}}-\frac {1}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {128}{315}-\frac {\csc \left (d x +c \right )^{8}}{9}-\frac {8 \csc \left (d x +c \right )^{6}}{63}-\frac {16 \csc \left (d x +c \right )^{4}}{105}-\frac {64 \csc \left (d x +c \right )^{2}}{315}\right ) \cot \left (d x +c \right )}{d}\) | \(231\) |
default | \(\frac {a^{2} \left (-\frac {1}{9 \sin \left (d x +c \right )^{9} \cos \left (d x +c \right )}-\frac {10}{63 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )}-\frac {16}{63 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {32}{63 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {128}{63 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {256 \cot \left (d x +c \right )}{63}\right )+2 a^{2} \left (-\frac {1}{9 \sin \left (d x +c \right )^{9}}-\frac {1}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {128}{315}-\frac {\csc \left (d x +c \right )^{8}}{9}-\frac {8 \csc \left (d x +c \right )^{6}}{63}-\frac {16 \csc \left (d x +c \right )^{4}}{105}-\frac {64 \csc \left (d x +c \right )^{2}}{315}\right ) \cot \left (d x +c \right )}{d}\) | \(231\) |
risch | \(-\frac {4 i a^{2} \left (315 \,{\mathrm e}^{15 i \left (d x +c \right )}-1260 \,{\mathrm e}^{14 i \left (d x +c \right )}+525 \,{\mathrm e}^{13 i \left (d x +c \right )}+4200 \,{\mathrm e}^{12 i \left (d x +c \right )}-5817 \,{\mathrm e}^{11 i \left (d x +c \right )}-2772 \,{\mathrm e}^{10 i \left (d x +c \right )}+10161 \,{\mathrm e}^{9 i \left (d x +c \right )}-4944 \,{\mathrm e}^{8 i \left (d x +c \right )}-3919 \,{\mathrm e}^{7 i \left (d x +c \right )}+15532 \,{\mathrm e}^{6 i \left (d x +c \right )}-8633 \,{\mathrm e}^{5 i \left (d x +c \right )}-8472 \,{\mathrm e}^{4 i \left (d x +c \right )}+8973 \,{\mathrm e}^{3 i \left (d x +c \right )}+148 \,{\mathrm e}^{2 i \left (d x +c \right )}-2501 \,{\mathrm e}^{i \left (d x +c \right )}+704\right )}{315 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{9} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\) | \(259\) |
Input:
int(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
a^2*(35*cot(1/2*d*x+1/2*c)^9+460*cot(1/2*d*x+1/2*c)^7+63*tan(1/2*d*x+1/2*c )^7+3096*cot(1/2*d*x+1/2*c)^5+1092*tan(1/2*d*x+1/2*c)^5-80640*ln(tan(1/2*d *x+1/2*c)-1)*tan(1/2*d*x+1/2*c)^2+80640*ln(tan(1/2*d*x+1/2*c)+1)*tan(1/2*d *x+1/2*c)^2+16044*cot(1/2*d*x+1/2*c)^3+16800*tan(1/2*d*x+1/2*c)^3+80640*ln (tan(1/2*d*x+1/2*c)-1)-80640*ln(tan(1/2*d*x+1/2*c)+1)+119910*cot(1/2*d*x+1 /2*c)-238140*tan(1/2*d*x+1/2*c))/(40320*d*tan(1/2*d*x+1/2*c)^2-40320*d)
Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (185) = 370\).
Time = 0.14 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.02 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {1408 \, a^{2} \cos \left (d x + c\right )^{8} - 2186 \, a^{2} \cos \left (d x + c\right )^{7} - 3372 \, a^{2} \cos \left (d x + c\right )^{6} + 6200 \, a^{2} \cos \left (d x + c\right )^{5} + 2060 \, a^{2} \cos \left (d x + c\right )^{4} - 5784 \, a^{2} \cos \left (d x + c\right )^{3} + 268 \, a^{2} \cos \left (d x + c\right )^{2} + 1756 \, a^{2} \cos \left (d x + c\right ) - 315 \, {\left (a^{2} \cos \left (d x + c\right )^{7} - 2 \, a^{2} \cos \left (d x + c\right )^{6} - a^{2} \cos \left (d x + c\right )^{5} + 4 \, a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 315 \, {\left (a^{2} \cos \left (d x + c\right )^{7} - 2 \, a^{2} \cos \left (d x + c\right )^{6} - a^{2} \cos \left (d x + c\right )^{5} + 4 \, a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 315 \, a^{2}}{315 \, {\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{5} + 4 \, d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-1/315*(1408*a^2*cos(d*x + c)^8 - 2186*a^2*cos(d*x + c)^7 - 3372*a^2*cos(d *x + c)^6 + 6200*a^2*cos(d*x + c)^5 + 2060*a^2*cos(d*x + c)^4 - 5784*a^2*c os(d*x + c)^3 + 268*a^2*cos(d*x + c)^2 + 1756*a^2*cos(d*x + c) - 315*(a^2* cos(d*x + c)^7 - 2*a^2*cos(d*x + c)^6 - a^2*cos(d*x + c)^5 + 4*a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*lo g(sin(d*x + c) + 1)*sin(d*x + c) + 315*(a^2*cos(d*x + c)^7 - 2*a^2*cos(d*x + c)^6 - a^2*cos(d*x + c)^5 + 4*a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) - 315*a^2)/((d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^6 - d*cos(d*x + c)^5 + 4*d*cos(d*x + c)^4 - d*cos(d*x + c)^3 - 2*d*cos(d*x + c)^2 + d*cos(d*x + c))*sin(d*x + c))
Timed out. \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**10*(a+a*sec(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.01 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=-\frac {a^{2} {\left (\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{6} + 63 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{2} + 35\right )}}{\sin \left (d x + c\right )^{9}} - 315 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 5 \, a^{2} {\left (\frac {315 \, \tan \left (d x + c\right )^{8} + 210 \, \tan \left (d x + c\right )^{6} + 126 \, \tan \left (d x + c\right )^{4} + 45 \, \tan \left (d x + c\right )^{2} + 7}{\tan \left (d x + c\right )^{9}} - 63 \, \tan \left (d x + c\right )\right )} + \frac {{\left (315 \, \tan \left (d x + c\right )^{8} + 420 \, \tan \left (d x + c\right )^{6} + 378 \, \tan \left (d x + c\right )^{4} + 180 \, \tan \left (d x + c\right )^{2} + 35\right )} a^{2}}{\tan \left (d x + c\right )^{9}}}{315 \, d} \] Input:
integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/315*(a^2*(2*(315*sin(d*x + c)^8 + 105*sin(d*x + c)^6 + 63*sin(d*x + c)^ 4 + 45*sin(d*x + c)^2 + 35)/sin(d*x + c)^9 - 315*log(sin(d*x + c) + 1) + 3 15*log(sin(d*x + c) - 1)) + 5*a^2*((315*tan(d*x + c)^8 + 210*tan(d*x + c)^ 6 + 126*tan(d*x + c)^4 + 45*tan(d*x + c)^2 + 7)/tan(d*x + c)^9 - 63*tan(d* x + c)) + (315*tan(d*x + c)^8 + 420*tan(d*x + c)^6 + 378*tan(d*x + c)^4 + 180*tan(d*x + c)^2 + 35)*a^2/tan(d*x + c)^9)/d
Time = 0.20 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {63 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1155 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80640 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 80640 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 17955 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {80640 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {139545 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 19635 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3591 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 495 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}}}{40320 \, d} \] Input:
integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/40320*(63*a^2*tan(1/2*d*x + 1/2*c)^5 + 1155*a^2*tan(1/2*d*x + 1/2*c)^3 + 80640*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 80640*a^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + 17955*a^2*tan(1/2*d*x + 1/2*c) - 80640*a^2*tan(1/2*d* x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (139545*a^2*tan(1/2*d*x + 1/2*c) ^8 + 19635*a^2*tan(1/2*d*x + 1/2*c)^6 + 3591*a^2*tan(1/2*d*x + 1/2*c)^4 + 495*a^2*tan(1/2*d*x + 1/2*c)^2 + 35*a^2)/tan(1/2*d*x + 1/2*c)^9)/d
Time = 10.22 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{384\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{640\,d}+\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-699\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {1142\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {764\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {344\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}+\frac {92\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{63}+\frac {a^2}{9}}{d\,\left (128\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-128\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\right )}+\frac {57\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,d} \] Input:
int((a + a/cos(c + d*x))^2/sin(c + d*x)^10,x)
Output:
(11*a^2*tan(c/2 + (d*x)/2)^3)/(384*d) + (a^2*tan(c/2 + (d*x)/2)^5)/(640*d) + (4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - ((92*a^2*tan(c/2 + (d*x)/2)^2)/63 + (344*a^2*tan(c/2 + (d*x)/2)^4)/35 + (764*a^2*tan(c/2 + (d*x)/2)^6)/15 + (1142*a^2*tan(c/2 + (d*x)/2)^8)/3 - 699*a^2*tan(c/2 + (d*x)/2)^10 + a^2/9 )/(d*(128*tan(c/2 + (d*x)/2)^9 - 128*tan(c/2 + (d*x)/2)^11)) + (57*a^2*tan (c/2 + (d*x)/2))/(128*d)
Time = 0.18 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.19 \[ \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^{2} \left (-80640 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}+80640 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+80640 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-80640 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+63 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+1092 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+16800 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-238140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+119910 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+16044 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+3096 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+460 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+35\right )}{40320 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:
int(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x)
Output:
(a**2*( - 80640*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**11 + 80640*log (tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**9 + 80640*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**11 - 80640*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2) **9 + 63*tan((c + d*x)/2)**16 + 1092*tan((c + d*x)/2)**14 + 16800*tan((c + d*x)/2)**12 - 238140*tan((c + d*x)/2)**10 + 119910*tan((c + d*x)/2)**8 + 16044*tan((c + d*x)/2)**6 + 3096*tan((c + d*x)/2)**4 + 460*tan((c + d*x)/2 )**2 + 35))/(40320*tan((c + d*x)/2)**9*d*(tan((c + d*x)/2)**2 - 1))