\(\int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx\) [446]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 154 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {\left (3 a^4+24 a^2 b^2+8 b^4\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:

1/8*(3*a^4+24*a^2*b^2+8*b^4)*arctanh(sin(d*x+c))/d+4/3*a*b*(2*a^2+3*b^2)*t 
an(d*x+c)/d+1/8*a^2*(3*a^2+22*b^2)*sec(d*x+c)*tan(d*x+c)/d+5/6*a^3*b*sec(d 
*x+c)^2*tan(d*x+c)/d+1/4*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.69 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {24 b^4 \coth ^{-1}(\sin (c+d x))+a \left (9 a \left (a^2+8 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 a \left (a^2+8 b^2\right ) \sec (c+d x)+6 a^3 \sec ^3(c+d x)+32 b \left (3 \left (a^2+b^2\right )+a^2 \tan ^2(c+d x)\right )\right )\right )}{24 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5,x]
 

Output:

(24*b^4*ArcCoth[Sin[c + d*x]] + a*(9*a*(a^2 + 8*b^2)*ArcTanh[Sin[c + d*x]] 
 + Tan[c + d*x]*(9*a*(a^2 + 8*b^2)*Sec[c + d*x] + 6*a^3*Sec[c + d*x]^3 + 3 
2*b*(3*(a^2 + b^2) + a^2*Tan[c + d*x]^2))))/(24*d)
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3271, 3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

\(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 3271

\(\displaystyle \frac {1}{4} \int (a+b \cos (c+d x)) \left (10 b a^2+3 \left (a^2+4 b^2\right ) \cos (c+d x) a+b \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (10 b a^2+3 \left (a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{4} \left (\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 \left (3 a^2+22 b^2\right ) a^2+16 b \left (2 a^2+3 b^2\right ) \cos (c+d x) a+3 b^2 \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)\right )dx\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (3 \left (3 a^2+22 b^2\right ) a^2+16 b \left (2 a^2+3 b^2\right ) \cos (c+d x) a+3 b^2 \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)dx+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 \left (3 a^2+22 b^2\right ) a^2+16 b \left (2 a^2+3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 b^2 \left (a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (32 a b \left (2 a^2+3 b^2\right )+3 \left (3 a^4+24 b^2 a^2+8 b^4\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {32 a b \left (2 a^2+3 b^2\right )+3 \left (3 a^4+24 b^2 a^2+8 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (32 a b \left (2 a^2+3 b^2\right ) \int \sec ^2(c+d x)dx+3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \int \sec (c+d x)dx\right )+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (32 a b \left (2 a^2+3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {32 a b \left (2 a^2+3 b^2\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {32 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{d}\right )+\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac {1}{4} \left (\frac {10 a^3 b \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {3 a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {32 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{d}+\frac {3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \text {arctanh}(\sin (c+d x))}{d}\right )\right )\right )\)

Input:

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5,x]
 

Output:

(a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((10*a^3* 
b*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a^2*(3*a^2 + 22*b^2)*Sec[c + d* 
x]*Tan[c + d*x])/(2*d) + ((3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*ArcTanh[Sin[c + 
d*x]])/d + (32*a*b*(2*a^2 + 3*b^2)*Tan[c + d*x])/d)/2)/3)/4
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3271
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3510
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [A] (verified)

Time = 16.81 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95

method result size
derivativedivides \(\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a^{3} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 \tan \left (d x +c \right ) a \,b^{3}+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(147\)
default \(\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 a^{3} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 \tan \left (d x +c \right ) a \,b^{3}+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(147\)
parts \(\frac {a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{3} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d}\) \(164\)
parallelrisch \(\frac {-36 \left (a^{4}+8 a^{2} b^{2}+\frac {8}{3} b^{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (a^{4}+8 a^{2} b^{2}+\frac {8}{3} b^{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (256 a^{3} b +192 a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (18 a^{4}+144 a^{2} b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (64 a^{3} b +96 a \,b^{3}\right ) \sin \left (4 d x +4 c \right )+\left (66 a^{4}+144 a^{2} b^{2}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) \(228\)
risch \(-\frac {i a \left (9 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+72 b^{2} a \,{\mathrm e}^{7 i \left (d x +c \right )}-96 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+33 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+72 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-192 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-288 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{3}-72 b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}-256 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-288 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{3} {\mathrm e}^{i \left (d x +c \right )}-72 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}-64 a^{2} b -96 b^{3}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{4}}{d}-\frac {3 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{4}}{d}\) \(353\)
norman \(\frac {\frac {a \left (5 a^{3}-32 a^{2} b +24 b^{2} a -32 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}+\frac {a \left (5 a^{3}+32 a^{2} b +24 b^{2} a +32 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (45 a^{3}-32 a^{2} b +24 b^{2} a +96 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a \left (45 a^{3}+32 a^{2} b +24 b^{2} a -96 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {a \left (69 a^{3}-224 a^{2} b +216 b^{2} a -96 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {a \left (69 a^{3}+224 a^{2} b +216 b^{2} a +96 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {a \left (165 a^{3}-32 a^{2} b -360 b^{2} a -288 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 d}+\frac {a \left (165 a^{3}+32 a^{2} b -360 b^{2} a +288 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (3 a^{4}+24 a^{2} b^{2}+8 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a^{4}+24 a^{2} b^{2}+8 b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) \(424\)

