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

Optimal result

Integrand size = 28, antiderivative size = 73 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2815, 2753, 2752} \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a^2 c^4 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 5.94 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 a^2 c \sec (e+f x) (1+\sin (e+f x))^3 (-9+5 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{35 f} \]

[In]

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

[Out]

(-2*a^2*c*Sec[e + f*x]*(1 + Sin[e + f*x])^3*(-9 + 5*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(35*f)

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (\sin \left (f x +e \right )+1\right )^{3} a^{2} \left (5 \sin \left (f x +e \right )-9\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(61\)
parts	\(\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (\sin \left (f x +e \right )+1\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (\sin \left (f x +e \right )+1\right ) \left (15 \left (\sin ^{3}\left (f x +e \right )\right )-39 \left (\sin ^{2}\left (f x +e \right )\right )+52 \sin \left (f x +e \right )-104\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {4 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (\sin \left (f x +e \right )+1\right ) \left (\sin ^{2}\left (f x +e \right )-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(202\)

[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*(sin(f*x+e)+1)^3*a^2*(5*sin(f*x+e)-9)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (65) = 130\).

Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.08 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (5 \, a^{2} c \cos \left (f x + e\right )^{4} - a^{2} c \cos \left (f x + e\right )^{3} + 2 \, a^{2} c \cos \left (f x + e\right )^{2} - 8 \, a^{2} c \cos \left (f x + e\right ) - 16 \, a^{2} c - {\left (5 \, a^{2} c \cos \left (f x + e\right )^{3} + 6 \, a^{2} c \cos \left (f x + e\right )^{2} + 8 \, a^{2} c \cos \left (f x + e\right ) + 16 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{35 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[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)

Sympy [F]

\[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=a^{2} \left (\int c \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx\right ) \]

[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))

Maxima [F]

\[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (65) = 130\).

Time = 0.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.86 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (105 \, a^{2} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 35 \, a^{2} c \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 7 \, a^{2} c \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, a^{2} c \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{140 \, f} \]

[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

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]

[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)