\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) [18]

Optimal result

Integrand size = 38, antiderivative size = 97 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/8*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/c/f/(c-c*sin(f*x+e))^(9/2)+1/48*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/c^
2/f/(c-c*sin(f*x+e))^(7/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 97, 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, 2822, 2821} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*a*c*f*(c - c*Sin[e + f*x])^(9/2)) + (Cos[e + f*x]*(a + a*Sin[e +
f*x])^(5/2))/(48*a*c^2*f*(c - c*Sin[e + f*x])^(7/2))

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 2822

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] + Dist[(m + n + 1)/(a*(2*m
+ 1)), 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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 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))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}+\frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{8 a c^2} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.60 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} (5-3 \cos (2 (e+f x))+4 \sin (e+f x))}{12 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-1/12*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(5 - 3*Cos[2*(e + f*x)] + 4*Sin[e
+ f*x]))/(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 [A] (verified)

Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.23

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \left (\cos ^{3}\left (f x +e \right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-5 \cos \left (f x +e \right )-10 \tan \left (f x +e \right )+4 \sec \left (f x +e \right )\right )}{6 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\)	\(119\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/f*(a*(1+sin(f*x+e)))^(1/2)*a/(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*(cos(f*x+e)^3+4*cos(f*x+e)*sin(f*x+e)-5*cos(f*x+e)-10*tan(f*x+e)+4*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{6} f \cos \left (f x + e\right )^{5} - 8 \, c^{6} f \cos \left (f x + e\right )^{3} + 8 \, c^{6} f \cos \left (f x + e\right ) + 4 \, {\left (c^{6} f \cos \left (f x + e\right )^{3} - 2 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/6*(3*a*cos(f*x + e)^2 - 2*a*sin(f*x + e) - 4*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^6*f*c
os(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. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (6 \, a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 8 \, a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{48 \, c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

-1/48*(6*a*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)^4 - 8*a*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 + 3*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
)*sqrt(a)/(c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)

Mupad [B] (verification not implemented)

Time = 14.61 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.43 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {40\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^6\,f}-\frac {8\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^6\,f}+\frac {32\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^6\,f}\right )}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(11/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((40*a*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^(1/2))/(3*c^6*f) - (8*a*exp(e*5i +
f*x*5i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(c^6*f) + (32*a*exp(e*5i + f*x*5i)*sin(e + f*x)*(a + a*si
n(e + f*x))^(1/2))/(3*c^6*f)))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 54*exp(e*5i + f*x*5i)*cos(3*e + 3*f*x) +
2*exp(e*5i + f*x*5i)*cos(5*e + 5*f*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x) + 16*exp(e*5i + f*x*5i)*sin(4*e
+ 4*f*x))