Integrand size = 15, antiderivative size = 94 \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=-\frac {3 \text {arctanh}(\cos (c+b x)) \cos (a-c)}{8 b}-\frac {3 \cos (a-c) \cot (c+b x) \csc (c+b x)}{8 b}-\frac {\cos (a-c) \cot (c+b x) \csc ^3(c+b x)}{4 b}-\frac {\csc ^5(c+b x) \sin (a-c)}{5 b} \]
-3/8*arctanh(cos(b*x+c))*cos(a-c)/b-3/8*cos(a-c)*cot(b*x+c)*csc(b*x+c)/b-1 /4*cos(a-c)*cot(b*x+c)*csc(b*x+c)^3/b-1/5*csc(b*x+c)^5*sin(a-c)/b
Time = 1.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.84 \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=-\frac {480 \text {arctanh}\left (\cos (c)-\sin (c) \tan \left (\frac {b x}{2}\right )\right ) \cos (a-c)+2 \csc ^5(c+b x) (64 \sin (a-c)+5 \cos (a-c) (14 \sin (2 (c+b x))-3 \sin (4 (c+b x))))}{640 b} \]
-1/640*(480*ArcTanh[Cos[c] - Sin[c]*Tan[(b*x)/2]]*Cos[a - c] + 2*Csc[c + b *x]^5*(64*Sin[a - c] + 5*Cos[a - c]*(14*Sin[2*(c + b*x)] - 3*Sin[4*(c + b* x)])))/b
Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5093, 3042, 25, 3086, 15, 4255, 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.
\(\displaystyle \int \sin (a+b x) \csc ^6(b x+c) \, dx\) |
\(\Big \downarrow \) 5093 |
\(\displaystyle \cos (a-c) \int \csc ^5(c+b x)dx+\sin (a-c) \int \cot (c+b x) \csc ^5(c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^5dx+\sin (a-c) \int -\sec \left (c+b x-\frac {\pi }{2}\right )^5 \tan \left (c+b x-\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^5dx-\sin (a-c) \int \sec \left (\frac {1}{2} (2 c-\pi )+b x\right )^5 \tan \left (\frac {1}{2} (2 c-\pi )+b x\right )dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^5dx-\frac {\sin (a-c) \int \csc ^4(c+b x)d\csc (c+b x)}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \cos (a-c) \int \csc (c+b x)^5dx-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \cos (a-c) \left (\frac {3}{4} \int \csc ^3(c+b x)dx-\frac {\cot (b x+c) \csc ^3(b x+c)}{4 b}\right )-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \left (\frac {3}{4} \int \csc (c+b x)^3dx-\frac {\cot (b x+c) \csc ^3(b x+c)}{4 b}\right )-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \cos (a-c) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc (c+b x)dx-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\cot (b x+c) \csc ^3(b x+c)}{4 b}\right )-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc (c+b x)dx-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\cot (b x+c) \csc ^3(b x+c)}{4 b}\right )-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \cos (a-c) \left (\frac {3}{4} \left (-\frac {\text {arctanh}(\cos (b x+c))}{2 b}-\frac {\cot (b x+c) \csc (b x+c)}{2 b}\right )-\frac {\cot (b x+c) \csc ^3(b x+c)}{4 b}\right )-\frac {\sin (a-c) \csc ^5(b x+c)}{5 b}\) |
Cos[a - c]*(-1/4*(Cot[c + b*x]*Csc[c + b*x]^3)/b + (3*(-1/2*ArcTanh[Cos[c + b*x]]/b - (Cot[c + b*x]*Csc[c + b*x])/(2*b)))/4) - (Csc[c + b*x]^5*Sin[a - c])/(5*b)
3.2.100.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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])
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[Csc[w_]^(n_.)*Sin[v_], x_Symbol] :> Simp[Sin[v - w] Int[Cot[w]*Csc[w] ^(n - 1), x], x] + Simp[Cos[v - w] Int[Csc[w]^(n - 1), x], x] /; GtQ[n, 0 ] && FreeQ[v - w, x] && NeQ[w, v]
Result contains complex when optimal does not.
