\(\int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) [1403]

Optimal result

Integrand size = 33, antiderivative size = 651 \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \] Output:

b*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*arctan(b^(1/2 
)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/(-a^2+b^2)^(7/4)/f/g^ 
(5/2)+b*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*arctan 
h(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/(-a^2+b^2)^(7 
/4)/f/g^(5/2)-2/3*b/a^2/f/g/(g*cos(f*x+e))^(3/2)-csc(f*x+e)/a/f/g/(g*cos(f 
*x+e))^(3/2)+5/3*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/a 
/f/g^2/(g*cos(f*x+e))^(1/2)+2/3*b^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f 
*x+1/2*e,2^(1/2))/a/(a^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)-b^4*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/(a^2 
-b^2)/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+e))^(1/2)-b^4*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 
/(a^2-b^2)/(a^2-b*(b+(-a^2+b^2)^(1/2)))/f/g^2/(g*cos(f*x+e))^(1/2)+5/3*sin 
(f*x+e)/a/f/g/(g*cos(f*x+e))^(3/2)-2/3*b^2*(b-a*sin(f*x+e))/a^2/(a^2-b^2)/ 
f/g/(g*cos(f*x+e))^(3/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 30.47 (sec) , antiderivative size = 2183, normalized size of antiderivative = 3.35 \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Cos[e + f*x]^(5/2)*((-2*(10*a^3 - 18*a*b^2)*(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 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I) 
*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*S 
qrt[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])) - ((-5*a^2*b + 3*b^3)*(-1 + Cos[e + f*x]^2)*(a + b*Sqr 
t[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 + (Sqr 
t[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]^(...
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 651, 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

\(\displaystyle \int \frac {1}{\sin (e+f x)^2 (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}+\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}\)

Input:

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

Output:

(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*g^(5/2)) - (b^(9/2)*ArcTan 
[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 
+ b^2)^(7/4)*f*g^(5/2)) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f 
*g^(5/2)) - (b^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^ 
(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(7/4)*f*g^(5/2)) - (2*b)/(3*a^2*f*g*(g* 
Cos[e + f*x])^(3/2)) - Csc[e + f*x]/(a*f*g*(g*Cos[e + f*x])^(3/2)) + (5*Sq 
rt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*a*f*g^2*Sqrt[g*Cos[e + f*x] 
]) + (2*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*a*(a^2 - b^2) 
*f*g^2*Sqrt[g*Cos[e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b 
 - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*(a^2 - b*(b - Sqrt[- 
a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*Ellipti 
cPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*(a^2 - b 
*(b + Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (5*Sin[e + f*x])/(3 
*a*f*g*(g*Cos[e + f*x])^(3/2)) - (2*b^2*(b - a*Sin[e + f*x]))/(3*a^2*(a^2 
- b^2)*f*g*(g*Cos[e + f*x])^(3/2))
 

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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1626\) vs. \(2(575)=1150\).

Time = 6.12 (sec) , antiderivative size = 1627, normalized size of antiderivative = 2.50

method	result	size
default	\(\text {Expression too large to display}\)	\(1627\)

Input:

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

Output:

(-1/g^2*b*(1/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))-2/(2+2^(1/2))^2/(-2+2^(1/2))^2/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))-2/(2+2^(1/2))^2/(-2+2^(1/2))^2/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))-1/3*2^(1/2)/(2+2^(1/2))/(-2+2^(1/2))/(a^2-b^2)* 
(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-2^(1/2)+cos(1/2*f*x+1/2*e))/g/(2*2^( 
1/2)*cos(1/2*f*x+1/2*e)+2*sin(1/2*f*x+1/2*e)^2-3)-1/3*2^(1/2)/(2+2^(1/2))/ 
(-2+2^(1/2))/(a^2-b^2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(2^(1/2)+cos(1/ 
2*f*x+1/2*e))/g/(2*2^(1/2)*cos(1/2*f*x+1/2*e)-2*sin(1/2*f*x+1/2*e)^2+3)-4* 
b^4/(a-b)/(a+b)/a^2*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(-ln((-(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)-2*g*cos(1/2*f*x+1/ 
2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+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)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^ 
2)^(1/2)+g))-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*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))/(16*a^2-16*b^2)/g)+1/24*(g*(-1+2*cos(1/2*f*x+ 
1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)/a/g^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)^(3/2)/(a^2-b^2)*(40*cos(1/2*f*x+1/ 
2*e)*(-1+2*sin(1/2*f*x+1/2*e)^2)^(3/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*Ell...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^2\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int(1/(sin(e + f*x)^2*(g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)
 

Output:

int(1/(sin(e + f*x)^2*(g*cos(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)
 

Reduce [F]

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

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

Output:

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