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

Optimal. Leaf size=112 \[ \frac{c 2^{\frac{1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]

[Out]

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

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Rubi [A]  time = 0.157636, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2745, 2689, 70, 69} \[ \frac{c 2^{\frac{1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2745

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

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [C]  time = 2.93898, size = 388, normalized size = 3.46 \[ \frac{2^{1-m} (2 m-3) \sin ^2\left (\frac{1}{8} (2 e+2 f x+3 \pi )\right ) \cos ^{1-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m F_1\left (\frac{1}{2}-m;-2 m,1;\frac{3}{2}-m;\tan ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right )\right ) (c-c \sin (e+f x))^{-m} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{2 m}}{f (2 m-1) \left (2 \sin ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right ) \left (2 m F_1\left (\frac{3}{2}-m;1-2 m,1;\frac{5}{2}-m;\tan ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right )\right )+F_1\left (\frac{3}{2}-m;-2 m,2;\frac{5}{2}-m;\tan ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right )\right )\right )+(2 m-3) \cos ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right ) F_1\left (\frac{1}{2}-m;-2 m,1;\frac{3}{2}-m;\tan ^2\left (\frac{1}{8} (-2 e-2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 e+2 f x-\pi )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(1 - m)*(-3 + 2*m)*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-2*e + Pi - 2*f*x)/8]^2, -Tan[(2*e - Pi + 2*f*x
)/8]^2]*Cos[(2*e + Pi + 2*f*x)/4]^(1 - 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*m)*(a*(1 + Sin[e + f*x]))
^m*Sin[(2*e + 3*Pi + 2*f*x)/8]^2)/(f*(-1 + 2*m)*(c - c*Sin[e + f*x])^m*((-3 + 2*m)*AppellF1[1/2 - m, -2*m, 1,
3/2 - m, Tan[(-2*e + Pi - 2*f*x)/8]^2, -Tan[(2*e - Pi + 2*f*x)/8]^2]*Cos[(2*e - Pi + 2*f*x)/8]^2 + 2*(2*m*Appe
llF1[3/2 - m, 1 - 2*m, 1, 5/2 - m, Tan[(-2*e + Pi - 2*f*x)/8]^2, -Tan[(2*e - Pi + 2*f*x)/8]^2] + AppellF1[3/2
- m, -2*m, 2, 5/2 - m, Tan[(-2*e + Pi - 2*f*x)/8]^2, -Tan[(2*e - Pi + 2*f*x)/8]^2])*Sin[(2*e - Pi + 2*f*x)/8]^
2))

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Maple [F]  time = 0.463, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{ \left ( c-c\sin \left ( fx+e \right ) \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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