\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\) [938]

Optimal result

Integrand size = 33, antiderivative size = 135 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {3}{2}+m,-\frac {1}{2},-n,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f (3+2 m) \sqrt {1-\sin (e+f x)}} \]

[Out]

2*AppellF1(3/2+m,-n,-1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(
c+d*sin(f*x+e))^n*2^(1/2)/a/f/(3+2*m)/(((c+d*sin(f*x+e))/(c-d))^n)/(1-sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2997, 145, 144, 143} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\frac {2 \sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} \operatorname {AppellF1}\left (m+\frac {3}{2},-\frac {1}{2},-n,m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)}} \]

[In]

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

[Out]

(2*Sqrt[2]*AppellF1[3/2 + m, -1/2, -n, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c + d*Sin[e + f*x])^n)/(a*f*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*((c + d*Sin
[e + f*x])/(c - d))^n)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 145

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*(b*(c/(b*c -
a*d)) + b*d*(x/(b*c - a*d)))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 2997

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(
x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]/(a^(p - 2)*f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Su
bst[Int[(a + b*x)^(m + p/2 - 1/2)*(a - b*x)^(p/2 - 1/2)*(c + d*x)^n, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[p/2] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) \text {Subst}\left (\int \sqrt {a-a x} (a+a x)^{\frac {1}{2}+m} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = \frac {\left (\sqrt {2} \cos (e+f x)\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{\frac {1}{2}+m} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {a+a \sin (e+f x)}} \\ & = \frac {\left (\sqrt {2} \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^n \, dx,x,\sin (e+f x)\right )}{f \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {a+a \sin (e+f x)}} \\ & = \frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {3}{2}+m,-\frac {1}{2},-n,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f (3+2 m) \sqrt {1-\sin (e+f x)}} \\ \end{align*}

Mathematica [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**n,x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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