3.215 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx\)
Optimal. Leaf size=158 \[ \frac{c 2^{\frac{1}{2}-m} (A+2 B m) \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m}}{f} \]
[Out]
(2^(1/2 - m)*c*(A + 2*B*m)*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)) - (B
*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(c - c*Sin[e + f*x])^m)
________________________________________________________________________________________
Rubi [A] time = 0.268144, antiderivative size = 158, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used =
{2973, 2745, 2689, 70, 69} \[ \frac{c 2^{\frac{1}{2}-m} (A+2 B m) \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m+1),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m}}{f} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^m,x]
[Out]
(2^(1/2 - m)*c*(A + 2*B*m)*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)) - (B
*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(c - c*Sin[e + f*x])^m)
Rule 2973
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]
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 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+(A+2 B m) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx\\ &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\left ((A+2 B m) \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{B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\frac{\left (c^2 (A+2 B m) \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{B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}+\frac{\left (2^{-\frac{1}{2}-m} c^2 (A+2 B m) \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 (A+2 B m) \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)}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m}}{f}\\ \end{align*}
Mathematica [C] time = 16.8999, size = 2552, normalized size = 16.15 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^m,x]
[Out]
(2^(2 - m)*((A + B)*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^
2] + 8*B*(-AppellF1[1/2 - m, -2*m, 2, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + Appe
llF1[1/2 - m, -2*m, 3, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]))*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^(2*m)*(a + a*Sin[e + f*x])^m*((A*Cos[(-e + Pi/2 - f*x)/2]^(2*m))/(Cos[Pi/4 + (e - Pi/2 + f*
x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^(2*m) + (B*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*Sin[e + f*x])/(Cos[Pi/4 + (e
- Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^(2*m))*Tan[(-e + Pi/2 - f*x)/4])/(f*(-1 + 2*m)*Sin[(-e + P
i/2 - f*x)/2]^(2*m)*(c - c*Sin[e + f*x])^m*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(-((2^(2 - m)*m*((A + B)*App
ellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*B*(-AppellF1[1/2
- m, -2*m, 2, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[1/2 - m, -2*m, 3,
3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]))*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*Sec[(-e + P
i/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1 - 2*m))/((-1 + 2*m)*Sin[(-e +
Pi/2 - f*x)/2]^(2*m))) - (((A + B)*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2] + 8*B*(-AppellF1[1/2 - m, -2*m, 2, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*
x)/4]^2] + AppellF1[1/2 - m, -2*m, 3, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]))*Cos[
(-e + Pi/2 - f*x)/2]^(2*m)*Sec[(-e + Pi/2 - f*x)/4]^2)/(2^m*(-1 + 2*m)*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(1 - Tan
[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + (2^(2 - m)*m*((A + B)*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f
*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*B*(-AppellF1[1/2 - m, -2*m, 2, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2,
-Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[1/2 - m, -2*m, 3, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/
2 - f*x)/4]^2]))*Cos[(-e + Pi/2 - f*x)/2]^(1 + 2*m)*Sin[(-e + Pi/2 - f*x)/2]^(-1 - 2*m)*Tan[(-e + Pi/2 - f*x)/
4])/((-1 + 2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + (2^(2 - m)*m*((A + B)*AppellF1[1/2 - m, -2*m, 1, 3/2
- m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*B*(-AppellF1[1/2 - m, -2*m, 2, 3/2 - m, Tan
[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[1/2 - m, -2*m, 3, 3/2 - m, Tan[(-e + Pi/2 - f
*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]))*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*Sin[(-e + Pi/2 - f*x)/2]^(1 - 2*m
)*Tan[(-e + Pi/2 - f*x)/4])/((-1 + 2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)) - (2^(2 - m)*Cos[(-e + Pi/2 -
f*x)/2]^(2*m)*Tan[(-e + Pi/2 - f*x)/4]*((A + B)*(-(((1/2 - m)*m*AppellF1[3/2 - m, 1 - 2*m, 1, 5/2 - m, Tan[(-e
+ Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(3/2 -
m)) - ((1/2 - m)*AppellF1[3/2 - m, -2*m, 2, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*
Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(3/2 - m))) + 8*B*(((1/2 - m)*m*AppellF1[3/2 - m, 1 -
2*m, 2, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e +
Pi/2 - f*x)/4])/(3/2 - m) - ((1/2 - m)*m*AppellF1[3/2 - m, 1 - 2*m, 3, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(3/2 - m) + ((1/2 - m)*Appell
F1[3/2 - m, -2*m, 3, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4
]^2*Tan[(-e + Pi/2 - f*x)/4])/(3/2 - m) - (3*(1/2 - m)*AppellF1[3/2 - m, -2*m, 4, 5/2 - m, Tan[(-e + Pi/2 - f*
x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(3/2 - m)))))/((
-1 + 2*m)*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))))
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Maple [F] time = 1.58, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) }{ \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*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x)
[Out]
int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/((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 (B \sin \left (f x + e\right ) + A\right )}{\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*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algorithm="maxima")
[Out]
integrate((B*sin(f*x + e) + A)*(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 (B \sin \left (f x + e\right ) + A\right )}{\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*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algorithm="fricas")
[Out]
integral((B*sin(f*x + e) + A)*(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*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))**m),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/((c-c*sin(f*x+e))^m),x, algorithm="giac")
[Out]
Exception raised: AttributeError