\(\int \sec ^3(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [311]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3669, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^p}{\cos (e+f x)^3}dx\)

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

(AppellF1[1/2, 2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sin[e 
+ f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/a)^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 3669

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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