\(\int \frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx\) [1372]

Optimal result

Integrand size = 33, antiderivative size = 544 \[ \int \frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {\sqrt {g} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}+\frac {b^2 \sqrt {g} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^{5/2} \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}-\frac {\sqrt {g} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {b^2 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{5/2} \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}+\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}-\frac {(g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f g}+\frac {b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {\cos (e+f x)}}-\frac {b^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {b^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \] Output:

1/4*g^(1/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f+b^2*g^(1/2)*arctan((g 
*cos(f*x+e))^(1/2)/g^(1/2))/a^3/f-b^(5/2)*g^(1/2)*arctan(b^(1/2)*(g*cos(f* 
x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/(-a^2+b^2)^(1/4)/f-1/4*g^(1/2)*a 
rctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f-b^2*g^(1/2)*arctanh((g*cos(f*x+e) 
)^(1/2)/g^(1/2))/a^3/f+b^(5/2)*g^(1/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2 
)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/(-a^2+b^2)^(1/4)/f+b*(g*cos(f*x+e))^(3/2)* 
csc(f*x+e)/a^2/f/g-1/2*(g*cos(f*x+e))^(3/2)*csc(f*x+e)^2/a/f/g+b*(g*cos(f* 
x+e))^(1/2)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/a^2/f/cos(f*x+e)^(1/2)-b 
^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))/a^2/(b-(-a^2+b^2)^(1/2))/f/(g*cos(f*x+e))^(1/2)-b^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))/a 
^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 = 29.83 (sec) , antiderivative size = 1582, normalized size of antiderivative = 2.91 \[ \int \frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[g*Cos[e + f*x]]*((b*Cot[e + f*x])/a^2 - (Cot[e + f*x]*Csc[e + f*x])/ 
(2*a)))/f - (Sqrt[g*Cos[e + f*x]]*((6*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2]) 
*((a*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*(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)*Sqrt[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]] + 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]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4))))/(Sqrt[1 - 
Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - ((-a^2 - 5*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]*Sq 
rt[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*Log[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)*Lo 
g[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...
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Sqrt[g]*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) + (b^2*Sqrt[g]*ArcT 
an[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) - (b^(5/2)*Sqrt[g]*ArcTan[(Sqrt[ 
b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^3*(-a^2 + b^2)^ 
(1/4)*f) - (Sqrt[g]*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) - (b^2* 
Sqrt[g]*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) + (b^(5/2)*Sqrt[g]* 
ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^3 
*(-a^2 + b^2)^(1/4)*f) + (b*(g*Cos[e + f*x])^(3/2)*Csc[e + f*x])/(a^2*f*g) 
 - ((g*Cos[e + f*x])^(3/2)*Csc[e + f*x]^2)/(2*a*f*g) + (b*Sqrt[g*Cos[e + f 
*x]]*EllipticE[(e + f*x)/2, 2])/(a^2*f*Sqrt[Cos[e + f*x]]) - (b^2*g*Sqrt[C 
os[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a^ 
2*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (b^2*g*Sqrt[Cos[e + f*x 
]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a^2*(b + Sqr 
t[-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 = 0.98 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.53

method	result	size
default	\(-\frac {g \left (-\frac {\sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\frac {\ln \left (\frac {-2 g +2 \sqrt {-g}\, \sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{\sqrt {-g}}-\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {\ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{2 \sqrt {g}}+\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {\ln \left (\frac {-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{2 \sqrt {g}}\right )}{4 a f}\)	\(287\)

Input:

int((g*cos(f*x+e))^(1/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

-1/4*g/a*(-1/2/g/cos(1/2*f*x+1/2*e)^2*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1 
/(-g)^(1/2)*ln((-2*g+2*(-g)^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2))/cos( 
1/2*f*x+1/2*e))-1/4/g/(cos(1/2*f*x+1/2*e)-1)*(-2*g*sin(1/2*f*x+1/2*e)^2+g) 
^(1/2)+1/2/g^(1/2)*ln((4*cos(1/2*f*x+1/2*e)*g+2*g^(1/2)*(-2*g*sin(1/2*f*x+ 
1/2*e)^2+g)^(1/2)-2*g)/(cos(1/2*f*x+1/2*e)-1))+1/4/g/(cos(1/2*f*x+1/2*e)+1 
)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+1/2/g^(1/2)*ln((-4*cos(1/2*f*x+1/2*e 
)*g+2*g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-2*g)/(cos(1/2*f*x+1/2*e) 
+1)))/f
 

Fricas [F]

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

integrate((g*cos(f*x+e))^(1/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((g*cos(f*x+e))^(1/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

integrate((g*cos(f*x+e))^(1/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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