\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [190]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 40, antiderivative size = 149 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {B c^2 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a+a \sin (e+f x))^{5/2}} \] Output:

-B*c^2*cos(f*x+e)*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f 
*x+e))^(1/2)-B*c*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(3 
/2)-1/4*(A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 11.75 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (B \cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-B \left (2+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+\left (A-3 B-4 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + 
 f*x])^(5/2),x]
 

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(B*Cos[2 
*(e + f*x)]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - B*(2 + 3*Log[Cos[(e 
 + f*x)/2] + Sin[(e + f*x)/2]]) + (A - 3*B - 4*B*Log[Cos[(e + f*x)/2] + Si 
n[(e + f*x)/2]])*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*( 
a*(1 + Sin[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 3451, 3042, 3218, 3042, 3216, 3042, 3146, 16}

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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3451

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

\(\displaystyle \frac {B \left (-\frac {c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + f*x]) 
^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3146
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3216
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x 
]]*Sqrt[c + d*Sin[e + f*x]]))   Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], 
x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
]
 

rule 3218
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 
2*n + 1)))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b 
^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && GtQ[2*m + 
n + 1, 0])
 

rule 3451
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(A*b - a*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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(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, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2* 
m + 1, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(135)=270\).

Time = 7.42 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.85

method result size
parts \(-\frac {A \sqrt {4}\, \left (5 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-3\right ) c \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}}{64 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}+\frac {2 B \left (\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c}{f \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a^{2} \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) \(425\)
default \(-\frac {A \sqrt {4}\, \left (5 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-3\right ) c \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}}{64 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}-\frac {2 B \left (\left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c}{f \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a^{2} \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) \(427\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

-1/64*A/f*4^(1/2)*(5*cos(1/4*Pi+1/2*f*x+1/2*e)^4+6*cos(1/4*Pi+1/2*f*x+1/2* 
e)^2-3)*c*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(a*sin(1/4*Pi+1/2*f*x+1/2* 
e)^2)^(1/2)/a^2*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)^3+2*B/ 
f*((4*cos(1/2*f*x+1/2*e)^4-4*cos(1/2*f*x+1/2*e)^2-4*sin(1/2*f*x+1/2*e)*cos 
(1/2*f*x+1/2*e)-1)*ln(-2*(cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1/2*e))/(cos(1/2* 
f*x+1/2*e)+1))+(-4*cos(1/2*f*x+1/2*e)^4+4*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1 
/2*e)+4*cos(1/2*f*x+1/2*e)^2+1)*ln(2/(cos(1/2*f*x+1/2*e)+1))+cos(1/2*f*x+1 
/2*e)*(4*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2+sin(1/2*f*x+1/2*e)))*(-(2 
*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)*c)^(1/2)*c/((2*sin(1/2*f*x+1/2*e 
)*cos(1/2*f*x+1/2*e)+1)*a)^(1/2)/a^2/(4*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1 
/2*e)+2*cos(1/2*f*x+1/2*e)^2-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)
 

Fricas [F]

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

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x 
, algorithm="fricas")
 

Output:

integral(-(B*c*cos(f*x + e)^2 - (A - B)*c*sin(f*x + e) + (A - B)*c)*sqrt(a 
*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 
 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)
 

Sympy [F]

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

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(5/2) 
,x)
 

Output:

Integral((-c*(sin(e + f*x) - 1))**(3/2)*(A + B*sin(e + f*x))/(a*(sin(e + f 
*x) + 1))**(5/2), x)
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x 
, algorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) 
 + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*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. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x)) 
^(5/2),x)
 

Output:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x)) 
^(5/2), x)
 

Reduce [F]

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

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x)
 

Output:

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