\(\int \cos ^5(e+f x) (a+b \sin ^4(e+f x))^p \, dx\) [351]

Optimal result

Integrand size = 23, antiderivative size = 197 \[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{1+p}}{b f (5+4 p)}-\frac {(a-b (5+4 p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{b f (5+4 p)}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{3 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.72 \[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p} \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right )-10 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin ^4(e+f x)\right )}{15 f} \] Input:

Integrate[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p,x]
 

Output:

(Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p*(15*Hypergeometric2F1[1/4, -p, 5/4, 
 -((b*Sin[e + f*x]^4)/a)] - 10*Hypergeometric2F1[3/4, -p, 7/4, -((b*Sin[e 
+ f*x]^4)/a)]*Sin[e + f*x]^2 + 3*Hypergeometric2F1[5/4, -p, 9/4, -((b*Sin[ 
e + f*x]^4)/a)]*Sin[e + f*x]^4))/(15*f*(1 + (b*Sin[e + f*x]^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3702, 1519, 25, 1516, 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.

\(\displaystyle \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^5 \left (a+b \sin (e+f x)^4\right )^pdx\)

\(\Big \downarrow \) 3702

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

\(\Big \downarrow \) 1519

\(\displaystyle \frac {\frac {\int -\left (\left (2 b (4 p+5) \sin ^2(e+f x)+a-b (4 p+5)\right ) \left (b \sin ^4(e+f x)+a\right )^p\right )d\sin (e+f x)}{b (4 p+5)}+\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{p+1}}{b (4 p+5)}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{p+1}}{b (4 p+5)}-\frac {\int \left (2 b (4 p+5) \sin ^2(e+f x)+a-5 b-4 b p\right ) \left (b \sin ^4(e+f x)+a\right )^pd\sin (e+f x)}{b (4 p+5)}}{f}\)

\(\Big \downarrow \) 1516

\(\displaystyle \frac {\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{p+1}}{b (4 p+5)}-\frac {\int \left (2 b (4 p+5) \sin ^2(e+f x) \left (b \sin ^4(e+f x)+a\right )^p+a \left (1-\frac {b (4 p+5)}{a}\right ) \left (b \sin ^4(e+f x)+a\right )^p\right )d\sin (e+f x)}{b (4 p+5)}}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^{p+1}}{b (4 p+5)}-\frac {(a-4 b p-5 b) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right )+\frac {2}{3} b (4 p+5) \sin ^3(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b \sin ^4(e+f x)}{a}\right )}{b (4 p+5)}}{f}\)

Input:

Int[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^4)^p,x]
 

Output:

((Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^(1 + p))/(b*(5 + 4*p)) - (((a - 5*b 
- 4*b*p)*Hypergeometric2F1[1/4, -p, 5/4, -((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*b*(5 + 4* 
p)*Hypergeometric2F1[3/4, -p, 7/4, -((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))/(b*(5 + 4*p)) 
)/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1516

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[Expa 
ndIntegrand[(d + e*x^2)*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && 
NeQ[c*d^2 + a*e^2, 0]
 

rule 1519

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Sim 
p[e^q*x^(2*q - 3)*((a + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[1/(c 
*(4*p + 2*q + 1))   Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d + e* 
x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x 
], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]
 

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]

\[\int \cos \left (f x +e \right )^{5} \left (a +b \sin \left (f x +e \right )^{4}\right )^{p}d x\]

Input:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x)
 

Output:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x)
 

Fricas [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(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*cos(f*x + e)^5, 
 x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**5*(a+b*sin(f*x+e)**4)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^4 + a)^p*cos(f*x + e)^5, x)
 

Giac [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e)^4 + a)^p*cos(f*x + e)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p \,d x \] Input:

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^4)^p,x)
 

Output:

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^4)^p, x)
 

Reduce [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int \left (\sin \left (f x +e \right )^{4} b +a \right )^{p} \cos \left (f x +e \right )^{5}d x \] Input:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^4)^p,x)
                                                                                    
                                                                                    
 

Output:

int((sin(e + f*x)**4*b + a)**p*cos(e + f*x)**5,x)