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

Optimal. Leaf size=94 \[ -\frac {\cos (e+f x) (3-\sin (e+f x))^{-m-1} \left (\frac {3-\sin (e+f x)}{\sin (e+f x)+1}\right )^{m+1} (\sin (e+f x)+1)^m \, _2F_1\left (\frac {1}{2},m+1;\frac {3}{2};-\frac {2 (1-\sin (e+f x))}{\sin (e+f x)+1}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 94, normalized size of antiderivative = 1.00, 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} \[ -\frac {\cos (e+f x) (3-\sin (e+f x))^{-m-1} \left (\frac {3-\sin (e+f x)}{\sin (e+f x)+1}\right )^{m+1} (\sin (e+f x)+1)^m \, _2F_1\left (\frac {1}{2},m+1;\frac {3}{2};-\frac {2 (1-\sin (e+f x))}{\sin (e+f x)+1}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 1.04, size = 182, normalized size = 1.94 \[ -\frac {2^{\frac {1}{2}-m} \cot \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (3-\sin (e+f x))^{-m} (\sin (e+f x)+1)^m \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{\frac {1}{2}-m} \cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{m-\frac {1}{2}} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};-\frac {4 \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sin (e+f x)-3}\right ) \left (-\frac {\cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sin (e+f x)-3}\right )^{\frac {1}{2}-m}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\sin \left (f x + e\right ) + 1\right )}^{m} {\left (-\sin \left (f x + e\right ) + 3\right )}^{-m - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\sin \left (f x + e\right ) + 1\right )}^{m} {\left (-\sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.87, size = 0, normalized size = 0.00 \[ \int \left (3-\sin \left (f x +e \right )\right )^{-1-m} \left (1+\sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\sin \left (f x + e\right ) + 1\right )}^{m} {\left (-\sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\sin \left (e+f\,x\right )+1\right )}^m}{{\left (3-\sin \left (e+f\,x\right )\right )}^{m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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