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

Optimal. Leaf size=188 \[ -\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 c f \sqrt {a \sin (e+f x)+a}}-\frac {4 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{35 c f}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f} \]

[Out]

-1/7*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/c/f-1/7*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c
*sin(f*x+e))^(7/2)/c/f-2/35*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/c/f/(a+a*sin(f*x+e))^(1/2)-4/35*a^2*cos(f*x+
e)*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2)/c/f

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Rubi [A]  time = 0.62, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac {4 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{35 c f}-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(35*c*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a^2*Cos[e + f*x]*Sqrt[
a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(35*c*f) - (a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*S
in[e + f*x])^(7/2))/(7*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2))/(7*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) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}+\frac {6 \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx}{7 c}\\ &=-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}+\frac {(4 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx}{7 c}\\ &=-\frac {4 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 c f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}+\frac {\left (8 a^2\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{35 c}\\ &=-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 c f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 c f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 c f}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 87, normalized size = 0.46 \[ \frac {a^2 c^2 (1225 \sin (e+f x)+245 \sin (3 (e+f x))+49 \sin (5 (e+f x))+5 \sin (7 (e+f x))) \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)}}{2240 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(1225*Sin[e + f*x] + 245*Sin[3*(e +
f*x)] + 49*Sin[5*(e + f*x)] + 5*Sin[7*(e + f*x)]))/(2240*f)

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fricas [A]  time = 0.49, size = 101, normalized size = 0.54 \[ \frac {{\left (5 \, a^{2} c^{2} \cos \left (f x + e\right )^{6} + 6 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 16 \, a^{2} c^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{35 \, f \cos \left (f x + e\right )} \]

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))^(5/2),x, algorithm="fricas")

[Out]

1/35*(5*a^2*c^2*cos(f*x + e)^6 + 6*a^2*c^2*cos(f*x + e)^4 + 8*a^2*c^2*cos(f*x + e)^2 + 16*a^2*c^2)*sqrt(a*sin(
f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(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))^(5/2),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)sqrt(2*a)*sqrt(2*c
)*(-4480*a^2*c^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))/(128*
f)^2-8064*a^2*c^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))/
(384*f)^2-4480*a^2*c^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(5*f*x+5*exp
(1))/(640*f)^2-896*a^2*c^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(7*f*x+7
*exp(1))/(896*f)^2)

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maple [A]  time = 0.37, size = 77, normalized size = 0.41 \[ \frac {\left (5 \left (\cos ^{6}\left (f x +e \right )\right )+6 \left (\cos ^{4}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right )+16\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{35 f \cos \left (f x +e \right )^{5}} \]

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))^(5/2),x)

[Out]

1/35/f*(5*cos(f*x+e)^6+6*cos(f*x+e)^4+8*cos(f*x+e)^2+16)*(-c*(sin(f*x+e)-1))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)
))^(5/2)/cos(f*x+e)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 11.80, size = 179, normalized size = 0.95 \[ \frac {\frac {1225\,a^2\,c^2\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {245\,a^2\,c^2\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {49\,a^2\,c^2\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {5\,a^2\,c^2\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}}{70\,f\,\cos \left (e+f\,x\right )} \]

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))^(5/2),x)

[Out]

((1225*a^2*c^2*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (245*a^2*c^2*sin(3*e +
3*f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (49*a^2*c^2*sin(5*e + 5*f*x)*(a + a*sin(e
+ f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (5*a^2*c^2*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*s
in(e + f*x))^(1/2))/32)/(70*f*cos(e + f*x))

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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))**(5/2),x)

[Out]

Timed out

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