Input:

int((a+cos(d*x+c)*b)^4*sec(d*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^4*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c) 
+tan(d*x+c)))-4*a^3*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+6*a^2*b^2*(1/2*se 
c(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4*tan(d*x+c)*a*b^3+b^4* 
ln(sec(d*x+c)+tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a^{3} b \cos \left (d x + c\right ) + 6 \, a^{4} + 32 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (a^{4} + 8 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="fricas")
 

Output:

1/48*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) 
- 3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2 
*(32*a^3*b*cos(d*x + c) + 6*a^4 + 32*(2*a^3*b + 3*a*b^3)*cos(d*x + c)^3 + 
9*(a^4 + 8*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**5,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} b - 3 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, a b^{3} \tan \left (d x + c\right )}{48 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="maxima")
 

Output:

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3*b - 3*a^4*(2*(3*sin(d*x + c 
)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin( 
d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 72*a^2*b^2*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*b^4 
*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 192*a*b^3*tan(d*x + c)) 
/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (144) = 288\).

Time = 0.59 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.34 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="giac")
 

Output:

1/24*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 
3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15* 
a^4*tan(1/2*d*x + 1/2*c)^7 - 96*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 72*a^2*b^2* 
tan(1/2*d*x + 1/2*c)^7 - 96*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*a^4*tan(1/2*d 
*x + 1/2*c)^5 + 160*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 72*a^2*b^2*tan(1/2*d*x 
+ 1/2*c)^5 + 288*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^4*tan(1/2*d*x + 1/2*c) 
^3 - 160*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 
- 288*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*a^4*tan(1/2*d*x + 1/2*c) + 96*a^3* 
b*tan(1/2*d*x + 1/2*c) + 72*a^2*b^2*tan(1/2*d*x + 1/2*c) + 96*a*b^3*tan(1/ 
2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 44.64 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.59 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx=\frac {\left (\frac {5\,a^4}{4}-8\,a^3\,b+6\,a^2\,b^2-8\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^4}{4}+\frac {40\,a^3\,b}{3}-6\,a^2\,b^2+24\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^4}{4}-\frac {40\,a^3\,b}{3}-6\,a^2\,b^2-24\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^4}{4}+8\,a^3\,b+6\,a^2\,b^2+8\,a\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^4}{4}+6\,a^2\,b^2+2\,b^4\right )}{d} \] Input:

int((a + b*cos(c + d*x))^4/cos(c + d*x)^5,x)
 

Output:

(tan(c/2 + (d*x)/2)*(8*a*b^3 + 8*a^3*b + (5*a^4)/4 + 6*a^2*b^2) - tan(c/2 
+ (d*x)/2)^7*(8*a*b^3 + 8*a^3*b - (5*a^4)/4 - 6*a^2*b^2) - tan(c/2 + (d*x) 
/2)^3*(24*a*b^3 + (40*a^3*b)/3 - (3*a^4)/4 + 6*a^2*b^2) + tan(c/2 + (d*x)/ 
2)^5*(24*a*b^3 + (40*a^3*b)/3 + (3*a^4)/4 - 6*a^2*b^2))/(d*(6*tan(c/2 + (d 
*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d* 
x)/2)^8 + 1)) + (atanh(tan(c/2 + (d*x)/2))*((3*a^4)/4 + 2*b^4 + 6*a^2*b^2) 
)/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 722, normalized size of antiderivative = 4.69 \[ \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x)
 

Output:

( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4 - 72*cos 
(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**2 - 24*cos(c + 
 d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**4 + 18*cos(c + d*x)*log 
(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 + 144*cos(c + d*x)*log(tan((c 
+ d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 + 48*cos(c + d*x)*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**2*b**4 - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1) 
*a**4 - 72*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**2 - 24*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*b**4 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 
 1)*sin(c + d*x)**4*a**4 + 72*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c 
 + d*x)**4*a**2*b**2 + 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d 
*x)**4*b**4 - 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a* 
*4 - 144*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 
- 48*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**4 + 9*cos(c 
 + d*x)*log(tan((c + d*x)/2) + 1)*a**4 + 72*cos(c + d*x)*log(tan((c + d*x) 
/2) + 1)*a**2*b**2 + 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**4 - 9*co 
s(c + d*x)*sin(c + d*x)**3*a**4 - 72*cos(c + d*x)*sin(c + d*x)**3*a**2*b** 
2 + 15*cos(c + d*x)*sin(c + d*x)*a**4 + 72*cos(c + d*x)*sin(c + d*x)*a**2* 
b**2 + 64*sin(c + d*x)**5*a**3*b + 96*sin(c + d*x)**5*a*b**3 - 160*sin(c + 
 d*x)**3*a**3*b - 192*sin(c + d*x)**3*a*b**3 + 96*sin(c + d*x)*a**3*b + 96 
*sin(c + d*x)*a*b**3)/(24*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d...