3.5.22 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx\) [422]

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

[Out]

2^(1/2-m)*c*cos(f*x+e)*hypergeom([1/2+m, 1/2+m],[3/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2+m)*(a+a*sin(f*
x+e))^m*(c-c*sin(f*x+e))^(-1-m)/f/(1+2*m)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 = {2824, 2768, 72, 71} \begin {gather*} \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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 2768

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 2824

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])

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 ) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 1.86, size = 388, normalized size = 3.46 \begin {gather*} \frac {2^{1-m} (-3+2 m) F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e+\pi -2 f x)\right ),-\tan ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )\right ) \cos ^{1-2 m}\left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{2 m} (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m} \sin ^2\left (\frac {1}{8} (2 e+3 \pi +2 f x)\right )}{f (-1+2 m) \left ((-3+2 m) F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e+\pi -2 f x)\right ),-\tan ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )\right ) \cos ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )+2 \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+\pi -2 f x)\right ),-\tan ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )\right )+F_1\left (\frac {3}{2}-m;-2 m,2;\frac {5}{2}-m;\tan ^2\left (\frac {1}{8} (-2 e+\pi -2 f x)\right ),-\tan ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )\right )\right ) \sin ^2\left (\frac {1}{8} (2 e-\pi +2 f x)\right )\right )} \end {gather*}

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))

________________________________________________________________________________________

Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-m}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m}\, dx \end {gather*}

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]

Integral((a*(sin(e + f*x) + 1))**m/(-c*(sin(e + f*x) - 1))**m, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________