\(\int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1396]

Optimal result

Integrand size = 31, antiderivative size = 507 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {2}{a f g \sqrt {g \cos (e+f x)}}+\frac {2 b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \] Output:

arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)-b^(5/2)*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)^(5/4)/f/g^(3/2)- 
arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)+b^(5/2)*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)^(5/4)/f/g^(3/2 
)+2/a/f/g/(g*cos(f*x+e))^(1/2)+2*b*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2* 
f*x+1/2*e),2^(1/2))/(a^2-b^2)/f/g^2/cos(f*x+e)^(1/2)+b^2*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^2)/ 
(b-(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)+b^2*cos(f*x+e)^(1/2)*Ellipti 
cPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/(a^2-b^2)/(b+(-a^ 
2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)+2*b*(b-a*sin(f*x+e))/a/(a^2-b^2)/f/ 
g/(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 = 28.46 (sec) , antiderivative size = 1587, normalized size of antiderivative = 3.13 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[Csc[e + f*x]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

-1/2*(Cos[e + f*x]^(3/2)*((8*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Appe 
llF1[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*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)* 
Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sq 
rt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 
+ I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + L 
og[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x] 
] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4))))/(Sqrt[1 - Cos[e + f 
*x]^2]*(b + a*Csc[e + f*x])) - ((-2*a^2 + b^2)*(-1 + Cos[e + f*x]^2)*(a + 
b*Sqrt[1 - Cos[e + f*x]^2])*Csc[e + f*x]*(6*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)] - 6 
*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)] + 12*(a^2 - b^2)*ArcTan[Sqrt[Cos[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) + 6*a^2*Log[1 - Sqrt[Cos[e + f*x]]] - 6*b^2*L 
og[1 - Sqrt[Cos[e + f*x]]] - 6*a^2*Log[1 + Sqrt[Cos[e + f*x]]] + 6*b^2*Log 
[1 + Sqrt[Cos[e + f*x]]] - 3*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]] + 3*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]]))/(12*...
 

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 507, 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.

\(\displaystyle \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x) (g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3377

\(\displaystyle \int \left (\frac {\csc (e+f x)}{a (g \cos (e+f x))^{3/2}}-\frac {b}{a (g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac {2 b E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{a f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}+\frac {2}{a f g \sqrt {g \cos (e+f x)}}\)

Input:

Int[Csc[e + f*x]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) - (b^(5/2)*ArcTan[(Sqrt 
[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^( 
5/4)*f*g^(3/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) + (b 
^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g]) 
])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)) + 2/(a*f*g*Sqrt[g*Cos[e + f*x]]) + (2* 
b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/((a^2 - b^2)*f*g^2*Sqrt[ 
Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + 
 b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b - Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Co 
s[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b 
^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[ 
e + f*x]]) + (2*b*(b - a*Sin[e + f*x]))/(a*(a^2 - b^2)*f*g*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 = 1.02 (sec) , antiderivative size = 436, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {4 g^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 \sqrt {-g}\, \ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} g +2 \sqrt {-g}\, \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} g -2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}}+4 \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}\, \sqrt {-g}\, \sqrt {g}-\ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, g -\ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}\, g}{2 g^{\frac {5}{2}} a \sqrt {-g}\, \left (-1+2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) f}\)	\(436\)

Input:

int(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBO 
SE)
 

Output:

-1/2/g^(5/2)/a/(-g)^(1/2)/(-1+2*sin(1/2*f*x+1/2*e)^2)*(4*g^(3/2)*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))*sin(1/2 
*f*x+1/2*e)^2+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))*sin(1/2*f*x+1/2*e)^2 
*g+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))*sin(1/2*f*x+1/2*e)^2*g-2*ln(2/co 
s(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^(3/ 
2)+4*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-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-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)/f
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

Integral(csc(e + f*x)/((g*cos(e + f*x))**(3/2)*(a + b*sin(e + f*x))), x)
 

Maxima [F]

\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="m 
axima")
 

Output:

integrate(csc(f*x + e)/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Giac [F]

\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

integrate(csc(f*x + e)/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int(1/(sin(e + f*x)*(g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x)
 

Output:

int(1/(sin(e + f*x)*(g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \csc \left (f x +e \right )}{\cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{2} a}d x \right )}{g^{2}} \] Input:

int(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

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