\(\int \sec ^3(e+f x) (a+b \sin ^4(e+f x))^p \, dx\) [355]

Optimal result

Integrand size = 23, antiderivative size = 239 \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{4},-p,2,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a},\sin ^4(e+f x)\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f}+\frac {2 \operatorname {AppellF1}\left (\frac {3}{4},-p,2,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a},\sin ^4(e+f x)\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-\frac {b \sin ^4(e+f x)}{a},\sin ^4(e+f x)\right ) \sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{5 f} \] Output:

AppellF1(1/4,2,-p,5/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)*(a+b*sin( 
f*x+e)^4)^p/f/((1+b*sin(f*x+e)^4/a)^p)+2/3*AppellF1(3/4,2,-p,7/4,sin(f*x+e 
)^4,-b*sin(f*x+e)^4/a)*sin(f*x+e)^3*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x+e 
)^4/a)^p)+1/5*AppellF1(5/4,2,-p,9/4,sin(f*x+e)^4,-b*sin(f*x+e)^4/a)*sin(f* 
x+e)^5*(a+b*sin(f*x+e)^4)^p/f/((1+b*sin(f*x+e)^4/a)^p)
 

Mathematica [F]

\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx \] Input:

Integrate[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p,x]
 

Output:

Integrate[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p, x]
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 234, 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, 3702, 1569, 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.

\(\displaystyle \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^4\right )^p}{\cos (e+f x)^3}dx\)

\(\Big \downarrow \) 3702

\(\displaystyle \frac {\int \frac {\left (b \sin ^4(e+f x)+a\right )^p}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 1569

\(\displaystyle \frac {\int \left (\frac {\sin ^4(e+f x) \left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}+\frac {2 \sin ^2(e+f x) \left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}+\frac {\left (b \sin ^4(e+f x)+a\right )^p}{\left (\sin ^4(e+f x)-1\right )^2}\right )d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},2,-p,\frac {5}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )+\frac {1}{5} \sin ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},2,-p,\frac {9}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )+\frac {2}{3} \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},2,-p,\frac {7}{4},\sin ^4(e+f x),-\frac {b \sin ^4(e+f x)}{a}\right )}{f}\)

Input:

Int[Sec[e + f*x]^3*(a + b*Sin[e + f*x]^4)^p,x]
 

Output:

((AppellF1[1/4, 2, -p, 5/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e 
 + f*x]*(a + b*Sin[e + f*x]^4)^p)/(1 + (b*Sin[e + f*x]^4)/a)^p + (2*Appell 
F1[3/4, 2, -p, 7/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^ 
3*(a + b*Sin[e + f*x]^4)^p)/(3*(1 + (b*Sin[e + f*x]^4)/a)^p) + (AppellF1[5 
/4, 2, -p, 9/4, Sin[e + f*x]^4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]^5*(a 
 + b*Sin[e + f*x]^4)^p)/(5*(1 + (b*Sin[e + f*x]^4)/a)^p))/f
 

Defintions of rubi rules used

rule 1569

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int 
[ExpandIntegrand[(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4) 
))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  ! 
IntegerQ[p] && ILtQ[q, 0]
 

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 3702

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x 
_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si 
mp[ff/f   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, 
 Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 
1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
 

Maple [F]

\[\int \sec \left (f x +e \right )^{3} \left (a +b \sin \left (f x +e \right )^{4}\right )^{p}d x\]

Input:

int(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)
 

Output:

int(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)
 

Fricas [F]

\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x, algorithm="fricas")
 

Output:

integral((b*cos(f*x + e)^4 - 2*b*cos(f*x + e)^2 + a + b)^p*sec(f*x + e)^3, 
 x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**3*(a+b*sin(f*x+e)**4)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^4 + a)^p*sec(f*x + e)^3, x)
 

Giac [F]

\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \sec \left (f x + e\right )^{3} \,d x } \] Input:

integrate(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e)^4 + a)^p*sec(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p}{{\cos \left (e+f\,x\right )}^3} \,d x \] Input:

int((a + b*sin(e + f*x)^4)^p/cos(e + f*x)^3,x)
 

Output:

int((a + b*sin(e + f*x)^4)^p/cos(e + f*x)^3, x)
 

Reduce [F]

\[ \int \sec ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \left (\sin \left (f x +e \right )^{4} b +a \right )^{p} \sec \left (f x +e \right )^{3}d x \] Input:

int(sec(f*x+e)^3*(a+b*sin(f*x+e)^4)^p,x)
 

Output:

int((sin(e + f*x)**4*b + a)**p*sec(e + f*x)**3,x)