\(\int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx\) [311]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 37, antiderivative size = 136 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {d} \sqrt {c+d} f} \]

[Out]

-(A-B)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/(c-d)/f/a^(1/2)-2*(-A*d+B*c)*arc
tanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/(c-d)/f/a^(1/2)/d^(1/2)/(c+d)^(1/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3064, 2728, 212, 2852, 214} \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}-\frac {2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} \sqrt {d} f (c-d) \sqrt {c+d}} \]

[In]

Int[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])),x]

[Out]

-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*f)) -
(2*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d
)*Sqrt[d]*Sqrt[c + d]*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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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 2852

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

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{c-d}+\frac {(B c-A d) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a (c-d)} \\ & = -\frac {(2 (A-B)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{(c-d) f}-\frac {(2 (B c-A d)) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{(c-d) f} \\ & = -\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {d} \sqrt {c+d} f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 3.09 (sec) , antiderivative size = 619, normalized size of antiderivative = 4.55 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {(-1)^{3/4} \left ((4+4 i) (A-B) \sqrt {d} \sqrt {c+d} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+\sqrt [4]{-1} (B c-A d) \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]-\sqrt [4]{-1} (B c-A d) \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 (c-d) \sqrt {d} \sqrt {c+d} f \sqrt {a (1+\sin (e+f x))}} \]

[In]

Integrate[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])),x]

[Out]

((-1)^(3/4)*((4 + 4*I)*(A - B)*Sqrt[d]*Sqrt[c + d]*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])] + (
-1)^(1/4)*(B*c - A*d)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]])
+ Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d]*L
og[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e +
f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] - (-1)^(1/4)*(B*c - A*
d)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d]*Sqrt[c +
d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f
*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*L
og[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(
2*(c - d)*Sqrt[d]*Sqrt[c + d]*f*Sqrt[a*(1 + Sin[e + f*x])])

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.46

method result size
default \(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a \left (c +d \right ) d}\, A -2 A \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sqrt {a}\, d -\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a \left (c +d \right ) d}\, B +2 B \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sqrt {a}\, c \right )}{\left (c -d \right ) \sqrt {a \left (c +d \right ) d}\, \sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) \(199\)

[In]

int((A+B*sin(f*x+e))/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(
c+d)*d)^(1/2)*A-2*A*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(1/2)*d-2^(1/2)*arctanh(1/2*(-a*(
sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*B+2*B*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^
(1/2))*a^(1/2)*c)/(c-d)/(a*(c+d)*d)^(1/2)/a^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (113) = 226\).

Time = 0.84 (sec) , antiderivative size = 744, normalized size of antiderivative = 5.47 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\left [\frac {\sqrt {a c d + a d^{2}} {\left (B c - A d\right )} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, \sqrt {a c d + a d^{2}} {\left (d \cos \left (f x + e\right )^{2} - {\left (c + 2 \, d\right )} \cos \left (f x + e\right ) + {\left (d \cos \left (f x + e\right ) + c + 3 \, d\right )} \sin \left (f x + e\right ) - c - 3 \, d\right )} \sqrt {a \sin \left (f x + e\right ) + a} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) + \frac {\sqrt {2} {\left ({\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\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 )}{\sqrt {a}}}{2 \, {\left (a c^{2} d - a d^{3}\right )} f}, -\frac {2 \, \sqrt {-a c d - a d^{2}} {\left (B c - A d\right )} \arctan \left (\frac {\sqrt {-a c d - a d^{2}} \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )}}{2 \, {\left (a c d + a d^{2}\right )} \cos \left (f x + e\right )}\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\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 )}{\sqrt {a}}}{2 \, {\left (a c^{2} d - a d^{3}\right )} f}\right ] \]

[In]

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

[Out]

[1/2*(sqrt(a*c*d + a*d^2)*(B*c - A*d)*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2
)*cos(f*x + e)^2 - 4*sqrt(a*c*d + a*d^2)*(d*cos(f*x + e)^2 - (c + 2*d)*cos(f*x + e) + (d*cos(f*x + e) + c + 3*
d)*sin(f*x + e) - c - 3*d)*sqrt(a*sin(f*x + e) + a) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*
x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (
2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f
*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + sqrt(2)*((A - B)*a*c*d + (A - B)*a*d^2)*log(-(cos(f*x + e)^2 - (
cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a)
+ 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a))/((a*c^2*
d - a*d^3)*f), -1/2*(2*sqrt(-a*c*d - a*d^2)*(B*c - A*d)*arctan(1/2*sqrt(-a*c*d - a*d^2)*sqrt(a*sin(f*x + e) +
a)*(d*sin(f*x + e) - c - 2*d)/((a*c*d + a*d^2)*cos(f*x + e))) - sqrt(2)*((A - B)*a*c*d + (A - B)*a*d^2)*log(-(
cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x
+ e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))
/sqrt(a))/((a*c^2*d - a*d^3)*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int { \frac {B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}} \,d x } \]

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (113) = 226\).

Time = 0.33 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.85 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\frac {2 \, \sqrt {2} {\left (B \sqrt {a} c - A \sqrt {a} d\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (\sqrt {2} a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \sqrt {2} a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {-c d - d^{2}}} - \frac {{\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {2} a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \sqrt {2} a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {{\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {2} a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \sqrt {2} a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{f} \]

[In]

integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-(2*sqrt(2)*(B*sqrt(a)*c - A*sqrt(a)*d)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((sq
rt(2)*a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(-c*d - d
^2)) - (A*sqrt(a) - B*sqrt(a))*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(2)*a*c*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)) - sqrt(2)*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (A*sqrt(a) - B*sqrt(a))*log(-sin(-1/4*pi + 1/2*
f*x + 1/2*e) + 1)/(sqrt(2)*a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1
/2*e))))/f

Mupad [F(-1)]

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

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))),x)

[Out]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))), x)