\(\int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx\) [1384]

Optimal result

Integrand size = 33, antiderivative size = 557 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {3 g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}+\frac {b^2 g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {\sqrt {b} \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 f}+\frac {3 g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {b^2 g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {\sqrt {b} \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 f}+\frac {b g (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f}-\frac {g (g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f}+\frac {b g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-3/4*g^(5/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f+b^2*g^(5/2)*arctan(( 
g*cos(f*x+e))^(1/2)/g^(1/2))/a^3/f-b^(1/2)*(-a^2+b^2)^(3/4)*g^(5/2)*arctan 
(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/f+3/4*g^(5/2)* 
arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f-b^2*g^(5/2)*arctanh((g*cos(f*x+e 
))^(1/2)/g^(1/2))/a^3/f+b^(1/2)*(-a^2+b^2)^(3/4)*g^(5/2)*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))^(3/ 
2)*csc(f*x+e)/a^2/f-1/2*g*(g*cos(f*x+e))^(3/2)*csc(f*x+e)^2/a/f+b*g^2*(g*c 
os(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)+(a^2-b^2)*g^3*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)+( 
a^2-b^2)*g^3*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 = 35.63 (sec) , antiderivative size = 1590, normalized size of antiderivative = 2.85 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

Integrate[((g*Cos[e + f*x])^(5/2)*Csc[e + f*x]^3)/(a + b*Sin[e + f*x]),x]
 

Output:

-1/4*((g*Cos[e + f*x])^(5/2)*((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])) - ((3*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]*Sqrt[C 
os[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)*Log[Sq 
rt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*C 
os[e + f*x]] + 3*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[Sqrt[a^2 - b^2] + S 
qrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))...
 

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 557, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x) (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(g \cos (e+f x))^{5/2}}{\sin (e+f x)^3 (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3377

\(\displaystyle \int \left (-\frac {b^3 (g \cos (e+f x))^{5/2}}{a^3 (a+b \sin (e+f x))}+\frac {b^2 \csc (e+f x) (g \cos (e+f x))^{5/2}}{a^3}-\frac {b \csc ^2(e+f x) (g \cos (e+f x))^{5/2}}{a^2}+\frac {\csc ^3(e+f x) (g \cos (e+f x))^{5/2}}{a}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[((g*Cos[e + f*x])^(5/2)*Csc[e + f*x]^3)/(a + b*Sin[e + f*x]),x]
 

Output:

(-3*g^(5/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) + (b^2*g^(5/2)*A 
rcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) - (Sqrt[b]*(-a^2 + b^2)^(3/4) 
*g^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g] 
)])/(a^3*f) + (3*g^(5/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) - 
(b^2*g^(5/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f) + (Sqrt[b]*(-a 
^2 + b^2)^(3/4)*g^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^ 
2)^(1/4)*Sqrt[g])])/(a^3*f) + (b*g*(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) + (b*g^2*Sqrt[g* 
Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(a^2*f*Sqrt[Cos[e + f*x]]) + ((a^ 
2 - b^2)*g^3*Sqrt[Cos[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]]) + ((a^ 
2 - b^2)*g^3*Sqrt[Cos[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]])
 

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 = 69.08 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {g^{3} \left (-\frac {\sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}-\frac {3 \ln \left (\frac {-2 g +2 \sqrt {-g}\, \sqrt {2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-g}}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{\sqrt {-g}}-\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {3 \ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{2 \sqrt {g}}+\frac {\sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}}{4 g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {3 \ln \left (\frac {-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{2 \sqrt {g}}\right )}{4 a f}\)	\(290\)

Input:

int((g*cos(f*x+e))^(5/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

-1/4*g^3/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) 
-3/(-g)^(1/2)*ln((-2*g+2*(-g)^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2))/co 
s(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)-3/2/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)-3/2/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]

\[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(5/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(5/2)*csc(f*x+e)**3/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((g*cos(f*x+e))^(5/2)*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]

\[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

integrate((g*cos(f*x+e))^(5/2)*csc(f*x+e)^3/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

integrate((g*cos(f*x + e))^(5/2)*csc(f*x + e)^3/(b*sin(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^3\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int((g*cos(e + f*x))^(5/2)/(sin(e + f*x)^3*(a + b*sin(e + f*x))),x)
 

Output:

int((g*cos(e + f*x))^(5/2)/(sin(e + f*x)^3*(a + b*sin(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right )^{2} \csc \left (f x +e \right )^{3}}{\sin \left (f x +e \right ) b +a}d x \right ) g^{2} \] Input:

int((g*cos(f*x+e))^(5/2)*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)**2*csc(e + f*x)**3)/(sin(e + 
f*x)*b + a),x)*g**2