3.74 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac{64 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7) \left (4 m^2+16 m+15\right ) \sqrt{c-c \sin (e+f x)}}+\frac{16 c \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+24 m+35\right )}+\frac{2 \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7)} \]

[Out]

(64*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(7 + 2*m)*(15 + 16*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]
) + (16*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c - c*Sin[e + f*x]])/(a*f*(35 + 24*m + 4*m^2)) + (2*C
os[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(3/2))/(a*f*(7 + 2*m))

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Rubi [A]  time = 0.467299, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2841, 2740, 2738} \[ \frac{64 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7) \left (4 m^2+16 m+15\right ) \sqrt{c-c \sin (e+f x)}}+\frac{16 c \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+24 m+35\right )}+\frac{2 \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(7 + 2*m)*(15 + 16*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]
) + (16*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c - c*Sin[e + f*x]])/(a*f*(35 + 24*m + 4*m^2)) + (2*C
os[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(3/2))/(a*f*(7 + 2*m))

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]

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 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)]

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}+\frac{8 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2} \, dx}{a (7+2 m)}\\ &=\frac{16 c \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)}}{a f \left (35+24 m+4 m^2\right )}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}+\frac{(32 c) \int (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)} \, dx}{a \left (35+24 m+4 m^2\right )}\\ &=\frac{64 c^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \left (35+24 m+4 m^2\right ) \sqrt{c-c \sin (e+f x)}}+\frac{16 c \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt{c-c \sin (e+f x)}}{a f \left (35+24 m+4 m^2\right )}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}\\ \end{align*}

Mathematica [A]  time = 3.11294, size = 149, normalized size = 0.87 \[ -\frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 (a (\sin (e+f x)+1))^m \left (4 \left (4 m^2+24 m+27\right ) \sin (e+f x)+\left (4 m^2+16 m+15\right ) \cos (2 (e+f x))-12 m^2-80 m-157\right )}{f (2 m+3) (2 m+5) (2 m+7) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(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]]*(-157 - 80*m -
12*m^2 + (15 + 16*m + 4*m^2)*Cos[2*(e + f*x)] + 4*(27 + 24*m + 4*m^2)*Sin[e + f*x]))/(f*(3 + 2*m)*(5 + 2*m)*(7
+ 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))

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Maple [F]  time = 0.349, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, 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))^(3/2),x)

[Out]

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

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Maxima [B]  time = 2.30023, size = 571, normalized size = 3.32 \begin{align*} -\frac{2 \,{\left ({\left (4 \, m^{2} + 32 \, m + 71\right )} a^{m} c^{\frac{3}{2}} - \frac{{\left (4 \, m^{2} - 105\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{{\left (12 \, m^{2} + 64 \, m - 91\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (12 \, m^{2} + 32 \, m + 245\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{{\left (12 \, m^{2} + 32 \, m + 245\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{{\left (12 \, m^{2} + 64 \, m - 91\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{{\left (4 \, m^{2} - 105\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{{\left (4 \, m^{2} + 32 \, m + 71\right )} a^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}\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 (8 \, m^{3} + 60 \, m^{2} + 142 \, m + \frac{2 \,{\left (8 \, m^{3} + 60 \, m^{2} + 142 \, m + 105\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (8 \, m^{3} + 60 \, m^{2} + 142 \, m + 105\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 105\right )} f{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{3}{2}}} \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))^(3/2),x, algorithm="maxima")

[Out]

-2*((4*m^2 + 32*m + 71)*a^m*c^(3/2) - (4*m^2 - 105)*a^m*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) - (12*m^2 + 64
*m - 91)*a^m*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (12*m^2 + 32*m + 245)*a^m*c^(3/2)*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + (12*m^2 + 32*m + 245)*a^m*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (12*m^2 + 64*m -
91)*a^m*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - (4*m^2 - 105)*a^m*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) +
1)^6 + (4*m^2 + 32*m + 71)*a^m*c^(3/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*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))/((8*m^3 + 60*m^2 + 142*m + 2*(8*m^3 + 60*m^2
+ 142*m + 105)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (8*m^3 + 60*m^2 + 142*m + 105)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 105)*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))

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Fricas [A]  time = 1.87571, size = 622, normalized size = 3.62 \begin{align*} \frac{2 \,{\left ({\left (4 \, c m^{2} + 16 \, c m + 15 \, c\right )} \cos \left (f x + e\right )^{4} +{\left (4 \, c m^{2} + 32 \, c m + 39 \, c\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (2 \, c m - c\right )} \cos \left (f x + e\right )^{2} + 32 \, c \cos \left (f x + e\right ) -{\left ({\left (4 \, c m^{2} + 16 \, c m + 15 \, c\right )} \cos \left (f x + e\right )^{3} - 8 \,{\left (2 \, c m + 3 \, c\right )} \cos \left (f x + e\right )^{2} - 32 \, c \cos \left (f x + e\right ) - 64 \, c\right )} \sin \left (f x + e\right ) + 64 \, c\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m +{\left (8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f\right )} \cos \left (f x + e\right ) -{\left (8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f\right )} \sin \left (f x + e\right ) + 105 \, f} \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))^(3/2),x, algorithm="fricas")

[Out]

2*((4*c*m^2 + 16*c*m + 15*c)*cos(f*x + e)^4 + (4*c*m^2 + 32*c*m + 39*c)*cos(f*x + e)^3 + 8*(2*c*m - c)*cos(f*x
+ e)^2 + 32*c*cos(f*x + e) - ((4*c*m^2 + 16*c*m + 15*c)*cos(f*x + e)^3 - 8*(2*c*m + 3*c)*cos(f*x + e)^2 - 32*
c*cos(f*x + e) - 64*c)*sin(f*x + e) + 64*c)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*m^3 + 60*f*m
^2 + 142*f*m + (8*f*m^3 + 60*f*m^2 + 142*f*m + 105*f)*cos(f*x + e) - (8*f*m^3 + 60*f*m^2 + 142*f*m + 105*f)*si
n(f*x + e) + 105*f)

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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))**(3/2),x)

[Out]

Timed out

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Giac [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))^(3/2),x, algorithm="giac")

[Out]

Timed out