Integrand size = 15, antiderivative size = 67 \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\frac {\text {arctanh}(\sin (c+b x)) \cos (a-c)}{2 b}-\frac {\sec ^3(c+b x) \sin (a-c)}{3 b}+\frac {\cos (a-c) \sec (c+b x) \tan (c+b x)}{2 b} \] Output:
1/2*arctanh(sin(b*x+c))*cos(a-c)/b-1/3*sec(b*x+c)^3*sin(a-c)/b+1/2*cos(a-c )*sec(b*x+c)*tan(b*x+c)/b
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\frac {12 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \cos (a-c)+\sec ^3(c+b x) (-4 \sin (a-c)+3 \cos (a-c) \sin (2 (c+b x)))}{12 b} \] Input:
Integrate[Cos[a + b*x]*Sec[c + b*x]^4,x]
Output:
(12*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Cos[a - c] + Sec[c + b*x]^3*(-4* Sin[a - c] + 3*Cos[a - c]*Sin[2*(c + b*x)]))/(12*b)
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {5094, 3042, 3086, 15, 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.
\(\displaystyle \int \cos (a+b x) \sec ^4(b x+c) \, dx\) |
\(\Big \downarrow \) 5094 |
\(\displaystyle \cos (a-c) \int \sec ^3(c+b x)dx-\sin (a-c) \int \sec ^3(c+b x) \tan (c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx-\sin (a-c) \int \sec (c+b x)^3 \tan (c+b x)dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \cos (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx-\frac {\sin (a-c) \int \sec ^2(c+b x)d\sec (c+b x)}{b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \cos (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx-\frac {\sin (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \cos (a-c) \left (\frac {1}{2} \int \sec (c+b x)dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )-\frac {\sin (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos (a-c) \left (\frac {1}{2} \int \csc \left (c+b x+\frac {\pi }{2}\right )dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )-\frac {\sin (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \cos (a-c) \left (\frac {\text {arctanh}(\sin (b x+c))}{2 b}+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )-\frac {\sin (a-c) \sec ^3(b x+c)}{3 b}\) |
Input:
Int[Cos[a + b*x]*Sec[c + b*x]^4,x]
Output:
-1/3*(Sec[c + b*x]^3*Sin[a - c])/b + Cos[a - c]*(ArcTanh[Sin[c + b*x]]/(2* b) + (Sec[c + b*x]*Tan[c + b*x])/(2*b))
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[Cos[v_]*Sec[w_]^(n_.), x_Symbol] :> Simp[-Sin[v - w] Int[Tan[w]*Sec[w ]^(n - 1), x], x] + Simp[Cos[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 = 9.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.87
method | result | size |
risch | \(\frac {i \left (-3 \,{\mathrm e}^{i \left (5 b x +7 a +4 c \right )}-3 \,{\mathrm e}^{i \left (5 b x +5 a +6 c \right )}+8 \,{\mathrm e}^{i \left (3 b x +7 a +2 c \right )}-8 \,{\mathrm e}^{i \left (3 b x +5 a +4 c \right )}+3 \,{\mathrm e}^{i \left (b x +7 a \right )}+3 \,{\mathrm e}^{i \left (b x +5 a +2 c \right )}\right )}{6 b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}\) | \(192\) |
default | \(\text {Expression too large to display}\) | \(2369\) |
Input:
int(cos(b*x+a)*sec(b*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/6*I/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))^3*(-3*exp(I*(5*b*x+7*a+4*c))-3*exp (I*(5*b*x+5*a+6*c))+8*exp(I*(3*b*x+7*a+2*c))-8*exp(I*(3*b*x+5*a+4*c))+3*ex p(I*(b*x+7*a))+3*exp(I*(b*x+5*a+2*c)))+1/2*ln(exp(I*(b*x+a))+I*exp(I*(a-c) ))/b*cos(a-c)-1/2*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))/b*cos(a-c)
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\frac {3 \, \cos \left (b x + c\right )^{3} \cos \left (-a + c\right ) \log \left (\sin \left (b x + c\right ) + 1\right ) - 3 \, \cos \left (b x + c\right )^{3} \cos \left (-a + c\right ) \log \left (-\sin \left (b x + c\right ) + 1\right ) + 6 \, \cos \left (b x + c\right ) \cos \left (-a + c\right ) \sin \left (b x + c\right ) + 4 \, \sin \left (-a + c\right )}{12 \, b \cos \left (b x + c\right )^{3}} \] Input:
integrate(cos(b*x+a)*sec(b*x+c)^4,x, algorithm="fricas")
Output:
1/12*(3*cos(b*x + c)^3*cos(-a + c)*log(sin(b*x + c) + 1) - 3*cos(b*x + c)^ 3*cos(-a + c)*log(-sin(b*x + c) + 1) + 6*cos(b*x + c)*cos(-a + c)*sin(b*x + c) + 4*sin(-a + c))/(b*cos(b*x + c)^3)
Timed out. \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)*sec(b*x+c)**4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1420 vs. \(2 (61) = 122\).
