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

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Rubi [A]  time = 0.35, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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)]

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 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]

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*}

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Mathematica [A]  time = 0.57, 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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fricas [A]  time = 0.49, size = 155, normalized size = 1.45 \[ \frac {2 \, {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, m - 1\right )} \cos \left (f x + e\right )^{2} + {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 8\right )} \sin \left (f x + e\right ) + 4 \, \cos \left (f x + e\right ) + 8\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 16 \, f m + {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}\,{d x} \]

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]

integrate(sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m*cos(f*x + e)^2, x)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}\, dx \]

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 = 0.53, size = 312, normalized size = 2.92 \[ -\frac {2 \, {\left (a^{m} \sqrt {c} {\left (2 \, m + 7\right )} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 7\right )} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 16 \, m + \frac {2 \, {\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} f \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \]

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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mupad [B]  time = 1.72, size = 104, normalized size = 0.97 \[ -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (25\,\cos \left (e+f\,x\right )+3\,\cos \left (3\,e+3\,f\,x\right )+8\,\sin \left (2\,e+2\,f\,x\right )+6\,m\,\cos \left (e+f\,x\right )+2\,m\,\cos \left (3\,e+3\,f\,x\right )\right )}{2\,f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (4\,m^2+16\,m+15\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(25*cos(e + f*x) + 3*cos(3*e + 3*f*x) + 8*sin(2*e + 2
*f*x) + 6*m*cos(e + f*x) + 2*m*cos(3*e + 3*f*x)))/(2*f*(sin(e + f*x) - 1)*(16*m + 4*m^2 + 15))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \cos ^{2}{\left (e + f x \right )}\, dx \]

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]

Integral((a*(sin(e + f*x) + 1))**m*sqrt(-c*(sin(e + f*x) - 1))*cos(e + f*x)**2, x)

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