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

Optimal. Leaf size=104 \[ \frac{2 c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{2 B c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.282108, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {2971, 2738} \[ \frac{2 c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{2 B c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx &=\frac{B \int (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{2 (A-B) c \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{2 B c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.443223, size = 116, normalized size = 1.12 \[ \frac{2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m (A (2 m+3)+B (2 m+1) \sin (e+f x)-2 B)}{f (2 m+1) (2 m+3) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.306, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \sqrt{c-c\sin \left ( fx+e \right ) }\, 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))^(1/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)

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Maxima [B]  time = 1.61937, size = 436, normalized size = 4.19 \begin{align*} -\frac{2 \,{\left (\frac{2 \,{\left (\frac{2 \, a^{m} \sqrt{c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, a^{m} \sqrt{c} m \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - a^{m} \sqrt{c} - \frac{a^{m} \sqrt{c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} B e^{\left (2 \, m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + \frac{{\left (4 \, m^{2} + 8 \, m + 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 3\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac{{\left (a^{m} \sqrt{c} + \frac{a^{m} \sqrt{c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A e^{\left (2 \, m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (2 \, m + 1\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \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))^(1/2),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)) + (a^m*sqrt(c) + a^m*sqrt(c)*sin(f*x + e)/(c
os(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 = 2.14303, size = 417, normalized size = 4.01 \begin{align*} -\frac{2 \,{\left ({\left (2 \, B m + B\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} m -{\left (2 \, A m + 3 \, A - 2 \, B\right )} \cos \left (f x + e\right ) -{\left (2 \,{\left (A + B\right )} m +{\left (2 \, B m + B\right )} \cos \left (f x + e\right ) + 3 \, A - B\right )} \sin \left (f x + e\right ) - 3 \, A + B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 8 \, f m +{\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right ) -{\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \sin \left (f x + e\right ) + 3 \, f} \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))^(1/2),x, algorithm="fricas")

[Out]

-2*((2*B*m + B)*cos(f*x + e)^2 - 2*(A + B)*m - (2*A*m + 3*A - 2*B)*cos(f*x + e) - (2*(A + B)*m + (2*B*m + B)*c
os(f*x + e) + 3*A - B)*sin(f*x + e) - 3*A + B)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(4*f*m^2 + 8*f
*m + (4*f*m^2 + 8*f*m + 3*f)*cos(f*x + e) - (4*f*m^2 + 8*f*m + 3*f)*sin(f*x + e) + 3*f)

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

[Out]

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

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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*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out