\(\int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx\) [445]

Optimal result

Integrand size = 21, antiderivative size = 115 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=b^4 x+\frac {2 a b \left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{3 d}+\frac {4 a^3 b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:

b^4*x+2*a*b*(a^2+2*b^2)*arctanh(sin(d*x+c))/d+1/3*a^2*(2*a^2+17*b^2)*tan(d 
*x+c)/d+4/3*a^3*b*sec(d*x+c)*tan(d*x+c)/d+1/3*a^2*(a+b*cos(d*x+c))^2*sec(d 
*x+c)^2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {3 b^4 d x+12 a b^3 \coth ^{-1}(\sin (c+d x))+6 a^3 b \text {arctanh}(\sin (c+d x))+3 a^4 \tan (c+d x)+18 a^2 b^2 \tan (c+d x)+6 a^3 b \sec (c+d x) \tan (c+d x)+a^4 \tan ^3(c+d x)}{3 d} \] Input:

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

Output:

(3*b^4*d*x + 12*a*b^3*ArcCoth[Sin[c + d*x]] + 6*a^3*b*ArcTanh[Sin[c + d*x] 
] + 3*a^4*Tan[c + d*x] + 18*a^2*b^2*Tan[c + d*x] + 6*a^3*b*Sec[c + d*x]*Ta 
n[c + d*x] + a^4*Tan[c + d*x]^3)/(3*d)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3271, 3042, 3510, 27, 3042, 3500, 27, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

\(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (3 \cos ^2(c+d x) b^3+8 a^2 b+a \left (2 a^2+9 b^2\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^3+8 a^2 b+a \left (2 a^2+9 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{3} \left (\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}-\frac {1}{2} \int -2 \left (3 \cos ^2(c+d x) b^4+6 a \left (a^2+2 b^2\right ) \cos (c+d x) b+a^2 \left (2 a^2+17 b^2\right )\right ) \sec ^2(c+d x)dx\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\int \left (3 \cos ^2(c+d x) b^4+6 a \left (a^2+2 b^2\right ) \cos (c+d x) b+a^2 \left (2 a^2+17 b^2\right )\right ) \sec ^2(c+d x)dx+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\int \frac {3 \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^4+6 a \left (a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b+a^2 \left (2 a^2+17 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{3} \left (\int 3 \left (\cos (c+d x) b^4+2 a \left (a^2+2 b^2\right ) b\right ) \sec (c+d x)dx+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (3 \int \left (\cos (c+d x) b^4+2 a \left (a^2+2 b^2\right ) b\right ) \sec (c+d x)dx+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (3 \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) b^4+2 a \left (a^2+2 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{3} \left (3 \left (2 a b \left (a^2+2 b^2\right ) \int \sec (c+d x)dx+b^4 x\right )+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (3 \left (2 a b \left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+b^4 x\right )+\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {4 a^3 b \tan (c+d x) \sec (c+d x)}{d}+3 \left (\frac {2 a b \left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+b^4 x\right )+\frac {a^2 \left (2 a^2+17 b^2\right ) \tan (c+d x)}{d}\right )\)

Input:

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

Output:

(a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (3*(b^4*x 
 + (2*a*b*(a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]])/d) + (a^2*(2*a^2 + 17*b^2)* 
Tan[c + d*x])/d + (4*a^3*b*Sec[c + d*x]*Tan[c + d*x])/d)/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

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 4257

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

Maple [A] (verified)

