\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\) [23]

Optimal result

Integrand size = 38, antiderivative size = 45 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/4*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.23 (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} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*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))^{7/2} \sqrt {c-c \sin (e+f x)} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 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. \(119\) vs. \(2(45)=90\).

Time = 7.41 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.64 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2} (-28 \cos (2 (e+f x))+\cos (4 (e+f x))+56 \sin (e+f x)-8 \sin (3 (e+f x)))}{32 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/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]))^(5/2)*(-28*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] +
56*Sin[e + f*x] - 8*Sin[3*(e + f*x)]))/(32*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sqrt[c - c*Sin[e + f*x]]
)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(150\) vs. \(2(39)=78\).

Time = 0.17 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.36

method	result	size
default	\(-\frac {\left (\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\left (\cos ^{4}\left (f x +e \right )\right )+3 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (\cos ^{3}\left (f x +e \right )\right )-7 \cos \left (f x +e \right ) \sin \left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right )-\sin \left (f x +e \right )-8 \cos \left (f x +e \right )+1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (1+\cos \left (f x +e \right )\right )}{4 f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(151\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f*(cos(f*x+e)^3*sin(f*x+e)-cos(f*x+e)^4+3*cos(f*x+e)^2*sin(f*x+e)+4*cos(f*x+e)^3-7*cos(f*x+e)*sin(f*x+e)+
4*cos(f*x+e)^2-sin(f*x+e)-8*cos(f*x+e)+1)*(a*(1+sin(f*x+e)))^(1/2)*a^2*(1+cos(f*x+e))/(cos(f*x+e)+sin(f*x+e)+1
)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (39) = 78\).

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.18 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {{\left (a^{2} \cos \left (f x + e\right )^{4} - 8 \, a^{2} \cos \left (f x + e\right )^{2} + 7 \, a^{2} - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, c f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/4*(a^2*cos(f*x + e)^4 - 8*a^2*cos(f*x + e)^2 + 7*a^2 - 4*(a^2*cos(f*x + e)^2 - 2*a^2)*sin(f*x + e))*sqrt(a*s
in(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c*f*cos(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {4 \, a^{\frac {5}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {c} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-4*a^(5/2)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*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 = 2.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (28\,\cos \left (e+f\,x\right )+27\,\cos \left (3\,e+3\,f\,x\right )-\cos \left (5\,e+5\,f\,x\right )-48\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{32\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

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

[Out]

-(a^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(28*cos(e + f*x) + 27*cos(3*e + 3*f*x) - cos(
5*e + 5*f*x) - 48*sin(2*e + 2*f*x) + 8*sin(4*e + 4*f*x)))/(32*c*f*(cos(2*e + 2*f*x) + 1))