3.3.100 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx\) [300]

Optimal. Leaf size=73 \[ \frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

8/35*a^2*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+2/7*a^2*c^3*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(3/2)

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Rubi [A]
time = 0.13, 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 = {2815, 2753, 2752} \begin {gather*} \frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*c^4*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) + (2*a^2*c^3*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e +
f*x])^(3/2))

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2753

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 2815

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))^2 (c-c \sin (e+f x))^{3/2} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}+\frac {1}{7} \left (4 a^2 c^3\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 84, normalized size = 1.15 \begin {gather*} -\frac {2 a^2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-9+5 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{35 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-9 + 5*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(35*f*(Cos[(
e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]
time = 1.94, size = 61, normalized size = 0.84

method result size
default \(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{3} a^{2} \left (5 \sin \left (f x +e \right )-9\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(sin(f*x+e)-1)*c^2*(1+sin(f*x+e))^3*a^2*(5*sin(f*x+e)-9)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (69) = 138\).
time = 0.35, size = 163, normalized size = 2.23 \begin {gather*} -\frac {2 \, {\left (5 \, a^{2} c \cos \left (f x + e\right )^{4} - a^{2} c \cos \left (f x + e\right )^{3} + 2 \, a^{2} c \cos \left (f x + e\right )^{2} - 8 \, a^{2} c \cos \left (f x + e\right ) - 16 \, a^{2} c - {\left (5 \, a^{2} c \cos \left (f x + e\right )^{3} + 6 \, a^{2} c \cos \left (f x + e\right )^{2} + 8 \, a^{2} c \cos \left (f x + e\right ) + 16 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{35 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(5*a^2*c*cos(f*x + e)^4 - a^2*c*cos(f*x + e)^3 + 2*a^2*c*cos(f*x + e)^2 - 8*a^2*c*cos(f*x + e) - 16*a^2*
c - (5*a^2*c*cos(f*x + e)^3 + 6*a^2*c*cos(f*x + e)^2 + 8*a^2*c*cos(f*x + e) + 16*a^2*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int c \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(c*sqrt(-c*sin(e + f*x) + c), x) + Integral(c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integ
ral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3,
x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (69) = 138\).
time = 0.57, size = 144, normalized size = 1.97 \begin {gather*} -\frac {\sqrt {2} {\left (105 \, a^{2} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 35 \, a^{2} c \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 7 \, a^{2} c \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, a^{2} c \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{140 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/140*sqrt(2)*(105*a^2*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 35*a^2*c*cos(-3
/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 7*a^2*c*cos(-5/4*pi + 5/2*f*x + 5/2*e)*sgn(sin(
-1/4*pi + 1/2*f*x + 1/2*e)) - 5*a^2*c*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt
(c)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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