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

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

[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)

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Rubi [A]
time = 0.17, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 = {2997, 145, 144, 143} \begin {gather*} \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)}} \end {gather*}

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 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*} \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) \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} 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*}

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Mathematica [A]
time = 0.57, size = 158, normalized size = 1.17 \begin {gather*} -\frac {4 F_1\left (\frac {3}{2};-\frac {1}{2}-m,-n;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \cos (e+f x) (a (1+\sin (e+f x)))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{-\frac {1}{2}-m}}{3 f} \end {gather*}

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 = 0.21, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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