\(\int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx\) [331]

Optimal result

Integrand size = 28, antiderivative size = 156 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\frac {15 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:

15/64*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/ 
2)/a/c^(5/2)/f+15/32*cos(f*x+e)/a/c/f/(c-c*sin(f*x+e))^(3/2)+1/4*sec(f*x+e 
)/a/c/f/(c-c*sin(f*x+e))^(3/2)-5/8*sec(f*x+e)/a/c^2/f/(c-c*sin(f*x+e))^(1/ 
2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\frac {1}{128}+\frac {i}{128}\right ) \cos (e+f x) \left (-60 \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(1-i) (-9+15 \cos (2 (e+f x))+40 \sin (e+f x))\right )}{a c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2)),x]
 

Output:

((1/128 + I/128)*Cos[e + f*x]*(-60*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4 
)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e 
+ f*x)/2] + Sin[(e + f*x)/2]) + (1 - I)*(-9 + 15*Cos[2*(e + f*x)] + 40*Sin 
[e + f*x])))/(a*c^2*f*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])*Sqrt[c - c* 
Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 3215, 3042, 3160, 3042, 3166, 3042, 3129, 3042, 3128, 219}

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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3215

\(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}}dx}{a c}\)

\(\Big \downarrow \) 3160

\(\displaystyle \frac {\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5 \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}}{a c}\)

Input:

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2)),x]
 

Output:

(Sec[e + f*x]/(4*f*(c - c*Sin[e + f*x])^(3/2)) + (5*(-(Sec[e + f*x]/(f*Sqr 
t[c - c*Sin[e + f*x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sq 
rt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[e + f*x]/(2*f*(c - c* 
Sin[e + f*x])^(3/2))))/2))/(8*c))/(a*c)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3129

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c 
+ d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n 
+ 1))   Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] 
&& EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3160

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^m/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1))   Int[ 
(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, 
f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & 
& IntegersQ[2*m, 2*p]
 

rule 3166

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. 
)*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S 
qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1)))   Int[(g 
*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e 
, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {30 c^{\frac {5}{2}} \sin \left (f x +e \right )^{2}-15 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right )^{2} c^{2}-40 c^{\frac {5}{2}} \sin \left (f x +e \right )+30 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c^{2}-6 c^{\frac {5}{2}}-15 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2}}{64 c^{\frac {9}{2}} a \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(210\)

Input:

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

Output:

1/64*(30*c^(5/2)*sin(f*x+e)^2-15*(c*(1+sin(f*x+e)))^(1/2)*arctanh(1/2*(c*( 
1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)^2*c^2-40*c^(5/2)* 
sin(f*x+e)+30*(c*(1+sin(f*x+e)))^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2 
)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c^2-6*c^(5/2)-15*(c*(1+sin(f*x+e)))^ 
(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2)/c 
^(9/2)/a/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, \cos \left (f x + e\right )^{2} + 20 \, \sin \left (f x + e\right ) - 12\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{128 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

1/128*(15*sqrt(2)*(cos(f*x + e)^3 + 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f* 
x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + 
c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f 
*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*si 
n(f*x + e) - cos(f*x + e) - 2)) - 4*(15*cos(f*x + e)^2 + 20*sin(f*x + e) - 
 12)*sqrt(-c*sin(f*x + e) + c))/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f* 
x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))
 

Sympy [F]

\[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )} - c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Timed out. \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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