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

Optimal. Leaf size=83 \[ -\frac {\left (\frac {5}{2}\right )^{-m-1} \cos (e+f x) (\sin (e+f x)+1)^{-m-1} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},m+1;\frac {3}{2};-\frac {a-a \sin (e+f x)}{5 (\sin (e+f x) a+a)}\right )}{f} \]

[Out]

-(5/2)^(-1-m)*cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],1/5*(-a+a*sin(f*x+e))/(a+a*sin(f*x+e)))*(1+sin(f*x+e))^(-1
-m)*(a+a*sin(f*x+e))^m/f

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Rubi [A]  time = 0.11, antiderivative size = 118, normalized size of antiderivative = 1.42, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2788, 132} \[ \frac {\sqrt {-\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (2 \sin (e+f x)+3)^{-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},-m;1-m;\frac {2 (2 \sin (e+f x)+3)}{5 (\sin (e+f x)+1)}\right )}{\sqrt {5} f m (1-\sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 0.62, size = 179, normalized size = 2.16 \[ -\frac {2\ 5^{-m-1} \cot \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (2 \sin (e+f x)+3)^{-m} \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{\frac {1}{2}-m} \cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{m-\frac {1}{2}} (a (\sin (e+f x)+1))^m \left ((2 \sin (e+f x)+3) \sec ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )^m \, _2F_1\left (\frac {1}{2},m+1;\frac {3}{2};-\frac {1}{5} \cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \sec ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*5^(-1 - 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 + m
, 3/2, -1/5*(Cos[(2*e + Pi + 2*f*x)/4]^2*Sec[(2*e - Pi + 2*f*x)/4]^2)]*(a*(1 + Sin[e + f*x]))^m*(Sec[(2*e - Pi
+ 2*f*x)/4]^2*(3 + 2*Sin[e + f*x]))^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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \left (3+2 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (2\,\sin \left (e+f\,x\right )+3\right )}^{m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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