Integrand size = 15, antiderivative size = 94 \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\frac {\cos (a-c) \sec ^5(c+b x)}{5 b}+\frac {3 \text {arctanh}(\sin (c+b x)) \sin (a-c)}{8 b}+\frac {3 \sec (c+b x) \sin (a-c) \tan (c+b x)}{8 b}+\frac {\sec ^3(c+b x) \sin (a-c) \tan (c+b x)}{4 b} \]
1/5*cos(a-c)*sec(b*x+c)^5/b+3/8*arctanh(sin(b*x+c))*sin(a-c)/b+3/8*sec(b*x +c)*sin(a-c)*tan(b*x+c)/b+1/4*sec(b*x+c)^3*sin(a-c)*tan(b*x+c)/b
Time = 1.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\frac {480 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \sin (a-c)+2 \sec ^5(c+b x) (64 \cos (a-c)+5 \sin (a-c) (14 \sin (2 (c+b x))+3 \sin (4 (c+b x))))}{640 b} \]
(480*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Sin[a - c] + 2*Sec[c + b*x]^5*( 64*Cos[a - c] + 5*Sin[a - c]*(14*Sin[2*(c + b*x)] + 3*Sin[4*(c + b*x)])))/ (640*b)
Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5091, 3042, 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.
\(\displaystyle \int \sin (a+b x) \sec ^6(b x+c) \, dx\) |
\(\Big \downarrow \) 5091 |
\(\displaystyle \sin (a-c) \int \sec ^5(c+b x)dx+\cos (a-c) \int \sec ^5(c+b x) \tan (c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^5dx+\cos (a-c) \int \sec (c+b x)^5 \tan (c+b x)dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\cos (a-c) \int \sec ^4(c+b x)d\sec (c+b x)}{b}+\sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^5dx+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \sin (a-c) \left (\frac {3}{4} \int \sec ^3(c+b x)dx+\frac {\tan (b x+c) \sec ^3(b x+c)}{4 b}\right )+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \left (\frac {3}{4} \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx+\frac {\tan (b x+c) \sec ^3(b x+c)}{4 b}\right )+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \sin (a-c) \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+b x)dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\tan (b x+c) \sec ^3(b x+c)}{4 b}\right )+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+b x+\frac {\pi }{2}\right )dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\tan (b x+c) \sec ^3(b x+c)}{4 b}\right )+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \sin (a-c) \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (b x+c))}{2 b}+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\tan (b x+c) \sec ^3(b x+c)}{4 b}\right )+\frac {\cos (a-c) \sec ^5(b x+c)}{5 b}\) |
(Cos[a - c]*Sec[c + b*x]^5)/(5*b) + Sin[a - c]*((Sec[c + b*x]^3*Tan[c + b* x])/(4*b) + (3*(ArcTanh[Sin[c + b*x]]/(2*b) + (Sec[c + b*x]*Tan[c + b*x])/ (2*b)))/4)
3.3.18.3.1 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[Sec[w_]^(n_.)*Sin[v_], x_Symbol] :> Simp[Cos[v - w] Int[Tan[w]*Sec[w] ^(n - 1), x], x] + Simp[Sin[v - w] Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0 ] && FreeQ[v - w, x] && NeQ[w, v]
Result contains complex when optimal does not.
