Integrand size = 39, antiderivative size = 544 \[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {2 b (A b-a B) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (A \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right )-B \left (a^2 c d-b^2 c d+a b \left (c^2-d^2\right )\right )\right ) E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{\sqrt {a+b} (c-d) \sqrt {c+d} (b c-a d)^3 f}+\frac {2 (A b c+b B c-a A d-2 A b d+a B d) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sec (e+f x) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sin (e+f x))}{(a-b) (c+d \sin (e+f x))}} (c+d \sin (e+f x))}{\sqrt {a+b} (c-d) \sqrt {c+d} (b c-a d)^2 f} \]
-2*(A*(a^2*d^2+b^2*(c^2-2*d^2))-B*(a^2*c*d-b^2*c*d+a*b*(c^2-d^2)))*Ellipti cE((c+d)^(1/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),( (a+b)*(c-d)/(a-b)/(c+d))^(1/2))*sec(f*x+e)*(c+d*sin(f*x+e))*((-a*d+b*c)*(1 -sin(f*x+e))/(a+b)/(c+d*sin(f*x+e)))^(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a- b)/(c+d*sin(f*x+e)))^(1/2)/(c-d)/(-a*d+b*c)^3/f/(a+b)^(1/2)/(c+d)^(1/2)+2* (-A*a*d+A*b*c-2*A*b*d+B*a*d+B*b*c)*EllipticF((c+d)^(1/2)*(a+b*sin(f*x+e))^ (1/2)/(a+b)^(1/2)/(c+d*sin(f*x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d))^(1/2))* sec(f*x+e)*(c+d*sin(f*x+e))*((-a*d+b*c)*(1-sin(f*x+e))/(a+b)/(c+d*sin(f*x+ e)))^(1/2)*(-(-a*d+b*c)*(1+sin(f*x+e))/(a-b)/(c+d*sin(f*x+e)))^(1/2)/(c-d) /(-a*d+b*c)^2/f/(a+b)^(1/2)/(c+d)^(1/2)+2*b*(A*b-B*a)*cos(f*x+e)/(a^2-b^2) /(-a*d+b*c)/f/(a+b*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2266\) vs. \(2(544)=1088\).
Time = 7.24 (sec) , antiderivative size = 2266, normalized size of antiderivative = 4.17 \[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]*((2*(A*b^3*Cos[e + f*x] - a*b^2*B*Cos[e + f*x]))/((a^2 - b^2)*(-(b*c) + a*d)^2*(a + b*Sin[e + f*x ])) - (2*(B*c*d^2*Cos[e + f*x] - A*d^3*Cos[e + f*x]))/((b*c - a*d)^2*(c^2 - d^2)*(c + d*Sin[e + f*x]))))/f + ((-4*(-(b*c) + a*d)*(a*A*b^2*c^3 - b^3* B*c^3 - 2*a^2*A*b*c^2*d + 2*A*b^3*c^2*d + a^3*A*c*d^2 - 2*a*A*b^2*c*d^2 + b^3*B*c*d^2 + 2*a^2*A*b*d^3 - 2*A*b^3*d^3 - a^3*B*d^3 + a*b^2*B*d^3)*Sqrt[ ((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcSin[Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + a*d)]/Sqrt [2]], (2*(-(b*c) + a*d))/((a + b)*(-c + d))]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + d)*Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x]))/ (-(b*c) + a*d)]*Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f *x]))/(-(b*c) + a*d)])/((a + b)*(c + d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - 4*(-(b*c) + a*d)*(A*b^3*c^3 - a*b^2*B*c^3 + a*A*b^2*c^2 *d - 2*a^2*b*B*c^2*d + b^3*B*c^2*d + a^2*A*b*c*d^2 - 2*A*b^3*c*d^2 - a^3*B *c*d^2 + 2*a*b^2*B*c*d^2 + a^3*A*d^3 - 2*a*A*b^2*d^3 + a^2*b*B*d^3)*((Sqrt [((c + d)*Cot[(-e + Pi/2 - f*x)/2]^2)/(-c + d)]*EllipticF[ArcSin[Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e + f*x]))/(-(b*c) + a*d)]/Sqr t[2]], (2*(-(b*c) + a*d))/((a + b)*(-c + d))]*Sec[e + f*x]*Sin[(-e + Pi/2 - f*x)/2]^4*Sqrt[((c + d)*Csc[(-e + Pi/2 - f*x)/2]^2*(a + b*Sin[e + f*x])) /(-(b*c) + a*d)]*Sqrt[((-a - b)*Csc[(-e + Pi/2 - f*x)/2]^2*(c + d*Sin[e...
Time = 1.73 (sec) , antiderivative size = 578, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {3042, 3479, 27, 3042, 3477, 3042, 3297, 3475}
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.
