\(\int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx\) [1369]

Optimal result

Integrand size = 31, antiderivative size = 341 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \sqrt [4]{-a^2+b^2} f}+\frac {a \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \sqrt [4]{-a^2+b^2} f}+\frac {2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-a*g^(1/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b 
^(3/2)/(-a^2+b^2)^(1/4)/f+a*g^(1/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/( 
-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(1/4)/f+2*(g*cos(f*x+e))^(1/2) 
*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b/f/cos(f*x+e)^(1/2)-a^2*g*cos(f*x+ 
e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b 
^2/(b-(-a^2+b^2)^(1/2))/f/(g*cos(f*x+e))^(1/2)-a^2*g*cos(f*x+e)^(1/2)*Elli 
pticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/b^2/(b+(-a^2+b 
^2)^(1/2))/f/(g*cos(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 16.77 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {\sqrt {g \cos (e+f x)} \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) f \sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \] Input:

Integrate[(Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]
 

Output:

-1/12*(Sqrt[g*Cos[e + f*x]]*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[e + 
 f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2] 
*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a 
^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 
- b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sq 
rt[Cos[e + f*x]] + b*Cos[e + f*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b] 
*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))*(a + b*Sqrt[Sin[ 
e + f*x]^2]))/(b^(3/2)*(-a^2 + b^2)*f*Sqrt[Cos[e + f*x]]*(a + b*Sin[e + f* 
x]))
 

Rubi [A] (warning: unable to verify)

Time = 1.51 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3346, 3042, 3121, 3042, 3119, 3180, 266, 827, 218, 221, 3042, 3286, 3042, 3284}

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.

\(\displaystyle \int \frac {\sin (e+f x) \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x) \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3346

\(\displaystyle \frac {\int \sqrt {g \cos (e+f x)}dx}{b}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{b \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\)

\(\Big \downarrow \) 3180

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {b g \int \frac {\sqrt {g \cos (e+f x)}}{b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2}d(g \cos (e+f x))}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \int \frac {g^2 \cos ^2(e+f x)}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\)

Input:

Int[(Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]
 

Output:

(2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b*f*Sqrt[Cos[e + f*x]] 
) - (a*((2*b*g*(ArcTan[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/ 
(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[g]) - ArcTanh[(Sqrt[b]*Sqrt[g]*Cos[e + 
f*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[g])))/f + (a* 
g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 
 2])/(b*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (a*g*Sqrt[Cos[e + 
 f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(b + S 
qrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])))/b
 

Defintions of rubi rules used

rule 218

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

rule 221

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3180

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ 
)]), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[a*(g/(2*b))   Int[1/(S 
qrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Simp[a*(g/(2*b))   In 
t[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Simp[b*(g/f)   Su 
bst[Int[Sqrt[x]/(g^2*(a^2 - b^2) + b^2*x^2), x], x, g*Cos[e + f*x]], x])] / 
; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 

rule 3284

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

rule 3286

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

rule 3346

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.41 (sec) , antiderivative size = 802, normalized size of antiderivative = 2.35

method	result	size
default	\(\text {Expression too large to display}\)	\(802\)

Input:

int((g*cos(f*x+e))^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x,method=_RETURNVERBO 
SE)
 

Output:

(-1/4*g*a/b^2/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e 
)^2-g-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2) 
+(g^2*(a^2-b^2)/b^2)^(1/2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2 
)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/ 
2)))+2*arctan(2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2- 
g)^(1/2)+1)-2*arctan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1 
/2*e)^2-g)^(1/2)+1))-8*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2 
)^(1/2)*g*b*(1/2*(sin(1/2*f*x+1/2*e)^2-1)/b^2*(sin(1/2*f*x+1/2*e)^2)^(1/2) 
*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+ 
1/2*e)^2))^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))-1/32/b^4*sum((-2*si 
n(1/2*f*x+1/2*e)^2*_alpha^2*b^2+sin(1/2*f*x+1/2*e)^2*a^2+2*b^2*_alpha^2-a^ 
2)/_alpha/(2*_alpha^2-1)*(2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2) 
*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2*cos(1/2*f* 
x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha^2* 
b^2+a^2-2*b^2)/b^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2 
))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2* 
cos(1/2*f*x+1/2*e)^2)^(1/2)/(-g*sin(1/2*f*x+1/2*e)^2*(-1+2*sin(1/2*f*x+1/2 
*e)^2))^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2 
))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2)))/sin(1/2*f*x+1/2*e)/(g*(-1+2 
*cos(1/2*f*x+1/2*e)^2))^(1/2))/f
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="m 
axima")
 

Output:

integrate(sqrt(g*cos(f*x + e))*sin(f*x + e)/(b*sin(f*x + e) + a), x)
 

Giac [F]

\[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

integrate(sqrt(g*cos(f*x + e))*sin(f*x + e)/(b*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {\sin \left (e+f\,x\right )\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \] Input:

int((sin(e + f*x)*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)),x)
 

Output:

int((sin(e + f*x)*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)), x)
 

Reduce [F]

\[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right ) b +a}d x \right ) \] Input:

int((g*cos(f*x+e))^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)),x)
 

Output:

sqrt(g)*int((sqrt(cos(e + f*x))*sin(e + f*x))/(sin(e + f*x)*b + a),x)