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

Optimal. Leaf size=77 \[ -\frac{2^{m+\frac{3}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},-m-\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{c f} \]

[Out]

-((2^(3/2 + m)*Cos[e + f*x]*Hypergeometric2F1[1/2, -1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1
/2 - m)*(a + a*Sin[e + f*x])^m)/(c*f))

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Rubi [A]  time = 0.164882, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {2840, 2652, 2651} \[ -\frac{2^{m+\frac{3}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},-m-\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(3/2 + m)*Cos[e + f*x]*Hypergeometric2F1[1/2, -1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1
/2 - m)*(a + a*Sin[e + f*x])^m)/(c*f))

Rule 2840

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

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx &=\frac{\int (a+a \sin (e+f x))^{1+m} \, dx}{a c}\\ &=\frac{\left ((1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^{1+m} \, dx}{c}\\ &=-\frac{2^{\frac{3}{2}+m} \cos (e+f x) \, _2F_1\left (\frac{1}{2},-\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m}{c f}\\ \end{align*}

Mathematica [C]  time = 19.9326, size = 6442, normalized size = 83.66 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 1.348, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{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)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c}\,{d x} \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)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c}, x\right ) \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)),x, algorithm="fricas")

[Out]

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

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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)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c}\,{d x} \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)),x, algorithm="giac")

[Out]

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