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

Optimal. Leaf size=197 \[ \frac{2 c (A (2 m+5)-B (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 c (2 B m+5 B+4 C m+2 C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)} \]

[Out]

(2*c*(C - 6*C*m + A*(5 + 2*m) - B*(5 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*(5 + 2*m)*Sqrt[
c - c*Sin[e + f*x]]) + (2*c*(5*B + 2*C + 2*B*m + 4*C*m)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 2
*m)*(5 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))
/(c*f*(5 + 2*m))

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Rubi [A]  time = 0.628587, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3039, 2971, 2738} \[ \frac{2 c (A (2 m+5)-B (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 c (2 B m+5 B+4 C m+2 C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(C - 6*C*m + A*(5 + 2*m) - B*(5 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*(5 + 2*m)*Sqrt[
c - c*Sin[e + f*x]]) + (2*c*(5*B + 2*C + 2*B*m + 4*C*m)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 2
*m)*(5 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))
/(c*f*(5 + 2*m))

Rule 3039

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

Rule 2971

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}-\frac{2 \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \left (-\frac{1}{2} a c (C (3-2 m)+A (5+2 m))-\frac{1}{2} a c (5 B+2 C+2 B m+4 C m) \sin (e+f x)\right ) \, dx}{a c (5+2 m)}\\ &=\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}+\frac{(5 B+2 C+2 B m+4 C m) \int (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)} \, dx}{a (5+2 m)}+\frac{(C-6 C m+A (5+2 m)-B (5+2 m)) \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx}{5+2 m}\\ &=\frac{2 c (C-6 C m+A (5+2 m)-B (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{2 c (5 B+2 C+2 B m+4 C m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}\\ \end{align*}

Mathematica [A]  time = 1.04685, size = 177, normalized size = 0.9 \[ \frac{\sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m \left (8 A m^2+32 A m+30 A+2 (2 m+1) (2 B m+5 B-4 C) \sin (e+f x)-8 B m-20 B-C \left (4 m^2+8 m+3\right ) \cos (2 (e+f x))+4 C m^2+8 C m+19 C\right )}{f (2 m+1) (2 m+3) (2 m+5) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]]*(30*A - 20*B + 19*C +
32*A*m - 8*B*m + 8*C*m + 8*A*m^2 + 4*C*m^2 - C*(3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] + 2*(1 + 2*m)*(5*B - 4*C +
2*B*m)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [F]  time = 0.715, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\sqrt{c-c\sin \left ( fx+e \right ) } \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))^(1/2)*(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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

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Maxima [B]  time = 1.78287, size = 869, normalized size = 4.41 \begin{align*} \text{result too large to display} \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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

-2*(2*(2*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a^m*sqrt(c)*m*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 -
a^m*sqrt(c) - a^m*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*B*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) +
1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((4*m^2 + 8*m + (4*m^2 + 8*m + 3)*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 3)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)) - 4*(4*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x +
e) + 1) - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*s
in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 4*a^m*sqrt(c)*m*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2*a^m*sqrt(c) - 2*a
^m*sqrt(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*C*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 36*m^2 + 46*m + 2*(8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + (8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15)*sqrt(sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 1)) + (a^m*sqrt(c) + a^m*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1))*A*e^(2*m*log(sin(
f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((2*m + 1)*sqrt(sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 1)))/f

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Fricas [A]  time = 1.86956, size = 786, normalized size = 3.99 \begin{align*} -\frac{2 \,{\left ({\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 4 \,{\left (A + B + C\right )} m^{2} +{\left (4 \,{\left (B + C\right )} m^{2} + 12 \, B m + 5 \, B - C\right )} \cos \left (f x + e\right )^{2} - 8 \,{\left (2 \, A + B\right )} m -{\left (4 \,{\left (A + C\right )} m^{2} + 4 \,{\left (4 \, A - B + 2 \, C\right )} m + 15 \, A - 10 \, B + 11 \, C\right )} \cos \left (f x + e\right ) -{\left (4 \,{\left (A + B + C\right )} m^{2} -{\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{2} + 8 \,{\left (2 \, A + B\right )} m +{\left (4 \, B m^{2} + 4 \,{\left (3 \, B - 2 \, C\right )} m + 5 \, B - 4 \, C\right )} \cos \left (f x + e\right ) + 15 \, A - 5 \, B + 7 \, C\right )} \sin \left (f x + e\right ) - 15 \, A + 5 \, B - 7 \, C\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m +{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) -{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

-2*((4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^3 - 4*(A + B + C)*m^2 + (4*(B + C)*m^2 + 12*B*m + 5*B - C)*cos(f*x +
e)^2 - 8*(2*A + B)*m - (4*(A + C)*m^2 + 4*(4*A - B + 2*C)*m + 15*A - 10*B + 11*C)*cos(f*x + e) - (4*(A + B + C
)*m^2 - (4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^2 + 8*(2*A + B)*m + (4*B*m^2 + 4*(3*B - 2*C)*m + 5*B - 4*C)*cos(f
*x + e) + 15*A - 5*B + 7*C)*sin(f*x + e) - 15*A + 5*B - 7*C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/
(8*f*m^3 + 36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 + 36*f*m^2 + 46*f*
m + 15*f)*sin(f*x + e) + 15*f)

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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))**(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)

[Out]

Timed out

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Giac [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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

Timed out