\(\int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx\) [9]

Optimal result

Integrand size = 25, antiderivative size = 247 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx=\frac {2 (a c-b d)^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} f}+\frac {d^3 (4 a c-b d) \text {arctanh}(\sin (e+f x))}{2 a^2 f}+\frac {d (2 a c-b d) \left (2 a^2 c^2-2 a b c d+b^2 d^2\right ) \text {arctanh}(\sin (e+f x))}{a^4 f}+\frac {d^4 \tan (e+f x)}{a f}+\frac {d^2 \left (6 a^2 c^2-4 a b c d+b^2 d^2\right ) \tan (e+f x)}{a^3 f}+\frac {d^3 (4 a c-b d) \sec (e+f x) \tan (e+f x)}{2 a^2 f}+\frac {d^4 \tan ^3(e+f x)}{3 a f} \] Output:

2*(a*c-b*d)^4*arctan((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a^4/(a-b) 
^(1/2)/(a+b)^(1/2)/f+1/2*d^3*(4*a*c-b*d)*arctanh(sin(f*x+e))/a^2/f+d*(2*a* 
c-b*d)*(2*a^2*c^2-2*a*b*c*d+b^2*d^2)*arctanh(sin(f*x+e))/a^4/f+d^4*tan(f*x 
+e)/a/f+d^2*(6*a^2*c^2-4*a*b*c*d+b^2*d^2)*tan(f*x+e)/a^3/f+1/2*d^3*(4*a*c- 
b*d)*sec(f*x+e)*tan(f*x+e)/a^2/f+1/3*d^4*tan(f*x+e)^3/a/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(526\) vs. \(2(247)=494\).

Time = 7.77 (sec) , antiderivative size = 526, normalized size of antiderivative = 2.13 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx=\frac {-\frac {24 (a c-b d)^4 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-6 d \left (8 a b^2 c d^2-2 b^3 d^3+4 a^3 c \left (2 c^2+d^2\right )-a^2 b d \left (12 c^2+d^2\right )\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-6 d \left (-8 a b^2 c d^2+2 b^3 d^3-4 a^3 c \left (2 c^2+d^2\right )+a^2 b d \left (12 c^2+d^2\right )\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {a^2 d^3 (-3 b d+a (12 c+d))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {2 a^3 d^4 \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {4 a d^2 \left (-12 a b c d+3 b^2 d^2+2 a^2 \left (9 c^2+d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {2 a^3 d^4 \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {a^2 d^3 (-3 b d+a (12 c+d))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {4 a d^2 \left (-12 a b c d+3 b^2 d^2+2 a^2 \left (9 c^2+d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}}{12 a^4 f} \] Input:

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

Output:

((-24*(a*c - b*d)^4*ArcTanh[((a - b)*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]])/ 
Sqrt[-a^2 + b^2] - 6*d*(8*a*b^2*c*d^2 - 2*b^3*d^3 + 4*a^3*c*(2*c^2 + d^2) 
- a^2*b*d*(12*c^2 + d^2))*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 6*d*( 
-8*a*b^2*c*d^2 + 2*b^3*d^3 - 4*a^3*c*(2*c^2 + d^2) + a^2*b*d*(12*c^2 + d^2 
))*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (a^2*d^3*(-3*b*d + a*(12*c + 
 d)))/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (2*a^3*d^4*Sin[(e + f*x)/2 
])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + (4*a*d^2*(-12*a*b*c*d + 3*b^2 
*d^2 + 2*a^2*(9*c^2 + d^2))*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + 
 f*x)/2]) + (2*a^3*d^4*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x) 
/2])^3 - (a^2*d^3*(-3*b*d + a*(12*c + d)))/(Cos[(e + f*x)/2] + Sin[(e + f* 
x)/2])^2 + (4*a*d^2*(-12*a*b*c*d + 3*b^2*d^2 + 2*a^2*(9*c^2 + d^2))*Sin[(e 
 + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(12*a^4*f)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3307, 3042, 3431, 2009}

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.

