3.75 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=107 \[ \frac{8 c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+16 m+15\right ) \sqrt{c-c \sin (e+f x)}}+\frac{2 \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+5)} \]

[Out]

(8*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(15 + 16*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (2*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c - c*Sin[e + f*x]])/(a*f*(5 + 2*m))

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Rubi [A]  time = 0.354111, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2841, 2740, 2738} \[ \frac{8 c \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+16 m+15\right ) \sqrt{c-c \sin (e+f x)}}+\frac{2 \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(15 + 16*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (2*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c - c*Sin[e + f*x]])/(a*f*(5 + 2*m))

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), 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, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 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 \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx &=\frac{\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)}}{a f (5+2 m)}+\frac{4 \int (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)} \, dx}{a (5+2 m)}\\ &=\frac{8 c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f \left (15+16 m+4 m^2\right ) \sqrt{c-c \sin (e+f x)}}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)}}{a f (5+2 m)}\\ \end{align*}

Mathematica [A]  time = 0.560205, size = 111, normalized size = 1.04 \[ -\frac{2 \sqrt{c-c \sin (e+f x)} ((2 m+3) \sin (e+f x)-2 m-7) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 (a (\sin (e+f x)+1))^m}{f (2 m+3) (2 m+5) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]]*(-7 - 2*m + (3 +
2*m)*Sin[e + f*x]))/(f*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [F]  time = 0.348, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\sqrt{c-c\sin \left ( fx+e \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.14321, size = 421, normalized size = 3.93 \begin{align*} -\frac{2 \,{\left (a^{m} \sqrt{c}{\left (2 \, m + 7\right )} + \frac{a^{m} \sqrt{c}{\left (2 \, m + 15\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{2 \, a^{m} \sqrt{c}{\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \, a^{m} \sqrt{c}{\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{a^{m} \sqrt{c}{\left (2 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{m} \sqrt{c}{\left (2 \, m + 7\right )} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} 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} + 16 \, m + \frac{2 \,{\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} f \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-2*(a^m*sqrt(c)*(2*m + 7) + a^m*sqrt(c)*(2*m + 15)*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^m*sqrt(c)*(2*m - 5)*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*a^m*sqrt(c)*(2*m - 5)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a^m*sqrt(c)
*(2*m + 15)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^m*sqrt(c)*(2*m + 7)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*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 + 16*
m + 2*(4*m^2 + 16*m + 15)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (4*m^2 + 16*m + 15)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 15)*f*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))

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Fricas [A]  time = 1.78459, size = 402, normalized size = 3.76 \begin{align*} \frac{2 \,{\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{3} +{\left (2 \, m - 1\right )} \cos \left (f x + e\right )^{2} +{\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 8\right )} \sin \left (f x + e\right ) + 4 \, \cos \left (f x + e\right ) + 8\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} + 16 \, f m +{\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) -{\left (4 \, f m^{2} + 16 \, 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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2*((2*m + 3)*cos(f*x + e)^3 + (2*m - 1)*cos(f*x + e)^2 + ((2*m + 3)*cos(f*x + e)^2 + 4*cos(f*x + e) + 8)*sin(f
*x + e) + 4*cos(f*x + e) + 8)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(4*f*m^2 + 16*f*m + (4*f*m^2 +
16*f*m + 15*f)*cos(f*x + e) - (4*f*m^2 + 16*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(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(1/2),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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out