Time = 49.84 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.76
method | result | size |
risch | \(\frac {-15 \,{\mathrm e}^{i \left (9 x b +11 a +8 c \right )}+15 \,{\mathrm e}^{i \left (9 x b +9 a +10 c \right )}-70 \,{\mathrm e}^{i \left (7 x b +11 a +6 c \right )}+70 \,{\mathrm e}^{i \left (7 x b +9 a +8 c \right )}+128 \,{\mathrm e}^{i \left (5 x b +11 a +4 c \right )}+128 \,{\mathrm e}^{i \left (5 x b +9 a +6 c \right )}+70 \,{\mathrm e}^{i \left (3 x b +11 a +2 c \right )}-70 \,{\mathrm e}^{i \left (3 x b +9 a +4 c \right )}+15 \,{\mathrm e}^{i \left (x b +11 a \right )}-15 \,{\mathrm e}^{i \left (x b +9 a +2 c \right )}}{40 b \left ({\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{8 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{8 b}\) | \(259\) |
default | \(\text {Expression too large to display}\) | \(6742\) |
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*e xp(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+1 1*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))-I*exp(I*(a-c)))/b*sin(a-c)+3/8*ln(exp(I*(b*x +a))+I*exp(I*(a-c)))/b*sin(a-c)
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.14 \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=-\frac {15 \, \cos \left (b x + c\right )^{5} \log \left (\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) - 15 \, \cos \left (b x + c\right )^{5} \log \left (-\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) + 10 \, {\left (3 \, \cos \left (b x + c\right )^{3} + 2 \, \cos \left (b x + c\right )\right )} \sin \left (b x + c\right ) \sin \left (-a + c\right ) - 16 \, \cos \left (-a + c\right )}{80 \, b \cos \left (b x + c\right )^{5}} \]
-1/80*(15*cos(b*x + c)^5*log(sin(b*x + c) + 1)*sin(-a + c) - 15*cos(b*x + c)^5*log(-sin(b*x + c) + 1)*sin(-a + c) + 10*(3*cos(b*x + c)^3 + 2*cos(b*x + c))*sin(b*x + c)*sin(-a + c) - 16*cos(-a + c))/(b*cos(b*x + c)^5)
Timed out. \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3096 vs. \(2 (86) = 172\).
Time = 0.52 (sec) , antiderivative size = 3096, normalized size of antiderivative = 32.94 \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \]
-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*( 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 + 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*c os(4*b*x + a + 4*c) + 5*cos(2*b*x + a + 2*c) + cos(a))*cos(5*b*x + 6*c) - 100*(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) + ...
Leaf count of result is larger than twice the leaf count of optimal. 756 vs. \(2 (86) = 172\).
Time = 0.32 (sec) , antiderivative size = 756, normalized size of antiderivative = 8.04 \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \]
1/20*(15*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c) + 1))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) - 15*(tan(1/2*a)^2*tan(1/2*c) - tan(1/ 2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c) - 1))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + 2*(2 5*tan(1/2*b*x + 1/2*c)^9*tan(1/2*a)^2*tan(1/2*c) - 25*tan(1/2*b*x + 1/2*c) ^9*tan(1/2*a)*tan(1/2*c)^2 - 20*tan(1/2*b*x + 1/2*c)^8*tan(1/2*a)^2*tan(1/ 2*c)^2 + 25*tan(1/2*b*x + 1/2*c)^9*tan(1/2*a) + 20*tan(1/2*b*x + 1/2*c)^8* tan(1/2*a)^2 - 25*tan(1/2*b*x + 1/2*c)^9*tan(1/2*c) - 80*tan(1/2*b*x + 1/2 *c)^8*tan(1/2*a)*tan(1/2*c) - 10*tan(1/2*b*x + 1/2*c)^7*tan(1/2*a)^2*tan(1 /2*c) + 20*tan(1/2*b*x + 1/2*c)^8*tan(1/2*c)^2 + 10*tan(1/2*b*x + 1/2*c)^7 *tan(1/2*a)*tan(1/2*c)^2 - 20*tan(1/2*b*x + 1/2*c)^8 - 10*tan(1/2*b*x + 1/ 2*c)^7*tan(1/2*a) + 10*tan(1/2*b*x + 1/2*c)^7*tan(1/2*c) - 40*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^2*tan(1/2*c)^2 + 40*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a )^2 - 160*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)*tan(1/2*c) + 10*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)^2*tan(1/2*c) + 40*tan(1/2*b*x + 1/2*c)^4*tan(1/2*c)^2 - 10*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a)*tan(1/2*c)^2 - 40*tan(1/2*b*x + 1/2 *c)^4 + 10*tan(1/2*b*x + 1/2*c)^3*tan(1/2*a) - 10*tan(1/2*b*x + 1/2*c)^3*t an(1/2*c) - 25*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^2*tan(1/2*c) + 25*tan(1/2*b *x + 1/2*c)*tan(1/2*a)*tan(1/2*c)^2 - 4*tan(1/2*a)^2*tan(1/2*c)^2 - 25*...
Timed out. \[ \int \sec ^6(c+b x) \sin (a+b x) \, dx=\text {Hanged} \]