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]

-1/7*2^(3/2+m)*a^3*c^2*cos(f*x+e)^7*hypergeom([7/2, -1/2-m],[9/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*
(a+a*sin(f*x+e))^(-3+m)/f

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Rubi [A]  time = 0.21, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 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]))

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 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 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])

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*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (c^{2} \cos \left (f x + e\right )^{4} + 2 \, c^{2} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 2 \, c^{2} \cos \left (f x + e\right )^{2}\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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

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)

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maple [F]  time = 7.57, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{2}\, dx \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

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 - c*sin(e + f*x))^2,x)

[Out]

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

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

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]

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

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