\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) [28]

Optimal result

Integrand size = 38, antiderivative size = 48 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

[Out]

1/8*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c/f/(c-c*sin(f*x+e))^(9/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 48, 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, 2821} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

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

Rule 2821

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*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -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 \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(117\) vs. \(2(48)=96\).

Time = 6.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.44 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} (-7 \sin (e+f x)+\sin (3 (e+f x)))}{4 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(-7*Sin[e + f*x] + Sin[3*(e + f*x)]))/
(4*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(42)=84\).

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.88

method	result	size
default	\(-\frac {\left (\cos ^{2}\left (f x +e \right )-2\right ) a^{2} \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\)	\(90\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*(cos(f*x+e)^2-2)*a^2*(a*(1+sin(f*x+e)))^(1/2)/(cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2-4*sin(f*x+e)+4)/(-c
*(sin(f*x+e)-1))^(1/2)/c^5*tan(f*x+e)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (42) = 84\).

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.65 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{c^{6} f \cos \left (f x + e\right )^{5} - 8 \, c^{6} f \cos \left (f x + e\right )^{3} + 8 \, c^{6} f \cos \left (f x + e\right ) + 4 \, {\left (c^{6} f \cos \left (f x + e\right )^{3} - 2 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-(a^2*cos(f*x + e)^2 - 2*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(c^6*f*cos(f*x +
e)^5 - 8*c^6*f*cos(f*x + e)^3 + 8*c^6*f*cos(f*x + e) + 4*(c^6*f*cos(f*x + e)^3 - 2*c^6*f*cos(f*x + e))*sin(f*
x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (42) = 84\).

Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 3.48 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (4 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 6 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 4 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{8 \, c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{11/2}} \,d x \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(11/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(11/2), x)