\(\int \cos ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [309]

Optimal result

Integrand size = 23, antiderivative size = 85 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},3,-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (\frac {a+b+b \tan ^2(e+f x)}{a+b}\right )^{-p}}{f} \] Output:

AppellF1(1/2,3,-p,3/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*tan(f*x+e)*(a+b 
+b*tan(f*x+e)^2)^p/f/(((a+b+b*tan(f*x+e)^2)/(a+b))^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1342\) vs. \(2(85)=170\).

Time = 14.90 (sec) , antiderivative size = 1342, normalized size of antiderivative = 15.79 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx =\text {Too large to display} \] Input:

Integrate[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

(AppellF1[1/2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] 
*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]*(3*(a + b)*AppellF1[ 
1/2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + 2*(b*p* 
AppellF1[3/2, 3, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b) 
)] - 3*(a + b)*AppellF1[3/2, 4, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x 
]^2)/(a + b))])*Tan[e + f*x]^2))/(f*(AppellF1[1/2, 3, -p, 3/2, -Tan[e + f* 
x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*(3*(a + b)*AppellF1[1/ 
2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + 2*(b*p*Ap 
pellF1[3/2, 3, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] 
 - 3*(a + b)*AppellF1[3/2, 4, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^ 
2)/(a + b))])*Tan[e + f*x]^2) - (2*a*p*AppellF1[1/2, 3, -p, 3/2, -Tan[e + 
f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[2*(e + f*x)]*Tan[e + f*x]*(3*(a 
 + b)*AppellF1[1/2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + 
 b))] + 2*(b*p*AppellF1[3/2, 3, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + 
f*x]^2)/(a + b))] - 3*(a + b)*AppellF1[3/2, 4, -p, 5/2, -Tan[e + f*x]^2, - 
((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2))/(a + 2*b + a*Cos[2*(e + f* 
x)]) + 2*(-3 + p)*AppellF1[1/2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + 
f*x]^2)/(a + b))]*Tan[e + f*x]^2*(3*(a + b)*AppellF1[1/2, 3, -p, 3/2, -Tan 
[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + 2*(b*p*AppellF1[3/2, 3, 1 - 
p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] - 3*(a + b)*App...
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 334, 333}

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.

\(\displaystyle \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^p}{\sec (e+f x)^4}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \frac {\left (b \tan ^2(e+f x)+a+b\right )^p}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \frac {\left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},3,-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{f}\)

Input:

Int[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

(AppellF1[1/2, 3, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] 
*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/(a 
+ b))^p)
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 

Maple [F]

\[\int \cos \left (f x +e \right )^{4} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e)^2 + a)^p*cos(f*x + e)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**4*(a+b*sec(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*cos(f*x + e)^4, x)
 

Giac [F]

\[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*cos(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \] Input:

int(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^p,x)
 

Output:

int(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \left (\sec \left (f x +e \right )^{2} b +a \right )^{p} \cos \left (f x +e \right )^{4}d x \] Input:

int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int((sec(e + f*x)**2*b + a)**p*cos(e + f*x)**4,x)