\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\) [13]

Optimal result

Integrand size = 38, antiderivative size = 45 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2920, 2817} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(111\) vs. \(2(45)=90\).

Time = 7.00 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.47 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2} (-6 \cos (2 (e+f x))+15 \sin (e+f x)-\sin (3 (e+f x)))}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2)*(-6*Cos[2*(e + f*x)] + 15*Sin[e + f*x] - S
in[3*(e + f*x)]))/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(120\) vs. \(2(39)=78\).

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.69

method	result	size
default	\(-\frac {\left (1+\cos \left (f x +e \right )\right ) \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\cos ^{3}\left (f x +e \right )-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )+2 \left (\cos ^{2}\left (f x +e \right )\right )-\sin \left (f x +e \right )-4 \cos \left (f x +e \right )+1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a}{3 f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(121\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {{\left (3 \, a \cos \left (f x + e\right )^{2} + {\left (a \cos \left (f x + e\right )^{2} - 4 \, a\right )} \sin \left (f x + e\right ) - 3 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {8 \, a^{\frac {3}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{3 \, \sqrt {c} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (6\,\cos \left (e+f\,x\right )+6\,\cos \left (3\,e+3\,f\,x\right )-14\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\right )}{12\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(6*cos(e + f*x) + 6*cos(3*e + 3*f*x) - 14*sin(2
*e + 2*f*x) + sin(4*e + 4*f*x)))/(12*c*f*(cos(2*e + 2*f*x) + 1))