\(\int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\) [83]

Optimal result

Integrand size = 28, antiderivative size = 103 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\left (a^4-6 a^2 b^2+b^4\right ) x-\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac {a b (a+b \tan (c+d x))^2}{d}+\frac {b (a+b \tan (c+d x))^3}{3 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {-3 i (a+i b)^4 \log (i-\tan (c+d x))+3 i (a-i b)^4 \log (i+\tan (c+d x))-6 b^2 \left (-6 a^2+b^2\right ) \tan (c+d x)+12 a b^3 \tan ^2(c+d x)+2 b^4 \tan ^3(c+d x)}{6 d} \] Input:

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

Output:

((-3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] + (3*I)*(a - I*b)^4*Log[I + Tan[ 
c + d*x]] - 6*b^2*(-6*a^2 + b^2)*Tan[c + d*x] + 12*a*b^3*Tan[c + d*x]^2 + 
2*b^4*Tan[c + d*x]^3)/(6*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3565, 3042, 3963, 3042, 4011, 3042, 4008, 3042, 3956}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^4}{\cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3565

\(\displaystyle \int (a+b \tan (c+d x))^4dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^4dx\)

\(\Big \downarrow \) 3963

\(\displaystyle \int (a+b \tan (c+d x))^2 \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {b (a+b \tan (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^2 \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {b (a+b \tan (c+d x))^3}{3 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int (a+b \tan (c+d x)) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x)) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

\(\Big \downarrow \) 4008

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3956

\(\displaystyle \frac {b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac {4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (a^4-6 a^2 b^2+b^4\right )+\frac {b (a+b \tan (c+d x))^3}{3 d}+\frac {a b (a+b \tan (c+d x))^2}{d}\)

Input:

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

Output:

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2 
*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*b*(a + b*Tan[c + d*x])^2)/d + (b*(a + 
b*Tan[c + d*x])^3)/(3*d)
 

Defintions of rubi rules used

rule 3042

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

rule 3565

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin 
[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[(a + b*Tan[c + d*x])^n, x] /; 
 FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b^2, 0 
]
 

rule 3956

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

rule 3963

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d 
*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 
+ b^2, 0] && GtQ[n, 1]
 

rule 4008

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

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {a^{4} \left (d x +c \right )-4 a^{3} b \ln \left (\cos \left (d x +c \right )\right )+6 a^{2} b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+4 b^{3} a \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(101\)
default	\(\frac {a^{4} \left (d x +c \right )-4 a^{3} b \ln \left (\cos \left (d x +c \right )\right )+6 a^{2} b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+4 b^{3} a \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(101\)
parts	\(\frac {a^{4} \left (d x +c \right )}{d}+\frac {b^{4} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}+\frac {4 a^{3} b \ln \left (\sec \left (d x +c \right )\right )}{d}+\frac {4 b^{3} a \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {6 a^{2} b^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(112\)
risch	\(4 i a^{3} b x -4 i x \,b^{3} a +a^{4} x -6 a^{2} b^{2} x +b^{4} x +\frac {8 i a^{3} b c}{d}-\frac {8 i b^{3} a c}{d}-\frac {4 i b^{2} \left (-9 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2}+2 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {4 b^{3} a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(218\)
parallelrisch	\(\frac {36 \left (a -b \right ) \left (a +b \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-36 \left (a -b \right ) \left (a +b \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-36 \left (a -b \right ) \left (a +b \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \left (a^{4} d x -6 a^{2} b^{2} d x +b^{4} d x -2 b^{3} a \right ) \cos \left (3 d x +3 c \right )+2 \left (9 a^{2} b^{2}-2 b^{4}\right ) \sin \left (3 d x +3 c \right )+3 \left (3 a^{4} d x -18 a^{2} b^{2} d x +3 b^{4} d x +2 b^{3} a \right ) \cos \left (d x +c \right )+18 a^{2} b^{2} \sin \left (d x +c \right )}{3 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(271\)
norman	\(\frac {\frac {8 b^{3} a}{d}+\left (-a^{4}+6 a^{2} b^{2}-b^{4}\right ) x +\frac {40 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\left (-3 a^{4}+18 a^{2} b^{2}-3 b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-3 a^{4}+18 a^{2} b^{2}-3 b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-a^{4}+6 a^{2} b^{2}-b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (a^{4}-6 a^{2} b^{2}+b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (a^{4}-6 a^{2} b^{2}+b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (3 a^{4}-18 a^{2} b^{2}+3 b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (3 a^{4}-18 a^{2} b^{2}+3 b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {8 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}-\frac {48 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {48 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {40 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {24 b^{2} \left (2 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 b^{2} \left (6 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b^{2} \left (6 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {2 b^{2} \left (18 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {2 b^{2} \left (18 a^{2}-19 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {4 b^{2} \left (18 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {4 b^{2} \left (18 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {4 a b \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {4 a b \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {4 a b \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(674\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.16 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 6 \, a b^{3} \cos \left (d x + c\right ) - 12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) + {\left (b^{4} + 2 \, {\left (9 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {3 \, {\left (d x + c\right )} a^{4} - 18 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b^{2} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} b^{4} - 6 \, a b^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 6 \, a^{3} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{3 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right ) - 3 \, b^{4} \tan \left (d x + c\right ) + 3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + 6 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{3 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 17.56 (sec) , antiderivative size = 546, normalized size of antiderivative = 5.30 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx =\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 552, normalized size of antiderivative = 5.36 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx =\text {Too large to display} \] Input:

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

Output:

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