\(\int (d \cos (e+f x))^m (a+b \sin ^2(e+f x))^p \, dx\) [306]

Optimal result

Integrand size = 25, antiderivative size = 115 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.67 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.98 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \tan (e+f x)}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-a (-1+m) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3-m}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \] Input:

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

Output:

(3*a*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2 
)/a)]*(d*Cos[e + f*x])^m*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(f*(3*a*Ap 
pellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + 
 (2*b*p*AppellF1[3/2, (1 - m)/2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + 
f*x]^2)/a)] - a*(-1 + m)*AppellF1[3/2, (3 - m)/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 = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3672, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3672

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

(d*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/ 
a)]*(d*Cos[e + f*x])^(-1 + m)*(Cos[e + f*x]^2)^((1 - m)/2)*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 3672

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[f 
f*d^(2*IntPart[(m - 1)/2] + 1)*((d*Cos[e + f*x])^(2*FracPart[(m - 1)/2])/(f 
*(Cos[e + f*x]^2)^FracPart[(m - 1)/2]))   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, d, e, 
f, m, p}, x] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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