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

Optimal. Leaf size=50 \[ \frac {2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

[Out]

2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(3+2*m)/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2841, 2738} \[ \frac {2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

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 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 \frac {\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} \sqrt {c-c \sin (e+f x)} \, dx}{a c}\\ &=\frac {2 \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*}

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Mathematica [A]  time = 0.37, size = 85, normalized size = 1.70 \[ \frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (\sin (e+f x)+1))^m}{f (2 m+3) \sqrt {c-c \sin (e+f x)}} \]

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

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fricas [B]  time = 0.46, size = 108, normalized size = 2.16 \[ -\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{2 \, c f m + 3 \, c f + {\left (2 \, c f m + 3 \, c f\right )} \cos \left (f x + e\right ) - {\left (2 \, c f m + 3 \, c f\right )} \sin \left (f x + e\right )} \]

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*(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x
+ e) + a)^m/(2*c*f*m + 3*c*f + (2*c*f*m + 3*c*f)*cos(f*x + e) - (2*c*f*m + 3*c*f)*sin(f*x + e))

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

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

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{\sqrt {c -c \sin \left (f x +e \right )}}\, dx \]

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 [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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mupad [B]  time = 0.90, size = 68, normalized size = 1.36 \[ -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (2\,\cos \left (e+f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\right )}{c\,f\,\left (2\,m+3\right )\,\left (\sin \left (e+f\,x\right )-1\right )} \]

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)*(2*cos(e + f*x) + sin(2*e + 2*f*x)))/(c*f*(2*m + 3)*(
sin(e + f*x) - 1))

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

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*cos(e + f*x)**2/sqrt(-c*(sin(e + f*x) - 1)), x)

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