\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {2 \int \frac {A d a^2-(A b c-b B d) a+b^2 (B c-2 A d)-(A b-a B) (b c+a d) \sin (e+f x)}{2 \sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\int \frac {A d a^2-(A b c-b B d) a+b^2 (B c-2 A d)-(A b-a B) (b c+a d) \sin (e+f x)}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\int \frac {A d a^2-(A b c-b B d) a+b^2 (B c-2 A d)-(A b-a B) (b c+a d) \sin (e+f x)}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\frac {\left (a^2 (-A) d^2+a^2 B c d+a b B \left (c^2-d^2\right )-A b^2 \left (c^2-2 d^2\right )-b^2 B c d\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {(a-b) (-a A d+a B d+A b c-2 A b d+b B c) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\frac {\left (a^2 (-A) d^2+a^2 B c d+a b B \left (c^2-d^2\right )-A b^2 \left (c^2-2 d^2\right )-b^2 B c d\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {(a-b) (-a A d+a B d+A b c-2 A b d+b B c) \int \frac {1}{\sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}dx}{c-d}}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3297 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\frac {\left (a^2 (-A) d^2+a^2 B c d+a b B \left (c^2-d^2\right )-A b^2 \left (c^2-2 d^2\right )-b^2 B c d\right ) \int \frac {\sin (e+f x)+1}{\sqrt {a+b \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}dx}{c-d}-\frac {2 (a-b) \sqrt {a+b} \sec (e+f x) (-a A d+a B d+A b c-2 A b d+b B c) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}}{\left (a^2-b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3475 |
\(\displaystyle \frac {2 b (A b-a B) \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) \sqrt {a+b \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {-\frac {2 (a-b) \sqrt {a+b} \sec (e+f x) \left (a^2 (-A) d^2+a^2 B c d+a b B \left (c^2-d^2\right )-A b^2 \left (c^2-2 d^2\right )-b^2 B c d\right ) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} E\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)^2}-\frac {2 (a-b) \sqrt {a+b} \sec (e+f x) (-a A d+a B d+A b c-2 A b d+b B c) (c+d \sin (e+f x)) \sqrt {\frac {(b c-a d) (1-\sin (e+f x))}{(a+b) (c+d \sin (e+f x))}} \sqrt {-\frac {(b c-a d) (\sin (e+f x)+1)}{(a-b) (c+d \sin (e+f x))}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c+d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sin (e+f x)}}\right ),\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{f (c-d) \sqrt {c+d} (b c-a d)}}{\left (a^2-b^2\right ) (b c-a d)}\) |
(2*b*(A*b - a*B)*Cos[e + f*x])/((a^2 - b^2)*(b*c - a*d)*f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) - ((-2*(a - b)*Sqrt[a + b]*(a^2*B*c*d - b^2*B*c*d - a^2*A*d^2 - A*b^2*(c^2 - 2*d^2) + a*b*B*(c^2 - d^2))*Elliptic E[ArcSin[(Sqrt[c + d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Si n[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b* c - a*d)*(1 - Sin[e + f*x]))/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + f*x]))/((c - d)*Sqrt[c + d]*(b*c - a*d)^2*f) - (2*(a - b)*Sqrt[a + b]*(A* b*c + b*B*c - a*A*d - 2*A*b*d + a*B*d)*EllipticF[ArcSin[(Sqrt[c + d]*Sqrt[ a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sin[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*Sec[e + f*x]*Sqrt[((b*c - a*d)*(1 - Sin[e + f*x]) )/((a + b)*(c + d*Sin[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sin[e + f*x]))/ ((a - b)*(c + d*Sin[e + f*x])))]*(c + d*Sin[e + f*x]))/((c - d)*Sqrt[c + d ]*(b*c - a*d)*f))/((a^2 - b^2)*(b*c - a*d))
3.4.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ .) + (f_.)*(x_)]]), x_Symbol] :> Simp[2*((c + d*Sin[e + f*x])/(f*(b*c - a*d )*Rt[(c + d)/(a + b), 2]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 - Sin[e + f*x] )/((a + b)*(c + d*Sin[e + f*x])))]*Sqrt[(-(b*c - a*d))*((1 + Sin[e + f*x])/ ((a - b)*(c + d*Sin[e + f*x])))]*EllipticF[ArcSin[Rt[(c + d)/(a + b), 2]*(S qrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]])], (a + b)*((c - d)/((a - b)*(c + d)))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && N eQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/(a + b)]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_) + (b_.)*sin[(e_.) + (f_.) *(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[-2*A*(c - d)*((a + b*Sin[e + f*x])/(f*(b*c - a*d)^2*Rt[(a + b)/(c + d), 2 ]*Cos[e + f*x]))*Sqrt[(b*c - a*d)*((1 + Sin[e + f*x])/((c - d)*(a + b*Sin[e + f*x])))]*Sqrt[(-(b*c - a*d))*((1 - Sin[e + f*x])/((c + d)*(a + b*Sin[e + f*x])))]*EllipticE[ArcSin[Rt[(a + b)/(c + d), 2]*(Sqrt[c + d*Sin[e + f*x]] /Sqrt[a + b*Sin[e + f*x]])], (a - b)*((c + d)/((a + b)*(c - d)))], x] /; Fr eeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(a + b)/(c + d)]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
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*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Leaf count of result is larger than twice the leaf count of optimal. \(174461\) vs. \(2(510)=1020\).
Time = 18.71 (sec) , antiderivative size = 174462, normalized size of antiderivative = 320.70
method | result | size |
parts | \(\text {Expression too large to display}\) | \(174462\) |
default | \(\text {Expression too large to display}\) | \(177559\) |
\[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {B \sin \left (f x + e\right ) + A}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))/(a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(3/2),x , algorithm="fricas")
integral((B*sin(f*x + e) + A)*sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/(b^2*d^2*cos(f*x + e)^4 + 4*a*b*c*d + (a^2 + b^2)*c^2 + (a^2 + b^2)* d^2 - (b^2*c^2 + 4*a*b*c*d + (a^2 + 2*b^2)*d^2)*cos(f*x + e)^2 + 2*(a*b*c^ 2 + a*b*d^2 + (a^2 + b^2)*c*d - (b^2*c*d + a*b*d^2)*cos(f*x + e)^2)*sin(f* x + e)), x)
\[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {A + B \sin {\left (e + f x \right )}}{\left (a + b \sin {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {B \sin \left (f x + e\right ) + A}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))/(a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(3/2),x , algorithm="maxima")
\[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {B \sin \left (f x + e\right ) + A}{{\left (b \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e))/(a+b*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(3/2),x , algorithm="giac")
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+b \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]