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

Optimal. Leaf size=129 \[ -\frac{a 2^{m+\frac{1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)} \]

[Out]

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

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Rubi [A]  time = 0.148658, antiderivative size = 129, 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} \[ -\frac{a 2^{m+\frac{1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1/2 + m)*a*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, ((c - d)*(1 - Sin[e + f*x]))/(2*(c + d*Sin[
e + f*x]))]*(a + a*Sin[e + f*x])^(-1 + m)*(((c + d)*(1 + Sin[e + f*x]))/(c + d*Sin[e + f*x]))^(1/2 - m))/((c +
d)*f*(c + d*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 (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx &=\frac{\left (a^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^{-1-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{2^{\frac{1}{2}+m} a \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c+d) f}\\ \end{align*}

Mathematica [A]  time = 1.38097, size = 187, normalized size = 1.45 \[ -\frac{2 \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m+\frac{1}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right ) \left (\frac{(c+d) \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}\right )^{-m-\frac{1}{2}}}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.25, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{-1-m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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