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

Optimal. Leaf size=135 \[ \frac{2 \sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{3}{2};-\frac{1}{2},-n;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)}} \]

[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)

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Rubi [A]  time = 0.244927, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2918, 140, 139, 138} \[ \frac{2 \sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{3}{2};-\frac{1}{2},-n;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt{1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[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 2918

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]

Rule 140

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 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx &=\frac{\cos (e+f x) \operatorname{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 ) \operatorname{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 ) \operatorname{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} F_1\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 [A]  time = 0.676294, size = 158, normalized size = 1.17 \[ -\frac{4 \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos (e+f x) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{-m-\frac{1}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{3}{2};-m-\frac{1}{2},-n;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{3 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*AppellF1[3/2, -1/2 - m, -n, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)]*C
os[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(c + d*Sin[e + f*x])^n*Sin[(2*e - Pi + 2*f*x)/4]^2*(Sin[(2*e + Pi + 2*f*x
)/4]^2)^(-1/2 - m))/(3*f*((c + d*Sin[e + f*x])/(c + d))^n)

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Maple [F]  time = 0.437, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)