\(\int \cos ^5(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [307]

Optimal result

Integrand size = 23, antiderivative size = 207 \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {(3 a+10 b+4 b p) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2+2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \] Output:

-(4*b*p+3*a+10*b)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(p+1)/b^2/f/(3+2*p)/(5+2*p 
)+sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^(p+1)/b/f/(5+2*p)+(3*a^2+2*a*b*(5+2*p)+b 
^2*(4*p^2+16*p+15))*hypergeom([1/2, -p],[3/2],-b*sin(f*x+e)^2/a)*sin(f*x+e 
)*(a+b*sin(f*x+e)^2)^p/b^2/f/(3+2*p)/(5+2*p)/((1+b*sin(f*x+e)^2/a)^p)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.50 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.92 \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},-2,-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \cos ^4(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},-2,-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+2 \left (b p \operatorname {AppellF1}\left (\frac {3}{2},-2,1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-2 a \operatorname {AppellF1}\left (\frac {3}{2},-1,-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \] Input:

Integrate[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

(3*a*AppellF1[1/2, -2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*C 
os[e + f*x]^4*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(3*a*AppellF1[1/2, 
 -2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + 2*(b*p*AppellF1[3 
/2, -2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] - 2*a*AppellF 
1[3/2, -1, -p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)])*Sin[e + f*x] 
^2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3669, 318, 299, 238, 237}

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.

\(\displaystyle \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^5 \left (a+b \sin (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \int \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^pd\sin (e+f x)}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right )}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\)

Input:

Int[Cos[e + f*x]^5*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

(-((Sin[e + f*x]*(1 - Sin[e + f*x]^2)*(a + b*Sin[e + f*x]^2)^(1 + p))/(b*( 
5 + 2*p))) + (-(((3*a + b*(7 + 2*p))*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^( 
1 + p))/(b*(3 + 2*p))) + ((3*a^2 + 2*a*b*(5 + 2*p) + b^2*(15 + 16*p + 4*p^ 
2))*Hypergeometric2F1[1/2, -p, 3/2, -((b*Sin[e + f*x]^2)/a)]*Sin[e + f*x]* 
(a + b*Sin[e + f*x]^2)^p)/(b*(3 + 2*p)*(1 + (b*Sin[e + f*x]^2)/a)^p))/(b*( 
5 + 2*p)))/f
 

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

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]

\[\int \cos \left (f x +e \right )^{5} \left (a +b \sin \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x)
 

Output:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((-b*cos(f*x + e)^2 + a + b)^p*cos(f*x + e)^5, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**5*(a+b*sin(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e)^5, x)
 

Giac [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \] Input:

integrate(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \] Input:

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^2)^p,x)
 

Output:

int(cos(e + f*x)^5*(a + b*sin(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \left (\sin \left (f x +e \right )^{2} b +a \right )^{p} \cos \left (f x +e \right )^{5}d x \] Input:

int(cos(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x)
                                                                                    
                                                                                    
 

Output:

int((sin(e + f*x)**2*b + a)**p*cos(e + f*x)**5,x)