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

Optimal. Leaf size=72 \[ -\frac{\cos (e+f x) (\sin (e+f x)+1)^{-m-1} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},m+1;\frac{3}{2};\frac{4 (a-a \sin (e+f x))}{\sin (e+f x) a+a}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0980652, antiderivative size = 113, normalized size of antiderivative = 1.57, 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} \[ -\frac{\sqrt{\frac{1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (5 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},-m;1-m;-\frac{3-5 \sin (e+f x)}{\sin (e+f x)+1}\right )}{4 f m (1-\sin (e+f x))} \]

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

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)}{1+\sin (e+f x)}\right ) \sqrt{\frac{1-\sin (e+f x)}{1+\sin (e+f x)}} (-3+5 \sin (e+f x))^{-m} (a+a \sin (e+f x))^m}{4 f m (1-\sin (e+f x))}\\ \end{align*}

Mathematica [C]  time = 1.52856, size = 247, normalized size = 3.43 \[ \frac{2^{2 m-1} (\cosh (m \log (4))-\sinh (m \log (4))) (5 \sin (e+f x)-3)^{-m} (\sin (e+f x)+i \cos (e+f x)+1) (a (\sin (e+f x)+1))^m \left (\frac{2 \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )-\cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{2 \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )+\cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}\right )^m \, _2F_1\left (m+1,2 m+1;2 (m+1);\frac{2 \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )+2 \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}\right )}{f (2 m+1) ((1+2 i) \sin (e+f x)+(-2+i) \cos (e+f x)+(1-2 i))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.228, 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)