\(\displaystyle \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3307

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3431

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 (a c-b d)^4 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^4 f \sqrt {a-b} \sqrt {a+b}}+\frac {d^3 (4 a c-b d) \text {arctanh}(\sin (e+f x))}{2 a^2 f}+\frac {d^3 (4 a c-b d) \tan (e+f x) \sec (e+f x)}{2 a^2 f}+\frac {d (2 a c-b d) \left (2 a^2 c^2-2 a b c d+b^2 d^2\right ) \text {arctanh}(\sin (e+f x))}{a^4 f}+\frac {d^2 \left (6 a^2 c^2-4 a b c d+b^2 d^2\right ) \tan (e+f x)}{a^3 f}+\frac {d^4 \tan ^3(e+f x)}{3 a f}+\frac {d^4 \tan (e+f x)}{a f}\)

Input:

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

Output:

(2*(a*c - b*d)^4*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a^4* 
Sqrt[a - b]*Sqrt[a + b]*f) + (d^3*(4*a*c - b*d)*ArcTanh[Sin[e + f*x]])/(2* 
a^2*f) + (d*(2*a*c - b*d)*(2*a^2*c^2 - 2*a*b*c*d + b^2*d^2)*ArcTanh[Sin[e 
+ f*x]])/(a^4*f) + (d^4*Tan[e + f*x])/(a*f) + (d^2*(6*a^2*c^2 - 4*a*b*c*d 
+ b^2*d^2)*Tan[e + f*x])/(a^3*f) + (d^3*(4*a*c - b*d)*Sec[e + f*x]*Tan[e + 
 f*x])/(2*a^2*f) + (d^4*Tan[e + f*x]^3)/(3*a*f)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3307

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + 
 f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ 
[n]
 

rule 3431

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[Exp 
andTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])^n, x 
], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (Int 
egersQ[m, n] || IntegersQ[m, p] || IntegersQ[n, p]) && NeQ[p, 2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(483\) vs. \(2(232)=464\).

Time = 1.23 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.96

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a^{4} c^{4}-4 b \,a^{3} c^{3} d +6 b^{2} a^{2} c^{2} d^{2}-4 b^{3} a c \,d^{3}+b^{4} d^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{4}}{3 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {d \left (8 a^{3} c^{3}+4 a^{3} c \,d^{2}-12 a^{2} b \,c^{2} d -a^{2} b \,d^{3}+8 a \,b^{2} c \,d^{2}-2 b^{3} d^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a^{4}}-\frac {d^{2} \left (12 a^{2} c^{2}-4 d c \,a^{2}+2 d^{2} a^{2}-8 a b c d +d^{2} a b +2 b^{2} d^{2}\right )}{2 a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d^{3} \left (4 a c -a d -b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {d^{4}}{3 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {d \left (8 a^{3} c^{3}+4 a^{3} c \,d^{2}-12 a^{2} b \,c^{2} d -a^{2} b \,d^{3}+8 a \,b^{2} c \,d^{2}-2 b^{3} d^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a^{4}}-\frac {d^{2} \left (12 a^{2} c^{2}-4 d c \,a^{2}+2 d^{2} a^{2}-8 a b c d +d^{2} a b +2 b^{2} d^{2}\right )}{2 a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d^{3} \left (4 a c -a d -b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}}{f}\)	\(484\)
default	\(\frac {\frac {2 \left (a^{4} c^{4}-4 b \,a^{3} c^{3} d +6 b^{2} a^{2} c^{2} d^{2}-4 b^{3} a c \,d^{3}+b^{4} d^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{4}}{3 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {d \left (8 a^{3} c^{3}+4 a^{3} c \,d^{2}-12 a^{2} b \,c^{2} d -a^{2} b \,d^{3}+8 a \,b^{2} c \,d^{2}-2 b^{3} d^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a^{4}}-\frac {d^{2} \left (12 a^{2} c^{2}-4 d c \,a^{2}+2 d^{2} a^{2}-8 a b c d +d^{2} a b +2 b^{2} d^{2}\right )}{2 a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d^{3} \left (4 a c -a d -b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {d^{4}}{3 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {d \left (8 a^{3} c^{3}+4 a^{3} c \,d^{2}-12 a^{2} b \,c^{2} d -a^{2} b \,d^{3}+8 a \,b^{2} c \,d^{2}-2 b^{3} d^{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a^{4}}-\frac {d^{2} \left (12 a^{2} c^{2}-4 d c \,a^{2}+2 d^{2} a^{2}-8 a b c d +d^{2} a b +2 b^{2} d^{2}\right )}{2 a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d^{3} \left (4 a c -a d -b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}}{f}\)	\(484\)
risch	\(\text {Expression too large to display}\)	\(1324\)

