\(\int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [441]

Optimal result

Integrand size = 19, antiderivative size = 81 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{5/2} f}+\frac {2}{3 b f (b \sec (e+f x))^{3/2}} \] Output:

-arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(5/2)/f-arctanh((b*sec(f*x+e))^(1/ 
2)/b^(1/2))/b^(5/2)/f+2/3/b/f/(b*sec(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\left (-6 \arctan \left (\sqrt {\sec (e+f x)}\right )+3 \log \left (1-\sqrt {\sec (e+f x)}\right )-3 \log \left (1+\sqrt {\sec (e+f x)}\right )+\frac {4}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{6 b^2 f \sqrt {b \sec (e+f x)}} \] Input:

Integrate[Csc[e + f*x]/(b*Sec[e + f*x])^(5/2),x]
 

Output:

((-6*ArcTan[Sqrt[Sec[e + f*x]]] + 3*Log[1 - Sqrt[Sec[e + f*x]]] - 3*Log[1 
+ Sqrt[Sec[e + f*x]]] + 4/Sec[e + f*x]^(3/2))*Sqrt[Sec[e + f*x]])/(6*b^2*f 
*Sqrt[b*Sec[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3102, 25, 27, 264, 266, 756, 216, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 3102

\(\displaystyle \frac {\int -\frac {b^2}{(b \sec (e+f x))^{5/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{b f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {b^2}{(b \sec (e+f x))^{5/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{b f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {1}{(b \sec (e+f x))^{5/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{f}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {b \left (\frac {\int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{b^2}-\frac {2}{3 b^2 (b \sec (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {b \left (\frac {2 \int \frac {1}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}}{b^2}-\frac {2}{3 b^2 (b \sec (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}}{2 b}\right )}{b^2}-\frac {2}{3 b^2 (b \sec (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{b^2}-\frac {2}{3 b^2 (b \sec (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{b^2}-\frac {2}{3 b^2 (b \sec (e+f x))^{3/2}}\right )}{f}\)

Input:

Int[Csc[e + f*x]/(b*Sec[e + f*x])^(5/2),x]
 

Output:

-((b*((2*(ArcTan[Sqrt[b]*Sec[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Sec[e 
 + f*x]]/(2*b^(3/2))))/b^2 - 2/(3*b^2*(b*Sec[e + f*x])^(3/2))))/f)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3102

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S 
ymbol] :> Simp[1/(f*a^n)   Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 
2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 
)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(65)=130\).

Time = 2.02 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.30

method	result	size
default	\(\frac {\frac {\sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (-3-3 \sec \left (f x +e \right )\right )}{6}+\frac {\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (-3-3 \sec \left (f x +e \right )\right )}{6}+\frac {2 \cos \left (f x +e \right )}{3}}{f \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\)	\(186\)

Input:

int(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/6*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln((2*cos(f*x+e)*(-cos(f*x+e 
)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e 
)+1)/(cos(f*x+e)+1))*(-3-3*sec(f*x+e))+1/6*arctan(1/2/(-cos(f*x+e)/(cos(f* 
x+e)+1)^2)^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(-3-3*sec(f*x+e))+2 
/3*cos(f*x+e))/(b*sec(f*x+e))^(1/2)/b^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (65) = 130\).

Time = 0.23 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.94 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\left [\frac {8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + 6 \, \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) - 3 \, \sqrt {-b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{12 \, b^{3} f}, \frac {8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - 6 \, \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 3 \, \sqrt {b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{12 \, b^{3} f}\right ] \] Input:

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 + 6*sqrt(-b)*arctan(2*sqrt(-b 
)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) + b)) - 3*sqrt(-b)*log 
(-(b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/co 
s(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) 
)/(b^3*f), 1/12*(8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 - 6*sqrt(b)*arctan( 
2*sqrt(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) + 3*sqrt 
(b)*log(-(b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)*sqr 
t(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) 
 + 1)))/(b^3*f)]
 

Sympy [F]

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

integrate(csc(f*x+e)/(b*sec(f*x+e))**(5/2),x)
 

Output:

Integral(csc(e + f*x)/(b*sec(e + f*x))**(5/2), x)
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {b {\left (\frac {6 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} - \frac {3 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {7}{2}}} - \frac {4}{b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}}\right )}}{6 \, f} \] Input:

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

-1/6*b*(6*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/b^(7/2) - 3*log(-(sqrt(b) - 
 sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/b^(7/2) - 4/(b^2* 
(b/cos(f*x + e))^(3/2)))/f
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\frac {3 \, b \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 3 \, \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right ) + 2 \, \sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}{3 \, b^{3} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \] Input:

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

1/3*(3*b*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/sqrt(-b) + 3*sqrt(b)*arctan 
(sqrt(b*cos(f*x + e))/sqrt(b)) + 2*sqrt(b*cos(f*x + e))*cos(f*x + e))/(b^3 
*f*sgn(cos(f*x + e)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x)
 

Output:

(sqrt(b)*int((sqrt(sec(e + f*x))*csc(e + f*x))/sec(e + f*x)**3,x))/b**3