Time = 0.20 (sec) , antiderivative size = 1420, normalized size of antiderivative = 21.19 \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)*sec(b*x+c)^4,x, algorithm="maxima")
Output:
1/12*(2*(3*sin(5*b*x + 2*a + 4*c) + 3*sin(5*b*x + 6*c) - 8*sin(3*b*x + 2*a + 2*c) + 8*sin(3*b*x + 4*c) - 3*sin(b*x + 2*a) - 3*sin(b*x + 2*c))*cos(6* b*x + a + 6*c) - 6*(3*sin(4*b*x + a + 4*c) + 3*sin(2*b*x + a + 2*c) + sin( a))*cos(5*b*x + 2*a + 4*c) - 6*(3*sin(4*b*x + a + 4*c) + 3*sin(2*b*x + a + 2*c) + sin(a))*cos(5*b*x + 6*c) - 6*(8*sin(3*b*x + 2*a + 2*c) - 8*sin(3*b *x + 4*c) + 3*sin(b*x + 2*a) + 3*sin(b*x + 2*c))*cos(4*b*x + a + 4*c) + 16 *(3*sin(2*b*x + a + 2*c) + sin(a))*cos(3*b*x + 2*a + 2*c) - 16*(3*sin(2*b* x + a + 2*c) + sin(a))*cos(3*b*x + 4*c) - 18*(sin(b*x + 2*a) + sin(b*x + 2 *c))*cos(2*b*x + a + 2*c) - 3*(cos(6*b*x + a + 6*c)^2*cos(-a + c) + 9*cos( 4*b*x + a + 4*c)^2*cos(-a + c) + 9*cos(2*b*x + a + 2*c)^2*cos(-a + c) + 6* cos(2*b*x + a + 2*c)*cos(a)*cos(-a + c) + cos(-a + c)*sin(6*b*x + a + 6*c) ^2 + 9*cos(-a + c)*sin(4*b*x + a + 4*c)^2 + 9*cos(-a + c)*sin(2*b*x + a + 2*c)^2 + 6*cos(-a + c)*sin(2*b*x + a + 2*c)*sin(a) + 2*(3*cos(4*b*x + a + 4*c)*cos(-a + c) + 3*cos(2*b*x + a + 2*c)*cos(-a + c) + cos(a)*cos(-a + c) )*cos(6*b*x + a + 6*c) + 6*(3*cos(2*b*x + a + 2*c)*cos(-a + c) + cos(a)*co s(-a + c))*cos(4*b*x + a + 4*c) + (cos(a)^2 + sin(a)^2)*cos(-a + c) + 2*(3 *cos(-a + c)*sin(4*b*x + a + 4*c) + 3*cos(-a + c)*sin(2*b*x + a + 2*c) + c os(-a + c)*sin(a))*sin(6*b*x + a + 6*c) + 6*(3*cos(-a + c)*sin(2*b*x + a + 2*c) + cos(-a + c)*sin(a))*sin(4*b*x + a + 4*c))*log((cos(b*x + 2*c)^2 + cos(c)^2 - 2*cos(c)*sin(b*x + 2*c) + sin(b*x + 2*c)^2 + 2*cos(b*x + 2*c...
Leaf count of result is larger than twice the leaf count of optimal. 12158 vs. \(2 (61) = 122\).
