3.653 \(\int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=116 \[ -\frac{\cos (e+f x) \left (\frac{\sin (e+f x)+1}{1-\sin (e+f x)}\right )^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2}-m,-m;1-m;-\frac{2 \sin (e+f x)}{1-\sin (e+f x)}\right )}{d f m (\sin (e+f x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.189131, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2787, 2786, 2785, 132} \[ -\frac{\cos (e+f x) \left (\frac{\sin (e+f x)+1}{1-\sin (e+f x)}\right )^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2}-m,-m;1-m;-\frac{2 \sin (e+f x)}{1-\sin (e+f x)}\right )}{d f m (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2787

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 2786

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[((d/b
)^IntPart[n]*(d*Sin[e + f*x])^FracPart[n])/(b*Sin[e + f*x])^FracPart[n], Int[(a + b*Sin[e + f*x])^m*(b*Sin[e +
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] &&  !Gt
Q[d/b, 0]

Rule 2785

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

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

Mathematica [C]  time = 1.50715, size = 194, normalized size = 1.67 \[ \frac{(1-i) 2^m (\cosh (m \log (2))-\sinh (m \log (2))) (\cos (e+f x)-i (\sin (e+f x)+1)) (a (\sin (e+f x)+1))^m (d \sin (e+f x))^{-m} ((1-i) (-i \sin (e+f x)+\cos (e+f x)+1))^m ((1+i) (i \sin (e+f x)-\cos (e+f x)+1))^{-m} \, _2F_1\left (m+1,2 m+1;2 (m+1);\sqrt{2} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \csc \left (\frac{1}{2} (e+f x)\right )\right )}{d f (2 m+1) (-i \sin (e+f x)+\cos (e+f x)-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.24, size = 0, normalized size = 0. \begin{align*} \int \left ( d\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((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

[Out]

int((d*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 (d \sin \left (f x + e\right )\right )^{-m - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*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*(d*sin(f*x + e))^(-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 (d \sin \left (f x + e\right )\right )^{-m - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*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*(d*sin(f*x + e))^(-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((d*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 (d \sin \left (f x + e\right )\right )^{-m - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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