Time = 15.73 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 \tan \left (d x +c \right ) a^{2} b^{2}+4 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \left (d x +c \right )}{d}\)	\(109\)
default	\(\frac {-a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 \tan \left (d x +c \right ) a^{2} b^{2}+4 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \left (d x +c \right )}{d}\)	\(109\)
parts	\(-\frac {a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{4} \left (d x +c \right )}{d}+\frac {2 a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(124\)
risch	\(b^{4} x -\frac {4 i a^{2} \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-18 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-a^{2}-9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(196\)
parallelrisch	\(\frac {-18 a \left (a^{2}+2 b^{2}\right ) b \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 a \left (a^{2}+2 b^{2}\right ) b \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 b^{4} d x \cos \left (3 d x +3 c \right )+2 \left (a^{4}+9 a^{2} b^{2}\right ) \sin \left (3 d x +3 c \right )+12 \sin \left (2 d x +2 c \right ) a^{3} b +9 b^{4} d x \cos \left (d x +c \right )+6 a^{2} \sin \left (d x +c \right ) \left (a^{2}+3 b^{2}\right )}{3 d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\)	\(200\)
norman	\(\frac {b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}-b^{4} x -b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-3 b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-3 b^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {8 a^{2} \left (a^{2}-6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 a^{2} \left (a^{2}-2 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {2 a^{2} \left (a^{2}+2 a b +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a^{2} \left (5 a^{2}-12 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {4 a^{2} \left (5 a^{2}+12 a b +18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a^{2} \left (13 a^{2}-30 a b -18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {2 a^{2} \left (13 a^{2}+30 a b -18 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {2 a b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(439\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.20 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {3 \, b^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a^{3} b \cos \left (d x + c\right ) + a^{4} + 2 \, {\left (a^{4} + 9 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Sympy [F]

\[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{4} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} + 3 \, {\left (d x + c\right )} b^{4} - 3 \, a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )}{3 \, d} \] Input:

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

Output:

1/3*((tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 + 3*(d*x + c)*b^4 - 3*a^3*b*(2* 
sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + 
c) - 1)) + 6*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*a^ 
2*b^2*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (109) = 218\).

Time = 0.56 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.92 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} b^{4} + 6 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \] Input:

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

Output:

1/3*(3*(d*x + c)*b^4 + 6*(a^3*b + 2*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 
1)) - 6*(a^3*b + 2*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^4*ta 
n(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*b^2*tan(1/2 
*d*x + 1/2*c)^5 - 2*a^4*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*b^2*tan(1/2*d*x + 
1/2*c)^3 + 3*a^4*tan(1/2*d*x + 1/2*c) + 6*a^3*b*tan(1/2*d*x + 1/2*c) + 18* 
a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 41.57 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.61 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {2\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {8\,a\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a^3\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a^3\,b\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \] Input:

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

Output:

(2*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a^4*sin(c + d*x 
))/(3*d*cos(c + d*x)) + (a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (8*a*b^3 
*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*a^3*b*atanh(sin(c/2 
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*a^3*b*sin(c + d*x))/(d*cos(c + d*x) 
^2) + (6*a^2*b^2*sin(c + d*x))/(d*cos(c + d*x))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 356, normalized size of antiderivative = 3.10 \[ \int (a+b \cos (c+d x))^4 \sec ^4(c+d x) \, dx=\frac {-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3} b -12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}+6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} b +12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{3}+6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3} b +12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}-6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} b -12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{3}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{4} d x -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b -3 \cos \left (d x +c \right ) b^{4} d x +2 \sin \left (d x +c \right )^{3} a^{4}+18 \sin \left (d x +c \right )^{3} a^{2} b^{2}-3 \sin \left (d x +c \right ) a^{4}-18 \sin \left (d x +c \right ) a^{2} b^{2}}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

( - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b - 12*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + 6*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*a**3*b + 12*cos(c + d*x)*log(tan((c + d*x)/2 
) - 1)*a*b**3 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a 
**3*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3 - 
 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*b - 12*cos(c + d*x)*log(tan 
((c + d*x)/2) + 1)*a*b**3 + 3*cos(c + d*x)*sin(c + d*x)**2*b**4*d*x - 6*co 
s(c + d*x)*sin(c + d*x)*a**3*b - 3*cos(c + d*x)*b**4*d*x + 2*sin(c + d*x)* 
*3*a**4 + 18*sin(c + d*x)**3*a**2*b**2 - 3*sin(c + d*x)*a**4 - 18*sin(c + 
d*x)*a**2*b**2)/(3*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))