3.22 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=92 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/10*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f/(c-c*sin(f*x+e))^(1/2)+1/5*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c
*sin(f*x+e))^(1/2)/a/f

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Rubi [A]  time = 0.40, 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} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*a*f*Sqrt[c - c*Sin[e + f*x]]) + (Cos[e + f*x]*(a + a*Sin[e + f
*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(5*a*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) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx &=\frac {\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {2 \int (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx}{5 a}\\ &=\frac {c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{5 a f}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 92, normalized size = 1.00 \[ -\frac {a^2 \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)} (-70 \sin (e+f x)-5 \sin (3 (e+f x))+\sin (5 (e+f x))+20 \cos (2 (e+f x))+5 \cos (4 (e+f x)))}{80 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/80*(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(20*Cos[2*(e + f*x)] + 5*Cos[4*(e
+ f*x)] - 70*Sin[e + f*x] - 5*Sin[3*(e + f*x)] + Sin[5*(e + f*x)]))/f

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fricas [A]  time = 0.46, size = 96, normalized size = 1.04 \[ -\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{4} - 5 \, a^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/10*(5*a^2*cos(f*x + e)^4 - 5*a^2 + 2*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 - 4*a^2)*sin(f*x + e))*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*(a+a*sin(f*x+e))^(5/2)*(c-c*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)Unable to check sign: (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)*(-112*a^2*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))/(16*f)^2-288*a^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*
sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(3*f*x+3*exp(1))/(96*f)^2+160*a^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*s
ign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(5*f*x+5*exp(1))/(160*f)^2+16*a^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*si
gn(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(2*f*x+2*exp(1))/(16*f)^2+32*a^2*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))/(32*f)^2+16*a^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(c
os(1/2*(f*x+exp(1))-1/4*pi))*cos(-2*f*x-2*exp(1))/(-16*f)^2)

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maple [A]  time = 0.41, size = 106, normalized size = 1.15 \[ -\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (-2 \left (\cos ^{6}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (\cos ^{4}\left (f x +e \right )\right )+3 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+6 \sin \left (f x +e \right )-6\right )}{10 f \cos \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10/f*(-c*(sin(f*x+e)-1))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(-2*cos(f*x+e)^6+sin(f*x+e)*cos(f*x+e)^4
-2*cos(f*x+e)^4+3*cos(f*x+e)^2*sin(f*x+e)+6*sin(f*x+e)-6)/cos(f*x+e)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \cos \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e)^2, x)

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mupad [B]  time = 2.41, size = 108, normalized size = 1.17 \[ -\frac {a^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (20\,\cos \left (e+f\,x\right )+25\,\cos \left (3\,e+3\,f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )-75\,\sin \left (2\,e+2\,f\,x\right )-4\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )\right )}{80\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(20*cos(e + f*x) + 25*cos(3*e + 3*f*x) + 5*co
s(5*e + 5*f*x) - 75*sin(2*e + 2*f*x) - 4*sin(4*e + 4*f*x) + sin(6*e + 6*f*x)))/(80*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*(a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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