3.349 \(\int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=89 \[ -\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2740, 2738} \[ -\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx &=-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 f}+\frac {1}{2} a \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 137, normalized size = 1.54 \[ \frac {c^2 (\sin (e+f x)-1)^2 (a (\sin (e+f x)+1))^{3/2} \sqrt {c-c \sin (e+f x)} (8 (9 \sin (e+f x)+\sin (3 (e+f x)))+12 \cos (2 (e+f x))+3 \cos (4 (e+f x)))}{96 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

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fricas [A]  time = 0.45, size = 87, normalized size = 0.98 \[ \frac {{\left (3 \, a c^{2} \cos \left (f x + e\right )^{4} - 3 \, a c^{2} + 4 \, {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(3*a*c^2*cos(f*x + e)^4 - 3*a*c^2 + 4*(a*c^2*cos(f*x + e)^2 + 2*a*c^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((a+a*sin(f*x+e))^(3/2)*(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)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*a*c^2*f*sign(si
n(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*a*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))/(24*f)^2-32*a*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))*cos(2*f*x+2*exp(1))/(32*f)^2-64*a*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))*cos(4*f*x+4*exp(1))/(64*f)^2-32*a*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))*cos(-2*f*x-2*exp(1))/(-32*f)^2)

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maple [A]  time = 0.29, size = 90, normalized size = 1.01 \[ \frac {\left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+5 \sin \left (f x +e \right )+5\right )}{12 f \cos \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2), x)

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mupad [B]  time = 1.75, size = 100, normalized size = 1.12 \[ \frac {a\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )+80\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(5/2),x)

[Out]

(a*c^2*(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*c
os(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((a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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