\(\int \cos ^4(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [312]

Optimal result

Integrand size = 23, antiderivative size = 90 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(199\) vs. \(2(90)=180\).

Time = 1.05 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.21 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \cos ^3(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{2},1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-3 a \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \] Input:

Integrate[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

(3*a*AppellF1[1/2, -3/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] 
*Cos[e + f*x]^3*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(3*a*AppellF1[1/ 
2, -3/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + (2*b*p*Appell 
F1[3/2, -3/2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] - 3*a*A 
ppellF1[3/2, -1/2, -p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)])*Sin[ 
e + f*x]^2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 90, 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, 3671, 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.

\(\displaystyle \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^4 \left (a+b \sin (e+f x)^2\right )^pdx\)

\(\Big \downarrow \) 3671

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \left (1-\sin ^2(e+f x)\right )^{3/2} \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{f}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \left (1-\sin ^2(e+f x)\right )^{3/2} \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^pd\sin (e+f x)}{f}\)

\(\Big \downarrow \) 333

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

Input:

Int[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2)^p,x]
 

Output:

(AppellF1[1/2, -3/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqr 
t[Cos[e + f*x]^2]*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(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 3671

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*(Sqrt[ 
Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[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/2] &&  !IntegerQ[p]
 

Maple [F]

\[\int \cos \left (f x +e \right )^{4} \left (a +b \sin \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x)
 

Output:

int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(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)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**4*(a+b*sin(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(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)^4, x)
 

Giac [F]

\[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(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)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^4\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \] Input:

int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2)^p,x)
 

Output:

int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \left (\sin \left (f x +e \right )^{2} b +a \right )^{p} \cos \left (f x +e \right )^{4}d x \] Input:

int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2)^p,x)
 

Output:

int((sin(e + f*x)**2*b + a)**p*cos(e + f*x)**4,x)