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

Optimal. Leaf size=73 \[ \frac {8 a^3 c^5 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

8/63*a^3*c^5*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+2/9*a^3*c^4*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(5/2)

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Rubi [A]  time = 0.20, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac {2 a^3 c^4 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}+\frac {8 a^3 c^5 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^3*c^5*Cos[e + f*x]^7)/(63*f*(c - c*Sin[e + f*x])^(7/2)) + (2*a^3*c^4*Cos[e + f*x]^7)/(9*f*(c - c*Sin[e +
f*x])^(5/2))

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {2 a^3 c^4 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}+\frac {1}{9} \left (4 a^3 c^4\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac {8 a^3 c^5 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 3.09, size = 84, normalized size = 1.15 \[ -\frac {2 a^3 c (7 \sin (e+f x)-11) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^7}{63 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(-11 + 7*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(63*f*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2]))

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fricas [B]  time = 0.46, size = 179, normalized size = 2.45 \[ -\frac {2 \, {\left (7 \, a^{3} c \cos \left (f x + e\right )^{5} + 17 \, a^{3} c \cos \left (f x + e\right )^{4} - 2 \, a^{3} c \cos \left (f x + e\right )^{3} + 4 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 32 \, a^{3} c + {\left (7 \, a^{3} c \cos \left (f x + e\right )^{4} - 10 \, a^{3} c \cos \left (f x + e\right )^{3} - 12 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 32 \, a^{3} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{63 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(7*a^3*c*cos(f*x + e)^5 + 17*a^3*c*cos(f*x + e)^4 - 2*a^3*c*cos(f*x + e)^3 + 4*a^3*c*cos(f*x + e)^2 - 16
*a^3*c*cos(f*x + e) - 32*a^3*c + (7*a^3*c*cos(f*x + e)^4 - 10*a^3*c*cos(f*x + e)^3 - 12*a^3*c*cos(f*x + e)^2 -
16*a^3*c*cos(f*x + e) - 32*a^3*c)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) +
f)

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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*(c-c*sin(f*x+e))^(3/2),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*c)*(-2*a^3*c*sign(sin(1/2*(f*x+ex
p(1))-1/4*pi))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/f+8*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(2*f*x+2*
exp(1)+pi))/(4*f)^2-24*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(6*f*x+6*exp(1)-pi))/(12*f)^2+40*a^3
*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(10*f*x+10*exp(1)+pi))/(20*f)^2-56*a^3*c*f*sign(sin(1/2*(f*x+e
xp(1))-1/4*pi))*sin(1/4*(14*f*x+14*exp(1)-pi))/(28*f)^2+24*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*
(6*f*x+6*exp(1)+pi))/(12*f)^2-40*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(10*f*x+10*exp(1)-pi))/(20
*f)^2-224*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(14*f*x+14*exp(1)+pi))/(112*f)^2+288*a^3*c*f*sign
(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(18*f*x+18*exp(1)-pi))/(144*f)^2+48*a^3*c*f*sign(sin(1/2*(f*x+exp(1))-1
/4*pi))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/(8*f)^2)

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maple [A]  time = 0.83, size = 61, normalized size = 0.84 \[ \frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{4} a^{3} \left (7 \sin \left (f x +e \right )-11\right )}{63 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(sin(f*x+e)-1)*c^2*(1+sin(f*x+e))^4*a^3*(7*sin(f*x+e)-11)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int c \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int 2 c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- 2 c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(c*sqrt(-c*sin(e + f*x) + c), x) + Integral(2*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Int
egral(-2*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3, x) + Integral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)*
*4, x))

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