Input:

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

Output:

1/f*(2*(a^4*c^4-4*a^3*b*c^3*d+6*a^2*b^2*c^2*d^2-4*a*b^3*c*d^3+b^4*d^4)/a^4 
/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*f*x+1/2*e)/((a-b)*(a+b))^(1/2))- 
1/3*d^4/a/(tan(1/2*f*x+1/2*e)+1)^3+1/2*d*(8*a^3*c^3+4*a^3*c*d^2-12*a^2*b*c 
^2*d-a^2*b*d^3+8*a*b^2*c*d^2-2*b^3*d^3)/a^4*ln(tan(1/2*f*x+1/2*e)+1)-1/2*d 
^2*(12*a^2*c^2-4*a^2*c*d+2*a^2*d^2-8*a*b*c*d+a*b*d^2+2*b^2*d^2)/a^3/(tan(1 
/2*f*x+1/2*e)+1)-1/2*d^3*(4*a*c-a*d-b*d)/a^2/(tan(1/2*f*x+1/2*e)+1)^2-1/3* 
d^4/a/(tan(1/2*f*x+1/2*e)-1)^3-1/2*d*(8*a^3*c^3+4*a^3*c*d^2-12*a^2*b*c^2*d 
-a^2*b*d^3+8*a*b^2*c*d^2-2*b^3*d^3)/a^4*ln(tan(1/2*f*x+1/2*e)-1)-1/2*d^2*( 
12*a^2*c^2-4*a^2*c*d+2*a^2*d^2-8*a*b*c*d+a*b*d^2+2*b^2*d^2)/a^3/(tan(1/2*f 
*x+1/2*e)-1)+1/2*d^3*(4*a*c-a*d-b*d)/a^2/(tan(1/2*f*x+1/2*e)-1)^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (232) = 464\).

Time = 96.82 (sec) , antiderivative size = 1049, normalized size of antiderivative = 4.25 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/12*(6*(a^4*c^4 - 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c*d^3 + b 
^4*d^4)*sqrt(-a^2 + b^2)*cos(f*x + e)^3*log((2*a*b*cos(f*x + e) + (2*a^2 - 
 b^2)*cos(f*x + e)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(f*x + e) + b)*sin(f*x + e 
) - a^2 + 2*b^2)/(b^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + a^2)) - 3*(8*( 
a^5 - a^3*b^2)*c^3*d - 12*(a^4*b - a^2*b^3)*c^2*d^2 + 4*(a^5 + a^3*b^2 - 2 
*a*b^4)*c*d^3 - (a^4*b + a^2*b^3 - 2*b^5)*d^4)*cos(f*x + e)^3*log(sin(f*x 
+ e) + 1) + 3*(8*(a^5 - a^3*b^2)*c^3*d - 12*(a^4*b - a^2*b^3)*c^2*d^2 + 4* 
(a^5 + a^3*b^2 - 2*a*b^4)*c*d^3 - (a^4*b + a^2*b^3 - 2*b^5)*d^4)*cos(f*x + 
 e)^3*log(-sin(f*x + e) + 1) - 2*(2*(a^5 - a^3*b^2)*d^4 + 2*(18*(a^5 - a^3 
*b^2)*c^2*d^2 - 12*(a^4*b - a^2*b^3)*c*d^3 + (2*a^5 + a^3*b^2 - 3*a*b^4)*d 
^4)*cos(f*x + e)^2 + 3*(4*(a^5 - a^3*b^2)*c*d^3 - (a^4*b - a^2*b^3)*d^4)*c 
os(f*x + e))*sin(f*x + e))/((a^6 - a^4*b^2)*f*cos(f*x + e)^3), 1/12*(12*(a 
^4*c^4 - 4*a^3*b*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c*d^3 + b^4*d^4)*sqrt 
(a^2 - b^2)*arctan(-(a*cos(f*x + e) + b)/(sqrt(a^2 - b^2)*sin(f*x + e)))*c 
os(f*x + e)^3 + 3*(8*(a^5 - a^3*b^2)*c^3*d - 12*(a^4*b - a^2*b^3)*c^2*d^2 
+ 4*(a^5 + a^3*b^2 - 2*a*b^4)*c*d^3 - (a^4*b + a^2*b^3 - 2*b^5)*d^4)*cos(f 
*x + e)^3*log(sin(f*x + e) + 1) - 3*(8*(a^5 - a^3*b^2)*c^3*d - 12*(a^4*b - 
 a^2*b^3)*c^2*d^2 + 4*(a^5 + a^3*b^2 - 2*a*b^4)*c*d^3 - (a^4*b + a^2*b^3 - 
 2*b^5)*d^4)*cos(f*x + e)^3*log(-sin(f*x + e) + 1) + 2*(2*(a^5 - a^3*b^2)* 
d^4 + 2*(18*(a^5 - a^3*b^2)*c^2*d^2 - 12*(a^4*b - a^2*b^3)*c*d^3 + (2*a...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 607 vs. \(2 (232) = 464\).

