\(\int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) [1389]

Optimal result

Integrand size = 31, antiderivative size = 369 \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)+b^(3/2)*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)^(3/4)/f/g^(1/2) 
-arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)+b^(3/2)*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)^(3/4)/f/g^(1/ 
2)-b*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))/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/(g*cos(f*x+e))^(1/2)-b*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))/( 
a^2-b*(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 = 20.31 (sec) , antiderivative size = 698, normalized size of antiderivative = 1.89 \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {\cos (e+f x)} \left (-1+\cos ^2(e+f x)\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \csc (e+f x) \left (\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\sqrt {1-\cos ^2(e+f x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {-2 \sqrt {2} b^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt {2} b^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 \left (a^2-b^2\right )^{3/4} \arctan \left (\sqrt {\cos (e+f x)}\right )-2 \left (a^2-b^2\right )^{3/4} \log \left (1-\sqrt {\cos (e+f x)}\right )+2 \left (a^2-b^2\right )^{3/4} \log \left (1+\sqrt {\cos (e+f x)}\right )-\sqrt {2} b^{3/2} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\sqrt {2} b^{3/2} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )}{8 a \left (a^2-b^2\right )^{3/4}}\right )}{f \sqrt {g \cos (e+f x)} \left (1-\cos ^2(e+f x)\right ) (b+a \csc (e+f x))} \] Input:

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

Output:

(-2*Sqrt[Cos[e + f*x]]*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^ 
2])*Csc[e + f*x]*((5*b*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x] 
^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e 
 + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2* 
Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e 
+ f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 
3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f 
*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - (-2*Sqrt[2]*b^(3/2)*ArcTan[1 - 
 (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*Sqrt[2]*b^(3/ 
2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4* 
(a^2 - b^2)^(3/4)*ArcTan[Sqrt[Cos[e + f*x]]] - 2*(a^2 - b^2)^(3/4)*Log[1 - 
 Sqrt[Cos[e + f*x]]] + 2*(a^2 - b^2)^(3/4)*Log[1 + Sqrt[Cos[e + f*x]]] - S 
qrt[2]*b^(3/2)*Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqr 
t[Cos[e + f*x]] + b*Cos[e + f*x]] + Sqrt[2]*b^(3/2)*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]])/(8 
*a*(a^2 - b^2)^(3/4))))/(f*Sqrt[g*Cos[e + f*x]]*(1 - Cos[e + f*x]^2)*(b + 
a*Csc[e + f*x]))
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 369, 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

\(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g])) + (b^(3/2)*ArcTan[(S 
qrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2 
)^(3/4)*f*Sqrt[g]) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g]) + 
 (b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[ 
g])])/(a*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[ 
(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b*(b - Sqrt[-a^2 + 
b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b 
 + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b*(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 = 0.80 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sqrt {g}-\ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}-\ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}}{2 a \sqrt {-g}\, \sqrt {g}\, f}\)	\(186\)

Input:

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

Output:

1/2*(2*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))*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)-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))/a/(-g)^(1/2)/g^(1/2)/f
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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