3.24 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)
Optimal. Leaf size=232 \[ \frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{2 a f (2 m+3)}+\frac{(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}-\frac{C 2^{-m-\frac{1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]
[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))
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Rubi [A] time = 0.719757, antiderivative size = 232, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 50, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used =
{3035, 2972, 2745, 2689, 70, 69} \[ \frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{2 a f (2 m+3)}+\frac{(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}-\frac{C 2^{-m-\frac{1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]
Antiderivative was successfully verified.
[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 3035
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]))
Rule 2972
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 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))^{-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+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 ) \operatorname{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 ) \operatorname{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) \, _2F_1\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 [C] time = 21.8832, size = 1087, normalized size = 4.69 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[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]
-((2^(-5 - m)*(-3 + 2*m)*Cot[(-e + Pi/2 - f*x)/4]^3*(a + a*Sin[e + f*x])^m*(2*A + C + C*Cos[2*(-e + Pi/2 - f*x
)] + 2*B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-2 - m)*(-(((A + B + C)*Hypergeometric2F1[-3/2 - m, -2*m, -1/2 -
m, Tan[(-e + Pi/2 - f*x)/4]^2])/(3 + 2*m)) - ((3*A - 5*B - 13*C)*Hypergeometric2F1[-1/2 - m, -2*m, 1/2 - m, Ta
n[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2)/(1 + 2*m) + (64*C*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Ta
n[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^4)/(1 - 2*m) - (Tan[(-e + Pi/2
- f*x)/4]^4*((3*A - 5*B - 13*C)*(-3 + 2*m)*Hypergeometric2F1[1/2 - m, -2*m, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]
^2] + (A + B + C)*(-1 + 2*m)*Hypergeometric2F1[3/2 - m, -2*m, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + P
i/2 - f*x)/4]^2))/(3 - 8*m + 4*m^2)))/(f*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
(2*(-2 - m))*(64*C*(-3 + 2*m)*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]*Cos[(-e + Pi/2 - f*x)/4]^2*Sin[(-e + Pi/2 - f*x)/4]^4 + 256*C*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]*Sin[(-e + Pi/2 - f*x)/4]^6 + 128*C*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]*Sin[(-e + Pi/2 - f*x)/4]^6 -
3*A*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) + 3*C*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) + 2*A*m*(1 - Tan[(-e +
Pi/2 - f*x)/4]^2)^(2*m) - 2*C*m*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) - 3*B*Sin[e + f*x]*(1 - Tan[(-e + Pi/2
- f*x)/4]^2)^(2*m) - 6*C*Sin[e + f*x]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) + 2*B*m*Sin[e + f*x]*(1 - Tan[(-
e + Pi/2 - f*x)/4]^2)^(2*m) + 4*C*m*Sin[e + f*x]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))))
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Maple [F] time = 1.345, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-2-m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, 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))^(-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)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \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))^(-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)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}, 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))^(-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
)
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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))**(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \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))^(-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)