Time = 0.21 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.46 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(3*(8*a^3*c^3*d - 12*a^2*b*c^2*d^2 + 4*a^3*c*d^3 + 8*a*b^2*c*d^3 - a^2 
*b*d^4 - 2*b^3*d^4)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^4 - 3*(8*a^3*c^3* 
d - 12*a^2*b*c^2*d^2 + 4*a^3*c*d^3 + 8*a*b^2*c*d^3 - a^2*b*d^4 - 2*b^3*d^4 
)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^4 - 12*(a^4*c^4 - 4*a^3*b*c^3*d + 6 
*a^2*b^2*c^2*d^2 - 4*a*b^3*c*d^3 + b^4*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1 
/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*f*x + 1/ 
2*e))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4) - 2*(36*a^2*c^2*d^2*tan(1/2* 
f*x + 1/2*e)^5 - 12*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^5 - 24*a*b*c*d^3*tan(1/ 
2*f*x + 1/2*e)^5 + 6*a^2*d^4*tan(1/2*f*x + 1/2*e)^5 + 3*a*b*d^4*tan(1/2*f* 
x + 1/2*e)^5 + 6*b^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 72*a^2*c^2*d^2*tan(1/2*f 
*x + 1/2*e)^3 + 48*a*b*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*d^4*tan(1/2*f* 
x + 1/2*e)^3 - 12*b^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 36*a^2*c^2*d^2*tan(1/2* 
f*x + 1/2*e) + 12*a^2*c*d^3*tan(1/2*f*x + 1/2*e) - 24*a*b*c*d^3*tan(1/2*f* 
x + 1/2*e) + 6*a^2*d^4*tan(1/2*f*x + 1/2*e) - 3*a*b*d^4*tan(1/2*f*x + 1/2* 
e) + 6*b^2*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a^3)) 
/f
 

Mupad [B] (verification not implemented)

