\(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\) [24]

Optimal result

Integrand size = 50, antiderivative size = 232 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {2^{-\frac {1}{2}-m} C \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)} \]

[Out]

-2^(-1/2-m)*C*cos(f*x+e)^3*hypergeom([3/2+m, 3/2+m],[5/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2+m)*(a+a*si
n(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+2*m)+1/2*(A+B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(
-2-m)/a/f/(3+2*m)+1/2*(A-B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1-m)/c/f/(1+2*m)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3114, 3051, 2824, 2768, 72, 71} \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{2 a f (2 m+3)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}-\frac {C 2^{-m-\frac {1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]

[In]

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m)*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

-((2^(-1/2 - m)*C*Cos[e + f*x]^3*Hypergeometric2F1[(3 + 2*m)/2, (3 + 2*m)/2, (5 + 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])^(-2 - m))/(f*(3 + 2*m))) + ((A + B
+ C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(2*a*f*(3 + 2*m)) + ((A - B + C)
*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(2*c*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])

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(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)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rule 3114

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(2*b*c*f*(2*m + 1))), x] - Dist[1/(2*b*c*d*(2*m + 1)), Int[
(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n -
1) - C*(c^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m +
n + 2, 0] && NeQ[2*m + 1, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)}+\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \left (c^2 (A+B-C) (1+2 m)+2 c^2 C (1+2 m) \sin (e+f x)\right ) \, dx}{2 a c^2 (1+2 m)} \\ & = \frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)}-\frac {C \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-1-m} \, dx}{a c} \\ & = \frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)}-\left (C \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 (1+m)}(e+f x) (c-c \sin (e+f x))^{-2-2 m} \, dx \\ & = \frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)}-\frac {\left (c^2 C \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 (1+m))} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \text {Subst}\left (\int (c-c x)^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)}-\frac {\left (2^{-\frac {3}{2}-m} c C \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 (1+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 (1+m))}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-2-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f} \\ & = -\frac {2^{-\frac {1}{2}-m} C \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+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))^{-2-m}}{f (3+2 m)}+\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {\sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m} \left (2^{\frac {3}{2}+m} C (3+2 m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-m,-\frac {1}{2}-m,\frac {1}{2}-m,\frac {1}{2} (1-\sin (e+f x))\right ) (-1+\sin (e+f x)) \sqrt {1+\sin (e+f x)}+(1+\sin (e+f x))^{1+m} (2 A-B+2 C+2 A m+2 C m-(A+C-2 B (1+m)) \sin (e+f x))\right )}{c^2 f (1+2 m) (3+2 m) (-1+\sin (e+f x))} \]

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m)*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(2^(3/2 + m)*C*(3 + 2*m)*Hypergeometric2F1[-1/2 - m, -1/2 - m, 1/2 -
m, (1 - Sin[e + f*x])/2]*(-1 + Sin[e + f*x])*Sqrt[1 + Sin[e + f*x]] + (1 + Sin[e + f*x])^(1 + m)*(2*A - B + 2*
C + 2*A*m + 2*C*m - (A + C - 2*B*(1 + m))*Sin[e + f*x])))/(c^2*f*(1 + 2*m)*(3 + 2*m)*(-1 + Sin[e + f*x])*(1 +
Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m))

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]

[In]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

Fricas [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x
)

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 2} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]

[In]

integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 2)*(A + B*sin(e + f*x) + C*sin(e + f*x)**2),
x)

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)

Giac [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \]

[In]

int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c - c*sin(e + f*x))^(m + 2),x)

[Out]

int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c - c*sin(e + f*x))^(m + 2), x)