3.67 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=86 \[ -\frac{a^3 c^2 2^{m+\frac{3}{2}} \cos ^7(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{7}{2},-m-\frac{1}{2};\frac{9}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{7 f} \]

[Out]

-(2^(3/2 + m)*a^3*c^2*Cos[e + f*x]^7*Hypergeometric2F1[7/2, -1/2 - m, 9/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e +
f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(-3 + m))/(7*f)

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Rubi [A]  time = 0.209284, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2840, 2689, 70, 69} \[ -\frac{a^3 c^2 2^{m+\frac{3}{2}} \cos ^7(e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac{7}{2},-m-\frac{1}{2};\frac{9}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(3/2 + m)*a^3*c^2*Cos[e + f*x]^7*Hypergeometric2F1[7/2, -1/2 - m, 9/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e +
f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(-3 + m))/(7*f)

Rule 2840

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

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \cos ^6(e+f x) (a+a \sin (e+f x))^{-2+m} \, dx\\ &=\frac{\left (a^4 c^2 \cos ^7(e+f x)\right ) \operatorname{Subst}\left (\int (a-a x)^{5/2} (a+a x)^{\frac{1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{7/2} (a+a \sin (e+f x))^{7/2}}\\ &=\frac{\left (2^{\frac{1}{2}+m} a^4 c^2 \cos ^7(e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{2}+m} (a-a x)^{5/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{7/2}}\\ &=-\frac{2^{\frac{3}{2}+m} a^3 c^2 \cos ^7(e+f x) \, _2F_1\left (\frac{7}{2},-\frac{1}{2}-m;\frac{9}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^{-3+m}}{7 f}\\ \end{align*}

Mathematica [F]  time = 180.038, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [F]  time = 3.296, 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-c\sin \left ( fx+e \right ) \right ) ^{2}\, 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-c*sin(f*x+e))^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c \sin \left (f x + e\right ) - c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \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-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((c*sin(f*x + e) - c)^2*(a*sin(f*x + e) + a)^m*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 (c^{2} \cos \left (f x + e\right )^{4} + 2 \, c^{2} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 2 \, c^{2} \cos \left (f x + e\right )^{2}\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(-(c^2*cos(f*x + e)^4 + 2*c^2*cos(f*x + e)^2*sin(f*x + e) - 2*c^2*cos(f*x + e)^2)*(a*sin(f*x + e) + a)
^m, 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-c*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c \sin \left (f x + e\right ) - c\right )}^{2}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \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-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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