Time = 6.71 (sec) , antiderivative size = 9992, normalized size of antiderivative = 40.45 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((8*(4*a^13*c^4 - 8*a^12*b*c^4 - 2*a^12*b*d^4 + 8*a^13*c*d^3 + 1 
6*a^13*c^3*d + 4*a^11*b^2*c^4 - 4*a^8*b^5*d^4 + 6*a^9*b^4*d^4 - 2*a^10*b^3 
*d^4 + 2*a^11*b^2*d^4 + 16*a^9*b^4*c*d^3 - 24*a^10*b^3*c*d^3 + 8*a^11*b^2* 
c*d^3 + 16*a^11*b^2*c^3*d - 24*a^12*b*c^2*d^2 - 24*a^10*b^3*c^2*d^2 + 48*a 
^11*b^2*c^2*d^2 - 8*a^12*b*c*d^3 - 32*a^12*b*c^3*d))/a^9 - (8*tan(e/2 + (f 
*x)/2)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2)*(a^2*((b*d^4)/2 + 6*b*c^2*d^2) 
- a^3*(2*c*d^3 + 4*c^3*d) + b^3*d^4 - 4*a*b^2*c*d^3))/a^10)*(a^2*((b*d^4)/ 
2 + 6*b*c^2*d^2) - a^3*(2*c*d^3 + 4*c^3*d) + b^3*d^4 - 4*a*b^2*c*d^3))/a^4 
 - (8*tan(e/2 + (f*x)/2)*(4*a^9*c^8 - 8*b^9*d^8 - 4*a^8*b*c^8 + 16*a*b^8*d 
^8 - 16*a^2*b^7*d^8 + 16*a^3*b^6*d^8 - 13*a^4*b^5*d^8 + 7*a^5*b^4*d^8 - 3* 
a^6*b^3*d^8 + a^7*b^2*d^8 + 16*a^9*c^2*d^6 + 64*a^9*c^4*d^4 + 64*a^9*c^6*d 
^2 - 128*a^2*b^7*c*d^7 + 128*a^3*b^6*c*d^7 - 128*a^4*b^5*c*d^7 + 104*a^5*b 
^4*c*d^7 - 56*a^6*b^3*c*d^7 + 24*a^7*b^2*c*d^7 + 32*a^7*b^2*c^7*d - 48*a^8 
*b*c^2*d^6 - 112*a^8*b*c^3*d^5 - 192*a^8*b*c^4*d^4 - 192*a^8*b*c^5*d^3 - 1 
92*a^8*b*c^6*d^2 - 224*a^2*b^7*c^2*d^6 + 448*a^3*b^6*c^2*d^6 + 448*a^3*b^6 
*c^3*d^5 - 424*a^4*b^5*c^2*d^6 - 896*a^4*b^5*c^3*d^5 - 552*a^4*b^5*c^4*d^4 
 + 376*a^5*b^4*c^2*d^6 + 784*a^5*b^4*c^3*d^5 + 1096*a^5*b^4*c^4*d^4 + 416* 
a^5*b^4*c^5*d^3 - 280*a^6*b^3*c^2*d^6 - 560*a^6*b^3*c^3*d^5 - 880*a^6*b^3* 
c^4*d^4 - 800*a^6*b^3*c^5*d^3 - 176*a^6*b^3*c^6*d^2 + 136*a^7*b^2*c^2*d^6 
+ 336*a^7*b^2*c^3*d^5 + 464*a^7*b^2*c^4*d^4 + 576*a^7*b^2*c^5*d^3 + 304...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 2335, normalized size of antiderivative = 9.45 \[ \int \frac {(c+d \sec (e+f x))^4}{a+b \cos (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( 
a**2 - b**2))*cos(e + f*x)*sin(e + f*x)**2*a**4*c**4 - 48*sqrt(a**2 - b**2 
)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e 
+ f*x)*sin(e + f*x)**2*a**3*b*c**3*d + 72*sqrt(a**2 - b**2)*atan((tan((e + 
 f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*sin(e + f 
*x)**2*a**2*b**2*c**2*d**2 - 48*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a 
 - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*sin(e + f*x)**2*a*b 
**3*c*d**3 + 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x) 
/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*sin(e + f*x)**2*b**4*d**4 - 12*sqrt 
(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b 
**2))*cos(e + f*x)*a**4*c**4 + 48*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2) 
*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*a**3*b*c**3*d - 7 
2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a* 
*2 - b**2))*cos(e + f*x)*a**2*b**2*c**2*d**2 + 48*sqrt(a**2 - b**2)*atan(( 
tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*a 
*b**3*c*d**3 - 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f* 
x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*b**4*d**4 + 24*cos(e + f*x)**2*si 
n(e + f*x)*a**4*b*c*d**3 - 24*cos(e + f*x)**2*sin(e + f*x)*a**2*b**3*c*d** 
3 - 24*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a**5*c**3*d 
- 12*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a**5*c*d**3...