Time = 1.00 (sec) , antiderivative size = 12158, normalized size of antiderivative = 181.46 \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)*sec(b*x+c)^4,x, algorithm="giac")
Output:
-1/6*(3*(tan(1/2*a)^3*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c)^2 + tan(1/2*a )^2*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c) + 5*tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)^3 - 5*tan(1/2*a)^2*tan(1/2*c) + 5*ta n(1/2*a)*tan(1/2*c)^2 - tan(1/2*c)^3 - tan(1/2*a)^2 + 5*tan(1/2*a)*tan(1/2 *c) - tan(1/2*c)^2 - tan(1/2*a) + tan(1/2*c) + 1)*log(abs(-tan(1/2*b*x + 1 /2*a)*tan(1/2*a)*tan(1/2*c) + tan(1/2*b*x + 1/2*a)*tan(1/2*a) - tan(1/2*b* x + 1/2*a)*tan(1/2*c) + tan(1/2*a)*tan(1/2*c) - tan(1/2*b*x + 1/2*a) + tan (1/2*a) - tan(1/2*c) + 1))/(tan(1/2*a)^3*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1 /2*c)^2 + tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c) + tan(1/2*a) ^2*tan(1/2*c)^2 + tan(1/2*a)*tan(1/2*c)^3 - tan(1/2*a)^3 + tan(1/2*a)^2*ta n(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*c)^3 + tan(1/2*a)^2 + tan(1/2 *a)*tan(1/2*c) + tan(1/2*c)^2 - tan(1/2*a) + tan(1/2*c) + 1) - 3*(tan(1/2* a)^3*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c)^2 - tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c) + 5*tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1 /2*c)^3 - tan(1/2*a)^3 + 5*tan(1/2*a)^2*tan(1/2*c) - 5*tan(1/2*a)*tan(1/2* c)^2 + tan(1/2*c)^3 - tan(1/2*a)^2 + 5*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^ 2 + tan(1/2*a) - tan(1/2*c) + 1)*log(abs(-tan(1/2*b*x + 1/2*a)*tan(1/2*a)* tan(1/2*c) - tan(1/2*b*x + 1/2*a)*tan(1/2*a) + tan(1/2*b*x + 1/2*a)*tan(1/ 2*c) - tan(1/2*a)*tan(1/2*c) - tan(1/2*b*x + 1/2*a) + tan(1/2*a) - tan(1/2 *c) - 1))/(tan(1/2*a)^3*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c)^2 - tan(...
Timed out. \[ \int \cos (a+b x) \sec ^4(c+b x) \, dx=\text {Hanged} \] Input:
int(cos(a + b*x)/cos(c + b*x)^4,x)
Output:
\text{Hanged}
\[ \int \cos (a+b x) \sec ^4(c+b x) \, dx =\text {Too large to display} \] Input:
int(cos(b*x+a)*sec(b*x+c)^4,x)
Output:
( - 4*cos(b*x + c)**2*sin(a + b*x) + 8*cos(b*x + c)*int(cos(a + b*x)/(sin( b*x + c)**4 - 2*sin(b*x + c)**2 + 1),x)*sin(b*x + c)**2*b - 8*cos(b*x + c) *int(cos(a + b*x)/(sin(b*x + c)**4 - 2*sin(b*x + c)**2 + 1),x)*b - cos(b*x + c)*int((cos(a + b*x)*sin(b*x + c)**4)/(sin(b*x + c)**4 - 2*sin(b*x + c) **2 + 1),x)*sin(b*x + c)**2*b + cos(b*x + c)*int((cos(a + b*x)*sin(b*x + c )**4)/(sin(b*x + c)**4 - 2*sin(b*x + c)**2 + 1),x)*b - 4*cos(b*x + c)*int( (cos(a + b*x)*sin(b*x + c)**2)/(sin(b*x + c)**4 - 2*sin(b*x + c)**2 + 1),x )*sin(b*x + c)**2*b + 4*cos(b*x + c)*int((cos(a + b*x)*sin(b*x + c)**2)/(s in(b*x + c)**4 - 2*sin(b*x + c)**2 + 1),x)*b - 2*cos(b*x + c)*log(sin(b*x + c) - 1)*sin(b*x + c)**2 + 2*cos(b*x + c)*log(sin(b*x + c) - 1) + 2*cos(b *x + c)*log(sin(b*x + c) + 1)*sin(b*x + c)**2 - 2*cos(b*x + c)*log(sin(b*x + c) + 1) + 4*cos(b*x + c)*log(tan((b*x + c)/2) - 1)*sin(b*x + c)**2 - 4* cos(b*x + c)*log(tan((b*x + c)/2) - 1) - 4*cos(b*x + c)*log(tan((b*x + c)/ 2) + 1)*sin(b*x + c)**2 + 4*cos(b*x + c)*log(tan((b*x + c)/2) + 1) + cos(b *x + c)*sin(b*x + c)**2*sin(a + b*x) + cos(b*x + c)*sin(b*x + c)**2*a - 3* cos(b*x + c)*sin(a + b*x) - cos(b*x + c)*a - 4*cos(a + b*x)*sin(b*x + c) - 4*sin(b*x + c)**2*sin(a + b*x) + 4*sin(a + b*x))/(15*cos(b*x + c)*b*(sin( b*x + c)**2 - 1))