\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx\) [122]

Optimal result

Integrand size = 38, antiderivative size = 225 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {5 (7 A-B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}+\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:

5/128*(7*A-B)*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^(5/2)/f+5/64*(7*A-B)*cos(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^ 
(3/2)+1/24*(7*A-B)*sec(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)-5/48*(7*A-B)* 
sec(f*x+e)/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/3*(A-B)*sec(f*x+e)^3/a^2/c^2 
/f/(c-c*sin(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.20 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.91 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 (11 A+3 B) \cos ^3(e+f x)+16 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (-3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(15+15 i) \sqrt [4]{-1} (7 A-B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+6 (11 A+3 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{192 a^2 f (1+\sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \] Input:

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

Output:

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 
])*(3*(11*A + 3*B)*Cos[e + f*x]^3 + 16*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e 
 + f*x)/2])^4 + 24*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 12*(A + B)*(Cos[(e + f*x)/2] - Sin[( 
e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1 
/4)*(7*A - B)*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])^3 
+ 24*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 6* 
(11*A + 3*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^3))/(192*a^2*f*(1 + Sin[e + f*x])^2*(c - 
 c*Sin[e + f*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.90, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3446, 3042, 3334, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}}dx-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3160

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{6} c (7 A-B) \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {(A-B) \sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}}{a^2 c^2}\)

Input:

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

Output:

(-1/3*((A - B)*Sec[e + f*x]^3)/(f*Sqrt[c - c*Sin[e + f*x]]) + ((7*A - B)*c 
*(Sec[e + f*x]/(4*f*(c - c*Sin[e + f*x])^(3/2)) + (5*(-(Sec[e + f*x]/(f*Sq 
rt[c - c*Sin[e + f*x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*S 
qrt[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)))/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 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 3334

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

rule 3446

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(198)=396\).

Time = 0.37 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {-86 A \,c^{\frac {7}{2}}+122 B \,c^{\frac {7}{2}}+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \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}-15 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \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}+322 A \sin \left (f x +e \right ) c^{\frac {7}{2}}-46 B \sin \left (f x +e \right ) c^{\frac {7}{2}}-210 A \sin \left (f x +e \right )^{3} c^{\frac {7}{2}}+30 B \sin \left (f x +e \right )^{3} c^{\frac {7}{2}}+105 A \sin \left (f x +e \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, c^{2}-15 B \sin \left (f x +e \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, c^{2}-210 A \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, c^{2}+30 B \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, c^{2}+70 A \sin \left (f x +e \right )^{2} c^{\frac {7}{2}}-10 B \sin \left (f x +e \right )^{2} c^{\frac {7}{2}}}{384 c^{\frac {11}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(426\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-1/384/c^(11/2)/a^2*(-86*A*c^(7/2)+122*B*c^(7/2)+105*A*(c*(1+sin(f*x+e)))^ 
(3/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2-15 
*B*(c*(1+sin(f*x+e)))^(3/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c 
^(1/2))*2^(1/2)*c^2+322*A*sin(f*x+e)*c^(7/2)-46*B*sin(f*x+e)*c^(7/2)-210*A 
*sin(f*x+e)^3*c^(7/2)+30*B*sin(f*x+e)^3*c^(7/2)+105*A*sin(f*x+e)^2*arctanh 
(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(3/2)*2^ 
(1/2)*c^2-15*B*sin(f*x+e)^2*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c 
^(1/2))*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*c^2-210*A*sin(f*x+e)*arctanh(1/2* 
(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2) 
*c^2+30*B*sin(f*x+e)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2)) 
*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*c^2+70*A*sin(f*x+e)^2*c^(7/2)-10*B*sin(f 
*x+e)^2*c^(7/2))/(1+sin(f*x+e))/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e)) 
^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.24 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {15 \, \sqrt {2} {\left ({\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (5 \, {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} + 56 \, A - 8 \, B\right )} \sin \left (f x + e\right ) + 8 \, A - 56 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{768 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \] Input:

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

Output:

-1/768*(15*sqrt(2)*((7*A - B)*cos(f*x + e)^3*sin(f*x + e) - (7*A - B)*cos( 
f*x + e)^3)*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*c 
os(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*(5*(7*A - B)*cos(f*x + e)^2 - (15* 
(7*A - B)*cos(f*x + e)^2 + 56*A - 8*B)*sin(f*x + e) + 8*A - 56*B)*sqrt(-c* 
sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos( 
f*x + e)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(c)*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**5 - sin(e + f*x) 
**4 - 2*sin(e + f*x)**3 + 2*sin(e + f*x)**2 + sin(e + f*x) - 1),x)*a + int 
((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**5 - sin(e + f*x)* 
*4 - 2*sin(e + f*x)**3 + 2*sin(e + f*x)**2 + sin(e + f*x) - 1),x)*b))/(a** 
2*c**3)