\(\int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx\) [63]

Optimal result

Integrand size = 25, antiderivative size = 302 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx=-\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2} d}+\frac {a e \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a e \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}} \] Output:

-e^(1/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^( 
1/2)/(-a^2+b^2)^(1/4)/d+e^(1/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2 
+b^2)^(1/4)/e^(1/2))/b^(1/2)/(-a^2+b^2)^(1/4)/d-a*e*EllipticPi(cos(1/2*c+1 
/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b/(b-(-a 
^2+b^2)^(1/2))/d/(e*sin(d*x+c))^(1/2)-a*e*EllipticPi(cos(1/2*c+1/4*Pi+1/2* 
d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b/(b+(-a^2+b^2)^(1 
/2))/d/(e*sin(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 3.24 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {2 \cos (c+d x) \left (a+b \sqrt {\cos ^2(c+d x)}\right ) \sqrt {e \sin (c+d x)} \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}\right )}{d \sqrt {\cos ^2(c+d x)} (a+b \cos (c+d x)) \sqrt {\sin (c+d x)}} \] Input:

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

Output:

(2*Cos[c + d*x]*(a + b*Sqrt[Cos[c + d*x]^2])*Sqrt[e*Sin[c + d*x]]*(((1/8 + 
 I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4 
)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] 
 - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + 
d*x]] + I*b*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + 
 b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2) 
^(1/4)) + (a*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^ 
2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))/(3*(a^2 - b^2))))/(d*Sqrt[Cos[c + d*x 
]^2]*(a + b*Cos[c + d*x])*Sqrt[Sin[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 1.08 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3180

\(\displaystyle -\frac {b e \int \frac {\sqrt {e \sin (c+d x)}}{b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2}d(e \sin (c+d x))}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {2 b e \int \frac {e^2 \sin ^2(c+d x)}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {2 b e \left (\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\)

\(\Big \downarrow \) 3286

\(\displaystyle -\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 b \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\)

Input:

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

Output:

(-2*b*e*(ArcTan[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3 
/2)*(-a^2 + b^2)^(1/4)*Sqrt[e]) - ArcTanh[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/( 
-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e])))/d + (a*e*Ellip 
ticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d* 
x]])/(b*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*e*EllipticPi[( 
2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b 
*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]])
 

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 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]
 

Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.64

method	result	size
default	\(\frac {-\frac {e \sqrt {2}\, \left (\ln \left (\frac {e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+\frac {a e \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \left (\operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, -\frac {b}{-b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}-\operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+\operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, -\frac {b}{-b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b +\operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b \right )}{2 b \left (-b +\sqrt {-a^{2}+b^{2}}\right ) \left (b +\sqrt {-a^{2}+b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(496\)

Input:

int((e*sin(d*x+c))^(1/2)/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
 

Output:

(-1/4/b*e/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((e*sin(d*x+c)-(e^2*(a^2-b^ 
2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*s 
in(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2 
-b^2)/b^2)^(1/2)))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c 
))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2 
)-1))+1/2*a*e*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2) 
/b*(EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))* 
(-a^2+b^2)^(1/2)-EllipticPi((1-sin(d*x+c))^(1/2),b/(b+(-a^2+b^2)^(1/2)),1/ 
2*2^(1/2))*(-a^2+b^2)^(1/2)+EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b 
^2)^(1/2)),1/2*2^(1/2))*b+EllipticPi((1-sin(d*x+c))^(1/2),b/(b+(-a^2+b^2)^ 
(1/2)),1/2*2^(1/2))*b)/(-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+ 
c)/(e*sin(d*x+c))^(1/2))/d
 

Fricas [F]

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

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

Output:

integral(sqrt(e*sin(d*x + c))/(b*cos(d*x + c) + a), x)
 

Sympy [F]

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

integrate((e*sin(d*x+c))**(1/2)/(a+b*cos(d*x+c)),x)
 

Output:

Integral(sqrt(e*sin(c + d*x))/(a + b*cos(c + d*x)), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(e*sin(d*x + c))/(b*cos(d*x + c) + a), x)
 

Giac [F]

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

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

Output:

integrate(sqrt(e*sin(d*x + c))/(b*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(e)*int(sqrt(sin(c + d*x))/(cos(c + d*x)*b + a),x)