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(f*x+e)*hypergeom([1/2, -1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e
))^m/c/f

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Rubi [A]  time = 0.16, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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]

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 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])

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*}

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Mathematica [C]  time = 19.68, 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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c \sin \left (f x + e\right ) - c}, x\right ) \]

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

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)),x, algorithm="giac")

[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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maple [F]  time = 4.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{c -c \sin \left (f x +e \right )}\, dx \]

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

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)),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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{c-c\,\sin \left (e+f\,x\right )} \,d x \]

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)),x)

[Out]

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

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

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]

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

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