\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx\) [121]

Optimal result

Integrand size = 38, antiderivative size = 175 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(5 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}+\frac {(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3046, 2934, 2766, 2729, 2728, 212} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(5 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

((5*A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) + ((5
*A + B)*Cos[e + f*x])/(8*a^2*f*(c - c*Sin[e + f*x])^(3/2)) - ((5*A + B)*Sec[e + f*x])/(6*a^2*c*f*Sqrt[c - c*Si
n[e + f*x]]) - ((A - B)*Sec[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*c^2*f)

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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

Rule 2729

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] + Dist[(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 2766

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)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(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 2934

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] + Dist[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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx}{a^2 c^2} \\ & = -\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {(5 A+B) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{6 a^2 c} \\ & = -\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {(5 A+B) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac {(5 A+B) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 a^2 c} \\ & = \frac {(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f}-\frac {(5 A+B) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{8 a^2 c f} \\ & = \frac {(5 A+B) \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 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.37 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-12 A \cos ^2(e+f x)+4 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+3 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(3+3 i) \sqrt [4]{-1} (5 A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+6 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{24 a^2 f (1+\sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-12*A*Cos[e + f*x]^2 + 4*(-A + B
)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + 3*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^3 - (3 + 3*I)*(-1)^(1/4)*(5*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])^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 6*(A + B)*Sin[(e + f*x)/2]*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^3))/(24*a^2*f*(1 + Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(3/2))

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {15 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c +30 A \,c^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+3 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c +6 B \,c^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right )-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c A -20 A \,c^{\frac {5}{2}} \sin \left (f x +e \right )-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c B -4 B \,c^{\frac {5}{2}} \sin \left (f x +e \right )-4 A \,c^{\frac {5}{2}}-20 B \,c^{\frac {5}{2}}}{48 c^{\frac {7}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(280\)

[In]

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

[Out]

-1/48/c^(7/2)/a^2*(15*A*(c+c*sin(f*x+e))^(3/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*sin
(f*x+e)*c+30*A*c^(5/2)*cos(f*x+e)^2+3*B*(c+c*sin(f*x+e))^(3/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1
/2)/c^(1/2))*sin(f*x+e)*c+6*B*c^(5/2)*cos(f*x+e)^2-15*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/
2))*(c+c*sin(f*x+e))^(3/2)*c*A-20*A*c^(5/2)*sin(f*x+e)-3*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^
(1/2))*(c+c*sin(f*x+e))^(3/2)*c*B-4*B*c^(5/2)*sin(f*x+e)-4*A*c^(5/2)-20*B*c^(5/2))/(1+sin(f*x+e))/cos(f*x+e)/(
c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (5 \, A + B\right )} \sqrt {c} \cos \left (f x + e\right )^{3} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (3 \, {\left (5 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A + B\right )} \sin \left (f x + e\right ) - 2 \, A - 10 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \]

[In]

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

[Out]

1/96*(3*sqrt(2)*(5*A + B)*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*(3*(5*A + B)*cos(f*x + e)^2 - 2*(5*A +
B)*sin(f*x + e) - 2*A - 10*B)*sqrt(-c*sin(f*x + e) + c))/(a^2*c^2*f*cos(f*x + e)^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (152) = 304\).

Time = 0.45 (sec) , antiderivative size = 499, normalized size of antiderivative = 2.85 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {6 \, \sqrt {2} {\left (5 \, A \sqrt {c} + B \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a^{2} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3 \, \sqrt {2} {\left (\frac {A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}}{a^{2} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {3 \, \sqrt {2} {\left (A \sqrt {c} + B \sqrt {c} - \frac {10 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {2 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{a^{2} c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {16 \, \sqrt {2} {\left (7 \, A \sqrt {c} - B \sqrt {c} + \frac {12 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {9 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {3 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{a^{2} c^{2} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{192 \, f} \]

[In]

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

[Out]

1/192*(6*sqrt(2)*(5*A*sqrt(c) + B*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x +
1/2*e) + 1))/(a^2*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2
*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi +
1/2*f*x + 1/2*e) + 1))/(a^2*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 3*sqrt(2)*(A*sqrt(c) + B*sqrt(c) - 10*A
*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 2*B*sqrt(c)*(cos(-1/4*pi
+ 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*c^2*(c
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 16*sqrt(2)*(7*A*sqrt(c) - B*sqrt(c)
+ 12*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 9*A*sqrt(c)*(cos(-1
/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 3*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x +
1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^2*c^2*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4
*pi + 1/2*f*x + 1/2*e) + 1) + 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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