Time = 13.88 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.73
method | result | size |
risch | \(\frac {-15 \,{\mathrm e}^{i \left (9 x b +11 a +8 c \right )}-15 \,{\mathrm e}^{i \left (9 x b +9 a +10 c \right )}+70 \,{\mathrm e}^{i \left (7 x b +11 a +6 c \right )}+70 \,{\mathrm e}^{i \left (7 x b +9 a +8 c \right )}+128 \,{\mathrm e}^{i \left (5 x b +11 a +4 c \right )}-128 \,{\mathrm e}^{i \left (5 x b +9 a +6 c \right )}-70 \,{\mathrm e}^{i \left (3 x b +11 a +2 c \right )}-70 \,{\mathrm e}^{i \left (3 x b +9 a +4 c \right )}+15 \,{\mathrm e}^{i \left (x b +11 a \right )}+15 \,{\mathrm e}^{i \left (x b +9 a +2 c \right )}}{40 b \left (-{\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{8 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{8 b}\) | \(257\) |
default | \(\text {Expression too large to display}\) | \(6750\) |
1/40/b/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^5*(-15*exp(I*(9*b*x+11*a+8*c))-15* exp(I*(9*b*x+9*a+10*c))+70*exp(I*(7*b*x+11*a+6*c))+70*exp(I*(7*b*x+9*a+8*c ))+128*exp(I*(5*b*x+11*a+4*c))-128*exp(I*(5*b*x+9*a+6*c))-70*exp(I*(3*b*x+ 11*a+2*c))-70*exp(I*(3*b*x+9*a+4*c))+15*exp(I*(b*x+11*a))+15*exp(I*(b*x+9* a+2*c)))+3/8*ln(exp(I*(b*x+a))-exp(I*(a-c)))/b*cos(a-c)-3/8*ln(exp(I*(b*x+ a))+exp(I*(a-c)))/b*cos(a-c)
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (86) = 172\).
Time = 0.26 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.10 \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=-\frac {15 \, {\left (\cos \left (b x + c\right )^{4} \cos \left (-a + c\right ) - 2 \, \cos \left (b x + c\right )^{2} \cos \left (-a + c\right ) + \cos \left (-a + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) - 15 \, {\left (\cos \left (b x + c\right )^{4} \cos \left (-a + c\right ) - 2 \, \cos \left (b x + c\right )^{2} \cos \left (-a + c\right ) + \cos \left (-a + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + c\right ) + \frac {1}{2}\right ) \sin \left (b x + c\right ) - 10 \, {\left (3 \, \cos \left (b x + c\right )^{3} \cos \left (-a + c\right ) - 5 \, \cos \left (b x + c\right ) \cos \left (-a + c\right )\right )} \sin \left (b x + c\right ) - 16 \, \sin \left (-a + c\right )}{80 \, {\left (b \cos \left (b x + c\right )^{4} - 2 \, b \cos \left (b x + c\right )^{2} + b\right )} \sin \left (b x + c\right )} \]
-1/80*(15*(cos(b*x + c)^4*cos(-a + c) - 2*cos(b*x + c)^2*cos(-a + c) + cos (-a + c))*log(1/2*cos(b*x + c) + 1/2)*sin(b*x + c) - 15*(cos(b*x + c)^4*co s(-a + c) - 2*cos(b*x + c)^2*cos(-a + c) + cos(-a + c))*log(-1/2*cos(b*x + c) + 1/2)*sin(b*x + c) - 10*(3*cos(b*x + c)^3*cos(-a + c) - 5*cos(b*x + c )*cos(-a + c))*sin(b*x + c) - 16*sin(-a + c))/((b*cos(b*x + c)^4 - 2*b*cos (b*x + c)^2 + b)*sin(b*x + c))
Timed out. \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3879 vs. \(2 (86) = 172\).
