\(\int \sec ^3(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [299]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},2+p,-p,\frac {3}{2},\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right ) \cos ^2(e+f x)^p \sin (e+f x) \left (\frac {a+b-a \sin ^2(e+f x)}{a+b}\right )^{-p} \left (\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )\right )^p}{f} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1392\) vs. \(2(105)=210\).

Time = 15.47 (sec) , antiderivative size = 1392, normalized size of antiderivative = 13.26 \[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx =\text {Too large to display} \] Input:

Integrate[Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

(AppellF1[1/2, -1/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b 
))]*Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p*Tan[e + f*x]*(3*(a + b)*Appell 
F1[1/2, -1/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + ( 
2*b*p*AppellF1[3/2, -1/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2 
)/(a + b))] + (a + b)*AppellF1[3/2, 1/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Ta 
n[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2))/(f*(AppellF1[1/2, -1/2, -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, -1/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b 
))] + (2*b*p*AppellF1[3/2, -1/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + 
 f*x]^2)/(a + b))] + (a + b)*AppellF1[3/2, 1/2, -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, -1/ 
2, -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, -1/2, -p, 3/2, -Tan[e + f*x]^2, - 
((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, -1/2, 1 - p, 5/2, -Ta 
n[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*AppellF1[3/2, 1/2, 
-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*(1/2 + p)*AppellF1[1/2, -1/2, -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, -1/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b 
))] + (2*b*p*AppellF1[3/2, -1/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[...
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4636, 2057, 2058, 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.

\(\displaystyle \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 4636

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

\(\Big \downarrow \) 2057

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

Int[Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

(AppellF1[1/2, 2 + p, -p, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] 
*Sin[e + f*x]*(1 - Sin[e + f*x]^2)^p*((a + b - a*Sin[e + f*x]^2)/(1 - Sin[ 
e + f*x]^2))^p)/(f*(1 - (a*Sin[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 2057

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* 
((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
 

rule 2058

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ 
(r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + 
 b*x^n)^(p*q)*(c + d*x^n)^(p*r))]   Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* 
r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
 

rule 3042

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

rule 4636

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

Maple [F]

\[\int \sec \left (f x +e \right )^{3} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^3, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**3*(a+b*sec(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^3, x)
 

Giac [F]

\[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^3} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^3,x)
 

Output:

int((a + b/cos(e + f*x)^2)^p/cos(e + f*x)^3, x)
 

Reduce [F]

\[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \left (\sec \left (f x +e \right )^{2} b +a \right )^{p} \sec \left (f x +e \right )^{3}d x \] Input:

int(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x)
                                                                                    
                                                                                    
 

Output:

int((sec(e + f*x)**2*b + a)**p*sec(e + f*x)**3,x)