\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\) [29]

Optimal result

Integrand size = 38, antiderivative size = 97 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 a c^2 f (c-c \sin (e+f x))^{9/2}} \] Output:

1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c/f/(c-c*sin(f*x+e))^(11/2)+1/80* 
cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c^2/f/(c-c*sin(f*x+e))^(9/2)
 

Mathematica [A] (verified)

Time = 9.75 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} (14-10 \cos (2 (e+f x))+35 \sin (e+f x)-5 \sin (3 (e+f x)))}{40 c^6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^6 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x]) 
^(13/2),x]
 

Output:

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(1 
4 - 10*Cos[2*(e + f*x)] + 35*Sin[e + f*x] - 5*Sin[3*(e + f*x)]))/(40*c^6*f 
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^6*Sqrt[c - c*Si 
n[e + f*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3320, 3042, 3222, 3042, 3221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^2(e+f x) (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (e+f x)^2 (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{11/2}}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{11/2}}dx}{a c}\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{9/2}}dx}{10 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{9/2}}dx}{10 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\)

\(\Big \downarrow \) 3221

\(\displaystyle \frac {\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\)

Input:

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(13/2 
),x]
 

Output:

((Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*(c - c*Sin[e + f*x])^(11/ 
2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(80*c*f*(c - c*Sin[e + f*x 
])^(9/2)))/(a*c)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3221

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] && Ne 
Q[m, -2^(-1)]
 

rule 3222

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] + Simp[(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] /; Free 
Q[{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 3320

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

Maple [F]

\[\int \frac {\cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {13}{2}}}d x\]

Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x)
 

Output:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=-\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 6 \, a^{2} + 5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, {\left (5 \, c^{7} f \cos \left (f x + e\right )^{5} - 20 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right ) - {\left (c^{7} f \cos \left (f x + e\right )^{5} - 12 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, a 
lgorithm="fricas")
 

Output:

-1/10*(5*a^2*cos(f*x + e)^2 - 6*a^2 + 5*(a^2*cos(f*x + e)^2 - 2*a^2)*sin(f 
*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^7*f*cos(f 
*x + e)^5 - 20*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e) - (c^7*f*cos(f 
*x + e)^5 - 12*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e))*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(13/2),x 
)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, a 
lgorithm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, a 
lgorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 24.44 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.27 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,112{}\mathrm {i}}{5\,c^7\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,56{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{c^7\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \] Input:

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(13/2 
),x)
 

Output:

((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^ 
(1/2)*112i)/(5*c^7*f) + (a^2*exp(e*6i + f*x*6i)*sin(e + f*x)*(a + a*sin(e 
+ f*x))^(1/2)*56i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a + 
 a*sin(e + f*x))^(1/2)*16i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*sin(3*e + 3* 
f*x)*(a + a*sin(e + f*x))^(1/2)*8i)/(c^7*f)))/(cos(e + f*x)*exp(e*6i + f*x 
*6i)*264i - exp(e*6i + f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)* 
cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i 
 + f*x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)
 

Reduce [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{7}-7 \sin \left (f x +e \right )^{6}+21 \sin \left (f x +e \right )^{5}-35 \sin \left (f x +e \right )^{4}+35 \sin \left (f x +e \right )^{3}-21 \sin \left (f x +e \right )^{2}+7 \sin \left (f x +e \right )-1}d x \right )-2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{7}-7 \sin \left (f x +e \right )^{6}+21 \sin \left (f x +e \right )^{5}-35 \sin \left (f x +e \right )^{4}+35 \sin \left (f x +e \right )^{3}-21 \sin \left (f x +e \right )^{2}+7 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{7}-7 \sin \left (f x +e \right )^{6}+21 \sin \left (f x +e \right )^{5}-35 \sin \left (f x +e \right )^{4}+35 \sin \left (f x +e \right )^{3}-21 \sin \left (f x +e \right )^{2}+7 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{7}} \] Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(13/2),x)
 

Output:

(sqrt(c)*sqrt(a)*a**2*( - int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) 
 + 1)*cos(e + f*x)**2*sin(e + f*x)**2)/(sin(e + f*x)**7 - 7*sin(e + f*x)** 
6 + 21*sin(e + f*x)**5 - 35*sin(e + f*x)**4 + 35*sin(e + f*x)**3 - 21*sin( 
e + f*x)**2 + 7*sin(e + f*x) - 1),x) - 2*int((sqrt(sin(e + f*x) + 1)*sqrt( 
 - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x))/(sin(e + f*x)**7 - 7*si 
n(e + f*x)**6 + 21*sin(e + f*x)**5 - 35*sin(e + f*x)**4 + 35*sin(e + f*x)* 
*3 - 21*sin(e + f*x)**2 + 7*sin(e + f*x) - 1),x) - int((sqrt(sin(e + f*x) 
+ 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2)/(sin(e + f*x)**7 - 7*sin(e 
 + f*x)**6 + 21*sin(e + f*x)**5 - 35*sin(e + f*x)**4 + 35*sin(e + f*x)**3 
- 21*sin(e + f*x)**2 + 7*sin(e + f*x) - 1),x)))/c**7