Time = 0.43 (sec) , antiderivative size = 3879, normalized size of antiderivative = 41.27 \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \]
1/80*(2*(15*cos(9*b*x + 2*a + 8*c) + 15*cos(9*b*x + 10*c) - 70*cos(7*b*x + 2*a + 6*c) - 70*cos(7*b*x + 8*c) - 128*cos(5*b*x + 2*a + 4*c) + 128*cos(5 *b*x + 6*c) + 70*cos(3*b*x + 2*a + 2*c) + 70*cos(3*b*x + 4*c) - 15*cos(b*x + 2*a) - 15*cos(b*x + 2*c))*cos(10*b*x + a + 10*c) - 30*(5*cos(8*b*x + a + 8*c) - 10*cos(6*b*x + a + 6*c) + 10*cos(4*b*x + a + 4*c) - 5*cos(2*b*x + a + 2*c) + cos(a))*cos(9*b*x + 2*a + 8*c) - 30*(5*cos(8*b*x + a + 8*c) - 10*cos(6*b*x + a + 6*c) + 10*cos(4*b*x + a + 4*c) - 5*cos(2*b*x + a + 2*c) + cos(a))*cos(9*b*x + 10*c) + 10*(70*cos(7*b*x + 2*a + 6*c) + 70*cos(7*b* x + 8*c) + 128*cos(5*b*x + 2*a + 4*c) - 128*cos(5*b*x + 6*c) - 70*cos(3*b* x + 2*a + 2*c) - 70*cos(3*b*x + 4*c) + 15*cos(b*x + 2*a) + 15*cos(b*x + 2* c))*cos(8*b*x + a + 8*c) - 140*(10*cos(6*b*x + a + 6*c) - 10*cos(4*b*x + a + 4*c) + 5*cos(2*b*x + a + 2*c) - cos(a))*cos(7*b*x + 2*a + 6*c) - 140*(1 0*cos(6*b*x + a + 6*c) - 10*cos(4*b*x + a + 4*c) + 5*cos(2*b*x + a + 2*c) - cos(a))*cos(7*b*x + 8*c) - 20*(128*cos(5*b*x + 2*a + 4*c) - 128*cos(5*b* x + 6*c) - 70*cos(3*b*x + 2*a + 2*c) - 70*cos(3*b*x + 4*c) + 15*cos(b*x + 2*a) + 15*cos(b*x + 2*c))*cos(6*b*x + a + 6*c) + 256*(10*cos(4*b*x + a + 4 *c) - 5*cos(2*b*x + a + 2*c) + cos(a))*cos(5*b*x + 2*a + 4*c) - 256*(10*co s(4*b*x + a + 4*c) - 5*cos(2*b*x + a + 2*c) + cos(a))*cos(5*b*x + 6*c) - 1 00*(14*cos(3*b*x + 2*a + 2*c) + 14*cos(3*b*x + 4*c) - 3*cos(b*x + 2*a) - 3 *cos(b*x + 2*c))*cos(4*b*x + a + 4*c) + 140*(5*cos(2*b*x + a + 2*c) - c...
Leaf count of result is larger than twice the leaf count of optimal. 8035 vs. \(2 (86) = 172\).
Time = 0.36 (sec) , antiderivative size = 8035, normalized size of antiderivative = 85.48 \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \]
1/320*(120*(tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/ 2*c) - tan(1/2*c)^2 + 1)*log(abs(tan(1/2*b*x + 1/2*c)))/(tan(1/2*a)^2*tan( 1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) - (4*tan(1/2*b*x + 1/2*c)^5*ta n(1/2*a)^10*tan(1/2*c)^9 - 4*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^9*tan(1/2*c )^10 - 5*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^10*tan(1/2*c)^10 + 16*tan(1/2*b *x + 1/2*c)^5*tan(1/2*a)^10*tan(1/2*c)^7 - 12*tan(1/2*b*x + 1/2*c)^5*tan(1 /2*a)^9*tan(1/2*c)^8 - 15*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^10*tan(1/2*c)^ 8 + 12*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^8*tan(1/2*c)^9 - 20*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^9*tan(1/2*c)^9 + 20*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a) ^10*tan(1/2*c)^9 - 16*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^7*tan(1/2*c)^10 - 15*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^8*tan(1/2*c)^10 - 20*tan(1/2*b*x + 1/ 2*c)^3*tan(1/2*a)^9*tan(1/2*c)^10 - 40*tan(1/2*b*x + 1/2*c)^2*tan(1/2*a)^1 0*tan(1/2*c)^10 + 24*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^10*tan(1/2*c)^5 - 8 *tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^9*tan(1/2*c)^6 - 10*tan(1/2*b*x + 1/2*c )^4*tan(1/2*a)^10*tan(1/2*c)^6 + 48*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a)^8*ta n(1/2*c)^7 - 80*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^9*tan(1/2*c)^7 + 80*tan( 1/2*b*x + 1/2*c)^3*tan(1/2*a)^10*tan(1/2*c)^7 - 48*tan(1/2*b*x + 1/2*c)^5* tan(1/2*a)^7*tan(1/2*c)^8 - 45*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^8*tan(1/2 *c)^8 - 60*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^9*tan(1/2*c)^8 - 120*tan(1/2* b*x + 1/2*c)^2*tan(1/2*a)^10*tan(1/2*c)^8 + 8*tan(1/2*b*x + 1/2*c)^5*ta...
Timed out. \[ \int \csc ^6(c+b x) \sin (a+b x) \, dx=\text {Hanged} \]