\(\int \frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx\) [1376]

Optimal result

Integrand size = 31, antiderivative size = 439 \[ \int \frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {g^{3/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {g^{3/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \sqrt {b} f}-\frac {2 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \] Output:

-g^(3/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f+(-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/b^(1/2)/f 
-g^(3/2)*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f+(-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/b^(1/2) 
/f-2*g^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b/f/(g*co 
s(f*x+e))^(1/2)+(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))/b/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/(g*c 
os(f*x+e))^(1/2)+(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))/b/(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 = 16.59 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.10 \[ \int \frac {(g \cos (e+f x))^{3/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{3/2} \csc (e+f x) \left (8 a b^{3/2} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {5}{2}}(e+f x)-5 \left (a^2-b^2\right ) \left (2 \sqrt {2} \sqrt [4]{a^2-b^2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \sqrt {2} \sqrt [4]{a^2-b^2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 \sqrt {b} \arctan \left (\sqrt {\cos (e+f x)}\right )-2 \sqrt {b} \log \left (1-\sqrt {\cos (e+f x)}\right )+2 \sqrt {b} \log \left (1+\sqrt {\cos (e+f x)}\right )+\sqrt {2} \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )-\sqrt {2} \sqrt [4]{a^2-b^2} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{20 a \sqrt {b} \left (a^2-b^2\right ) f \cos ^{\frac {3}{2}}(e+f x) (b+a \csc (e+f x))} \] Input:

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

Output:

((g*Cos[e + f*x])^(3/2)*Csc[e + f*x]*(8*a*b^(3/2)*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) - 
 5*(a^2 - b^2)*(2*Sqrt[2]*(a^2 - b^2)^(1/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sq 
rt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 2*Sqrt[2]*(a^2 - b^2)^(1/4)*ArcTan[ 
1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4*Sqrt[b]*Ar 
cTan[Sqrt[Cos[e + f*x]]] - 2*Sqrt[b]*Log[1 - Sqrt[Cos[e + f*x]]] + 2*Sqrt[ 
b]*Log[1 + Sqrt[Cos[e + f*x]]] + Sqrt[2]*(a^2 - b^2)^(1/4)*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]] - Sqrt[2]*(a^2 - b^2)^(1/4)*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]]))*(a + b*Sqrt[Sin[e + f 
*x]^2]))/(20*a*Sqrt[b]*(a^2 - b^2)*f*Cos[e + f*x]^(3/2)*(b + a*Csc[e + f*x 
]))
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((g^(3/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a*f)) + ((-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*Sqrt[b]*f) - (g^(3/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a* 
f) + ((-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*Sqrt[b]*f) - (2*g^2*Sqrt[Cos[e + f*x]]*Ell 
ipticF[(e + f*x)/2, 2])/(b*f*Sqrt[g*Cos[e + f*x]]) + ((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])/( 
b*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) + ((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])/(b*(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 = 0.92 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.49

method	result	size
default	\(-\frac {\sqrt {g}\, \left (-2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}}+\ln \left (-\frac {2 \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g -\sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, g -4 \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}\, \sqrt {-g}\, \sqrt {g}+\ln \left (\frac {4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) g +2 \sqrt {g}\, \sqrt {-2 g \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right ) \sqrt {-g}\, g \right )}{2 a \sqrt {-g}\, f}\)	\(214\)

Input:

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

Output:

-1/2*g^(1/2)/a/(-g)^(1/2)*(-2*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*si 
n(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*g^(3/2)+ln(-2/(cos(1/2*f*x+1/2*e)+1)*(2*co 
s(1/2*f*x+1/2*e)*g-g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+g))*(-g)^(1 
/2)*g-4*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-g)^(1/2)*g^(1/2)+ln(2/(cos(1 
/2*f*x+1/2*e)-1)*(2*cos(1/2*f*x+1/2*e)*g+g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^ 
2+g)^(1/2)-g))*(-g)^(1/2)*g)/f
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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