3.627 \(\int (3-2 \sin (e+f x))^{-1-m} (1+\sin (e+f x))^m \, dx\)

Optimal. Leaf size=114 \[ \frac{\sqrt{-\frac{1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (3-2 \sin (e+f x))^{-m} (\sin (e+f x)+1)^m \, _2F_1\left (\frac{1}{2},-m;1-m;\frac{2 (3-2 \sin (e+f x))}{\sin (e+f x)+1}\right )}{\sqrt{5} f m (1-\sin (e+f x))} \]

[Out]

(Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, (2*(3 - 2*Sin[e + f*x]))/(1 + Sin[e + f*x])]*Sqrt[-((1 - Sin[e
+ f*x])/(1 + Sin[e + f*x]))]*(1 + Sin[e + f*x])^m)/(Sqrt[5]*f*m*(3 - 2*Sin[e + f*x])^m*(1 - Sin[e + f*x]))

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Rubi [A]  time = 0.099271, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2788, 132} \[ \frac{\sqrt{-\frac{1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (3-2 \sin (e+f x))^{-m} (\sin (e+f x)+1)^m \, _2F_1\left (\frac{1}{2},-m;1-m;\frac{2 (3-2 \sin (e+f x))}{\sin (e+f x)+1}\right )}{\sqrt{5} f m (1-\sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(3 - 2*Sin[e + f*x])^(-1 - m)*(1 + Sin[e + f*x])^m,x]

[Out]

(Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, (2*(3 - 2*Sin[e + f*x]))/(1 + Sin[e + f*x])]*Sqrt[-((1 - Sin[e
+ f*x])/(1 + Sin[e + f*x]))]*(1 + Sin[e + f*x])^m)/(Sqrt[5]*f*m*(3 - 2*Sin[e + f*x])^m*(1 - Sin[e + f*x]))

Rule 2788

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.884338, size = 177, normalized size = 1.55 \[ -\frac{2 \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (3-2 \sin (e+f x))^{-m} (\sin (e+f x)+1)^m \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{5 \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{3-2 \sin (e+f x)}\right ) \left (-\frac{\cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{2 \sin (e+f x)-3}\right )^{\frac{1}{2}-m}}{f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(3 - 2*Sin[e + f*x])^(-1 - m)*(1 + Sin[e + f*x])^m,x]

[Out]

(-2*(Cos[(2*e - Pi + 2*f*x)/4]^2)^(-1/2 + m)*Cot[(2*e + Pi + 2*f*x)/4]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (5
*Sin[(2*e - Pi + 2*f*x)/4]^2)/(3 - 2*Sin[e + f*x])]*(1 + Sin[e + f*x])^m*(-(Cos[(2*e - Pi + 2*f*x)/4]^2/(-3 +
2*Sin[e + f*x])))^(1/2 - m)*(Sin[(2*e + Pi + 2*f*x)/4]^2)^(1/2 - m))/(f*(3 - 2*Sin[e + f*x])^m)

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Maple [F]  time = 0.256, size = 0, normalized size = 0. \begin{align*} \int \left ( 3-2\,\sin \left ( fx+e \right ) \right ) ^{-1-m} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3-2*sin(f*x+e))^(-1-m)*(1+sin(f*x+e))^m,x)

[Out]

int((3-2*sin(f*x+e))^(-1-m)*(1+sin(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\sin \left (f x + e\right ) + 1\right )}^{m}{\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-2*sin(f*x+e))^(-1-m)*(1+sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((sin(f*x + e) + 1)^m*(-2*sin(f*x + e) + 3)^(-m - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\sin \left (f x + e\right ) + 1\right )}^{m}{\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-2*sin(f*x+e))^(-1-m)*(1+sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((sin(f*x + e) + 1)^m*(-2*sin(f*x + e) + 3)^(-m - 1), 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((3-2*sin(f*x+e))**(-1-m)*(1+sin(f*x+e))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\sin \left (f x + e\right ) + 1\right )}^{m}{\left (-2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-2*sin(f*x+e))^(-1-m)*(1+sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((sin(f*x + e) + 1)^m*(-2*sin(f*x + e) + 3)^(-m - 1), x)