\(\int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx\) [338]

Optimal result

Integrand size = 28, antiderivative size = 155 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}+\frac {5 \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f} \] Output:

5/16*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^2/c^(3/2)/f+5/8*cos(f*x+e)/a^2/f/(c-c*sin(f*x+e))^(3/2)-5/6*sec(f*x+e) 
/a^2/c/f/(c-c*sin(f*x+e))^(1/2)-1/3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/a^ 
2/c^2/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.73 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\frac {1}{96}+\frac {i}{96}\right ) \cos (e+f x) \left (60 \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1-i) (11+15 \cos (2 (e+f x))-20 \sin (e+f x))\right )}{a^2 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

((1/96 + I/96)*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])^2*(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2])^3 + (1 - I)*(11 + 15*Cos[2*(e + f*x)] - 20*Sin[ 
e + f*x])))/(a^2*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^2*Sqrt[c - c*S 
in[e + f*x]])
 

Rubi [A] (verified)

Time = 0.70 (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, 3154, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^4}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3154

\(\displaystyle \frac {\frac {5}{6} c \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} c \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

Input:

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

(-1/3*(Sec[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/f + (5*c*(-(Sec[e + f*x]/( 
f*Sqrt[c - c*Sin[e + f*x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[ 
2]*Sqrt[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))/6)/(a^2*c^2)
 

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 3154

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*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1)))   Int[(g* 
Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, 
e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ 
[m + 1/2, 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 = 0.46 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {-30 c^{\frac {5}{2}} \cos \left (f x +e \right )^{2}-15 \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c +20 c^{\frac {5}{2}} \sin \left (f x +e \right )+15 \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c +4 c^{\frac {5}{2}}}{48 c^{\frac {7}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(157\)

Input:

int(1/(a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*(-30*c^(5/2)*cos(f*x+e)^2-15*(c+c*sin(f*x+e))^(3/2)*arctanh(1/2*(c+c* 
sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c+20*c^(5/2)*sin(f*x 
+e)+15*(c+c*sin(f*x+e))^(3/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c 
^(1/2))*2^(1/2)*c+4*c^(5/2))/c^(7/2)/a^2/(1+sin(f*x+e))/cos(f*x+e)/(c-c*si 
n(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {15 \, \sqrt {2} \sqrt {c} \cos \left (f x + e\right )^{3} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, \cos \left (f x + e\right )^{2} - 10 \, \sin \left (f x + e\right ) - 2\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \] Input:

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas 
")
 

Output:

1/96*(15*sqrt(2)*sqrt(c)*cos(f*x + e)^3*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)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(15*cos(f*x + 
e)^2 - 10*sin(f*x + e) - 2)*sqrt(-c*sin(f*x + e) + c))/(a^2*c^2*f*cos(f*x 
+ e)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima 
")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

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

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

Output:

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