\(\int \frac {(g \cos (e+f x))^{3/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx\) [1378]

Optimal result

Integrand size = 33, antiderivative size = 574 \[ \int \frac {(g \cos (e+f x))^{3/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {g^{3/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {b^2 g^{3/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{3/2} \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 f}+\frac {g^{3/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {b^2 g^{3/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{3/2} \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 f}+\frac {b g \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f}-\frac {g \sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{2 a f}-\frac {b g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{a^2 f \sqrt {g \cos (e+f x)}}+\frac {b \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {b \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \] Output:

1/4*g^(3/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f-b^2*g^(3/2)*arctan((g 
*cos(f*x+e))^(1/2)/g^(1/2))/a^3/f+b^(3/2)*(-a^2+b^2)^(1/4)*g^(3/2)*arctan( 
b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/f+1/4*g^(3/2)*a 
rctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f-b^2*g^(3/2)*arctanh((g*cos(f*x+e) 
)^(1/2)/g^(1/2))/a^3/f+b^(3/2)*(-a^2+b^2)^(1/4)*g^(3/2)*arctanh(b^(1/2)*(g 
*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/f+b*g*(g*cos(f*x+e))^(1/2 
)*csc(f*x+e)/a^2/f-1/2*g*(g*cos(f*x+e))^(1/2)*csc(f*x+e)^2/a/f-b*g^2*cos(f 
*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/a^2/f/(g*cos(f*x+e))^(1 
/2)+b*(a^2-b^2)*g^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/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/(g*cos(f*x+e 
))^(1/2)+b*(a^2-b^2)*g^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/(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 = 34.83 (sec) , antiderivative size = 2129, normalized size of antiderivative = 3.71 \[ \int \frac {(g \cos (e+f x))^{3/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

((g*Cos[e + f*x])^(3/2)*((-2*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(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 + C 
os[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqr 
t[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]]))/(-a^2 + b^2)^(3/4)))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + 
 f*x])) - (b^2*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[ 
2*(e + f*x)]*Csc[e + f*x]*((-10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]* 
Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/ 
4)) + (10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f 
*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (20*ArcTan[Sqrt[ 
Cos[e + f*x]]])/a - (16*b*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2* 
Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(5/2))/(-a^2 + b^2) - (200*b...
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 574, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(g^(3/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) - (b^2*g^(3/2)*ArcT 
an[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) + (b^(3/2)*(-a^2 + b^2)^(1/4)*g^ 
(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])]) 
/(a^3*f) + (g^(3/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) - (b^2* 
g^(3/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) + (b^(3/2)*(-a^2 + 
b^2)^(1/4)*g^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1 
/4)*Sqrt[g])])/(a^3*f) + (b*g*Sqrt[g*Cos[e + f*x]]*Csc[e + f*x])/(a^2*f) - 
 (g*Sqrt[g*Cos[e + f*x]]*Csc[e + f*x]^2)/(2*a*f) - (b*g^2*Sqrt[Cos[e + f*x 
]]*EllipticF[(e + f*x)/2, 2])/(a^2*f*Sqrt[g*Cos[e + f*x]]) + (b*(a^2 - b^2 
)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x 
)/2, 2])/(a^2*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) + ( 
b*(a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2 
]), (e + f*x)/2, 2])/(a^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 = 1.00 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {g^{2} \left (\frac {\sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\frac {\ln \left (\frac {-2 g +2 \sqrt {-g}\, \sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{\sqrt {-g}}-\frac {\ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{2 \sqrt {g}}-\frac {\ln \left (\frac {-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{2 \sqrt {g}}-\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\right )}{4 a f}\)	\(289\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((g*cos(f*x+e))^(3/2)*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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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