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

Optimal result

Integrand size = 38, antiderivative size = 92 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 c f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 c f}-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

-1/6*(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin
[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(4*c*f)

Rule 2817

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 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-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] &&
EqQ[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 2920

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

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {c \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (12 \cos (2 (e+f x))+3 \cos (4 (e+f x))+8 (9 \sin (e+f x)+\sin (3 (e+f x))))}{96 f} \]

[In]

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

[Out]

(c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*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)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c \left (3 \left (\cos ^{3}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )+8 \tan \left (f x +e \right )-3 \sec \left (f x +e \right )\right )}{12 f}\)	\(73\)

[In]

int(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12/f*(a*(1+sin(f*x+e)))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)*c*(3*cos(f*x+e)^3+4*cos(f*x+e)*sin(f*x+e)+8*tan(f*x+
e)-3*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {{\left (3 \, c \cos \left (f x + e\right )^{4} + 4 \, {\left (c \cos \left (f x + e\right )^{2} + 2 \, c\right )} \sin \left (f x + e\right ) - 3 \, c\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*c*cos(f*x + e)^4 + 4*(c*cos(f*x + e)^2 + 2*c)*sin(f*x + e) - 3*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin
(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F]

\[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}} \cos ^{2}{\left (e + f x \right )}\, dx \]

[In]

integrate(cos(f*x+e)**2*(c-c*sin(f*x+e))**(3/2)*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(-c*(sin(e + f*x) - 1))**(3/2)*cos(e + f*x)**2, x)

Maxima [F]

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

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {4 \, {\left (3 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 4 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-4/3*(3*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*
e)^8 - 4*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2
*e)^6)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 1.87 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )+80\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(3/2),x)

[Out]

(c*(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*cos(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))