\(\int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx\) [1383]

Optimal result

Integrand size = 33, antiderivative size = 462 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {b g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \sqrt {b} f}+\frac {b g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \sqrt {b} f}-\frac {g (g \cos (e+f x))^{3/2} \csc (e+f x)}{a f}-\frac {g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a f \sqrt {\cos (e+f x)}}-\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a b \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a b \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-b*g^(5/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f+(-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^2/b^( 
1/2)/f+b*g^(5/2)*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f-(-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^2/b^(1/2)/f-g*(g*cos(f*x+e))^(3/2)*csc(f*x+e)/a/f-g^2*(g*cos(f*x+e))^(1 
/2)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/a/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/b/(b-(-a^2+b^2)^(1/2))/f/(g*cos(f*x+e))^(1/2)-(a^2-b^2)*g^3*co 
s(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/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 = 27.78 (sec) , antiderivative size = 1556, normalized size of antiderivative = 3.37 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((g*Cos[e + f*x])^(5/2)*((12*a*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Appell 
F1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Co 
s[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sq 
rt[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])) + (5*b*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Co 
s[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*AppellF 
1[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[C 
os[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] - Sq 
rt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] + 3*S 
qrt[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*(a^3 - a*b^2)...
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 462, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((b*g^(5/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f)) + ((-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^2*Sqrt[b]*f) + (b*g^(5/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[ 
g]])/(a^2*f) - ((-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^2*Sqrt[b]*f) - (g*(g*Cos[e + f*x 
])^(3/2)*Csc[e + f*x])/(a*f) - (g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f* 
x)/2, 2])/(a*f*Sqrt[Cos[e + f*x]]) - ((a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*E 
llipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*b*(b - Sqrt[-a 
^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - ((a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*E 
llipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 25.04 (sec) , antiderivative size = 1447, normalized size of antiderivative = 3.13

method	result	size
default	\(\text {Expression too large to display}\)	\(1447\)

Input:

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

Output:

(4*g^3*b*(1/4/a^2/(-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))+1/8/a^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 
))+1/8/a^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))-1/16/a^2*(a^2-b^2)/b^2/(g^ 
2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b 
^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b 
^2)^(1/2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos( 
1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan(2^( 
1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)-2*arc 
tan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+ 
1)))-2*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a*(- 
1/4/a^2/cos(1/2*f*x+1/2*e)/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e) 
^2)^(1/2)*(cos(1/2*f*x+1/2*e)*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f 
*x+1/2*e)^2)^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))-cos(1/2*f*x+1/2*e 
)*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE 
(cos(1/2*f*x+1/2*e),2^(1/2))-2*sin(1/2*f*x+1/2*e)^4+sin(1/2*f*x+1/2*e)^2)+ 
1/16/a^2/b^2*sum((-a^2+b^2)/_alpha*(2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/ 
b^2)^(1/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2* 
cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g...
 

Fricas [F]

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

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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