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

Optimal. Leaf size=92 \[ -\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 c f}-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a \sin (e+f x)+a}} \]

[Out]

-1/6*a*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/c/f/(a+a*sin(f*x+e))^(1/2)-1/4*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)*(a+a
*sin(f*x+e))^(1/2)/c/f

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Rubi [A]  time = 0.39, antiderivative size = 92, 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, 2740, 2738} \[ -\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 c f}-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(6*c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin[e
+ f*x]]*(c - c*Sin[e + f*x])^(5/2))/(4*c*f)

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 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 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) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=-\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 c f}+\frac {\int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx}{2 c}\\ &=-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 c f}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 83, normalized size = 0.90 \[ \frac {c \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)} (8 (9 \sin (e+f x)+\sin (3 (e+f x)))+12 \cos (2 (e+f x))+3 \cos (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(12*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)]
+ 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/(96*f)

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fricas [A]  time = 0.43, size = 75, normalized size = 0.82 \[ \frac {{\left (3 \, c \cos \left (f x + e\right )^{4} + 4 \, {\left (c \cos \left (f x + e\right )^{2} + 2 \, c\right )} \sin \left (f x + e\right ) - 3 \, c\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*c*cos(f*x + e)^4 + 4*(c*cos(f*x + e)^2 + 2*c)*sin(f*x + e) - 3*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin
(f*x + e) + c)/(f*cos(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*a)*sqrt(2*c)*(-24*c*f*sign(sin(1/
2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(f*x+exp(1))/(8*f)^2-24*c*f*sign(sin(1/2*(f*x+ex
p(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(3*f*x+3*exp(1))/(24*f)^2-32*c*f*sign(sin(1/2*(f*x+exp(1)
)-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(2*f*x+2*exp(1))/(32*f)^2-64*c*f*sign(sin(1/2*(f*x+exp(1))-1/
4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(4*f*x+4*exp(1))/(64*f)^2-32*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi
))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-2*f*x-2*exp(1))/(-32*f)^2)

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maple [A]  time = 0.40, size = 90, normalized size = 0.98 \[ \frac {\left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sin \left (f x +e \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+5 \sin \left (f x +e \right )+5\right )}{12 f \cos \left (f x +e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/f*(-c*(sin(f*x+e)-1))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(1/2)*(3*cos(f*x+e)^4+cos(f*x+e)^2*sin(f*x+e)+4
*cos(f*x+e)^2+5*sin(f*x+e)+5)/cos(f*x+e)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.71, size = 97, normalized size = 1.05 \[ \frac {c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )+80\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(12*cos(e + f*x) + 15*cos(3*e + 3*f*x) + 3*cos(5
*e + 5*f*x) + 80*sin(2*e + 2*f*x) + 8*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))

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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*(c-c*sin(f*x+e))**(3/2)*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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