\(\int \frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a+b \sin (e+f x)} \, dx\) [1370]

Optimal result

Integrand size = 31, antiderivative size = 355 \[ \int \frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {\sqrt {g} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\sqrt {b} \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {\sqrt {g} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt {b} \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt [4]{-a^2+b^2} f}-\frac {g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \] Output:

g^(1/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f-b^(1/2)*g^(1/2)*arctan(b^ 
(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(1/4)/f- 
g^(1/2)*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f+b^(1/2)*g^(1/2)*arctanh( 
b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(1/4)/ 
f-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-(-a^2+b^2)^(1/2))/f/(g*cos(f*x+e))^(1/2)-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+(-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 = 25.50 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {\sqrt {g \cos (e+f x)} \csc (e+f x) \left (8 a b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \left (2 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 a^2 \arctan \left (\sqrt {\cos (e+f x)}\right )-4 b^2 \arctan \left (\sqrt {\cos (e+f x)}\right )+2 a^2 \log \left (1-\sqrt {\cos (e+f x)}\right )-2 b^2 \log \left (1-\sqrt {\cos (e+f x)}\right )-2 a^2 \log \left (1+\sqrt {\cos (e+f x)}\right )+2 b^2 \log \left (1+\sqrt {\cos (e+f x)}\right )-\sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 a \left (a^2-b^2\right ) f \sqrt {\cos (e+f x)} (b+a \csc (e+f x))} \] Input:

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

Output:

(Sqrt[g*Cos[e + f*x]]*Csc[e + f*x]*(8*a*b*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*(2*Sq 
rt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f 
*x]])/(a^2 - b^2)^(1/4)] - 2*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + 
(Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4*a^2*ArcTan[Sqr 
t[Cos[e + f*x]]] - 4*b^2*ArcTan[Sqrt[Cos[e + f*x]]] + 2*a^2*Log[1 - Sqrt[C 
os[e + f*x]]] - 2*b^2*Log[1 - Sqrt[Cos[e + f*x]]] - 2*a^2*Log[1 + Sqrt[Cos 
[e + f*x]]] + 2*b^2*Log[1 + Sqrt[Cos[e + f*x]]] - Sqrt[2]*Sqrt[b]*(a^2 - b 
^2)^(3/4)*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]] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*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]))/(12*a*(a^2 - b^2)*f*Sqrt[Cos[e + 
 f*x]]*(b + a*Csc[e + f*x]))
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3042, 3377, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3377

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {\sqrt {g}\, \ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}+\sqrt {g}\, \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}+2 g \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{2 a \sqrt {-g}\, f}\)	\(185\)

Input:

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

Output:

-1/2/a/(-g)^(1/2)*(g^(1/2)*ln(-2/(cos(1/2*f*x+1/2*e)+1)*(2*cos(1/2*f*x+1/2 
*e)*g-g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+g))*(-g)^(1/2)+g^(1/2)*l 
n(2/(cos(1/2*f*x+1/2*e)-1)*(2*cos(1/2*f*x+1/2*e)*g+g^(1/2)*(-2*g*sin(1/2*f 
*x+1/2*e)^2+g)^(1/2)-g))*(-g)^(1/2)+2*g*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2 
)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-g)))/f
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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