3.652 \(\int (-3-5 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=115 \[ \frac{\sqrt{\frac{1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (-5 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},-m;1-m;\frac{5 \sin (e+f x)+3}{4 (\sin (e+f x)+1)}\right )}{4 f m (1-\sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.0982541, antiderivative size = 115, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{\frac{1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (-5 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},-m;1-m;\frac{5 \sin (e+f x)+3}{4 (\sin (e+f x)+1)}\right )}{4 f m (1-\sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(-3 - 5*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, (3 + 5*Sin[e + f*x])/(4*(1 + Sin[e + f*x]))]*Sqrt[(1 - Sin[e +
f*x])/(1 + Sin[e + f*x])]*(a + a*Sin[e + f*x])^m)/(4*f*m*(-3 - 5*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 (-3-5 \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{(-3-5 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{3+5 \sin (e+f x)}{4 (1+\sin (e+f x))}\right ) (-3-5 \sin (e+f x))^{-m} \sqrt{\frac{1-\sin (e+f x)}{1+\sin (e+f x)}} (a+a \sin (e+f x))^m}{4 f m (1-\sin (e+f x))}\\ \end{align*}

Mathematica [C]  time = 1.56132, size = 241, normalized size = 2.1 \[ -\frac{4^m (\cosh (m \log (4))-\sinh (m \log (4))) (-5 \sin (e+f x)-3)^{-m} (\sin (e+f x)+i \cos (e+f x)+1) (a (\sin (e+f x)+1))^m \left (-\frac{2 \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )+\cos \left (\frac{1}{4} (2 e+2 f x+\pi )\right )}{\sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )+2 \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}\right )^m \, _2F_1\left (m+1,2 m+1;2 (m+1);\frac{4 \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{2 \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )+\sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}\right )}{f (2 m+1) ((2+i) \sin (e+f x)+(-1+2 i) \cos (e+f x)+(2-i))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(-3 - 5*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

-((4^m*Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (4*Cos[(2*e - Pi + 2*f*x)/4])/(2*Cos[(2*e - Pi + 2*f*x)/4]
+ Sin[(2*e - Pi + 2*f*x)/4])]*(a*(1 + Sin[e + f*x]))^m*(1 + I*Cos[e + f*x] + Sin[e + f*x])*(-((2*Cos[(2*e - P
i + 2*f*x)/4] + Cos[(2*e + Pi + 2*f*x)/4])/(2*Cos[(2*e - Pi + 2*f*x)/4] + Sin[(2*e - Pi + 2*f*x)/4])))^m*(Cosh
[m*Log[4]] - Sinh[m*Log[4]]))/(f*(1 + 2*m)*(-3 - 5*Sin[e + f*x])^m*((2 - I) - (1 - 2*I)*Cos[e + f*x] + (2 + I)
*Sin[e + f*x])))

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Maple [F]  time = 0.263, size = 0, normalized size = 0. \begin{align*} \int \left ( -3-5\,\sin \left ( fx+e \right ) \right ) ^{-1-m} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3-5*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

[Out]

int((-3-5*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-5 \, \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((-3-5*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*(-5*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 (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-5 \, \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((-3-5*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*(-5*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((-3-5*sin(f*x+e))**(-1-m)*(a+a*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 (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-5 \, \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((-3-5*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*(-5*sin(f*x + e) - 3)^(-m - 1), x)