3.82 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx\)

Optimal. Leaf size=114 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^2 f \left (4 m^2+16 m+15\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c f (2 m+5)} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2743, 2742} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{a c^2 f \left (4 m^2+16 m+15\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-3}}{a c f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-4 - m),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-3 - m))/(a*c*f*(5 + 2*m)) + (Cos[e + f*x]*(a
+ a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(a*c^2*f*(15 + 16*m + 4*m^2))

Rule 2742

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

Rule 2743

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), 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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 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 \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx &=\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m} \, dx}{a c^2 (5+2 m)}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^2 f (3+2 m) (5+2 m)}\\ \end {align*}

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Mathematica [A]  time = 12.61, size = 142, normalized size = 1.25 \[ -\frac {2^{-m-1} \cos ^3\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sin ^{-2 m-5}\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (\sin (e+f x)-2 (m+2)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-4} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-m-4)}}{f (2 m+3) (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-4 - m),x]

[Out]

-((2^(-1 - m)*Cos[(-e + Pi/2 - f*x)/2]^3*Sin[(-e + Pi/2 - f*x)/2]^(-5 - 2*m)*(-2*(2 + m) + Sin[e + f*x])*(a +
a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-4 - m))/(f*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
(2*(-4 - m))))

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fricas [A]  time = 0.46, size = 76, normalized size = 0.67 \[ \frac {{\left (2 \, {\left (m + 2\right )} \cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4}}{4 \, f m^{2} + 16 \, f m + 15 \, f} \]

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))^(-4-m),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2}\,{d x} \]

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))^(-4-m),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 4)*cos(f*x + e)^2, x)

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maple [F]  time = 5.40, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-4-m}\, dx \]

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))^(-4-m),x)

[Out]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2}\,{d x} \]

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))^(-4-m),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 4)*cos(f*x + e)^2, x)

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mupad [B]  time = 10.60, size = 177, normalized size = 1.55 \[ -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+48\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+16\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+12\,m\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+4\,m\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )-32\right )}{c^4\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+16\,m+15\right )\,\left (56\,{\sin \left (e+f\,x\right )}^2-56\,\sin \left (e+f\,x\right )-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+8\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]

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))^(m + 4),x)

[Out]

-((a*(sin(e + f*x) + 1))^m*(2*sin(2*e + 2*f*x) + sin(4*e + 4*f*x) + 48*sin(e/2 + (f*x)/2)^2 + 16*sin((3*e)/2 +
(3*f*x)/2)^2 + 12*m*(2*sin(e/2 + (f*x)/2)^2 - 1) + 4*m*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1) - 32))/(c^4*f*(-c*(
sin(e + f*x) - 1))^m*(16*m + 4*m^2 + 15)*(8*sin(3*e + 3*f*x) - 56*sin(e + f*x) - 2*sin(2*e + 2*f*x)^2 + 56*sin
(e + f*x)^2 + 8))

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sympy [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))**(-4-m),x)

[Out]

Timed out

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