\(\int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [415]

Optimal result

Integrand size = 21, antiderivative size = 93 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f} \] Output:

-1/4*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(1/2)/f-1/4*arctanh((b*sec(f*x 
+e))^(1/2)/b^(1/2))/b^(1/2)/f-1/2*cot(f*x+e)^2*(b*sec(f*x+e))^(1/2)/b/f
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {\left (-2 \arctan \left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )-\frac {4 \csc ^2(e+f x)}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{8 f \sqrt {b \sec (e+f x)}} \] Input:

Integrate[Csc[e + f*x]^3/Sqrt[b*Sec[e + f*x]],x]
 

Output:

((-2*ArcTan[Sqrt[Sec[e + f*x]]] + Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sq 
rt[Sec[e + f*x]]] - (4*Csc[e + f*x]^2)/Sec[e + f*x]^(3/2))*Sqrt[Sec[e + f* 
x]])/(8*f*Sqrt[b*Sec[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3102, 27, 252, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc (e+f x)^3}{\sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[e + f*x]^3/Sqrt[b*Sec[e + f*x]],x]
 

Output:

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

Defintions of rubi rules used

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[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. \(280\) vs. \(2(73)=146\).

Time = 1.78 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.02

method	result	size
default	\(-\frac {\left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )-\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )\right ) \csc \left (f x +e \right )^{2}}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}}\)	\(281\)

Input:

int(csc(f*x+e)^3/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8/f*(4*cos(f*x+e)*(-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))*cos(f*x+e)+arctan(1/2/(-cos(f*x+e)/(cos(f 
*x+e)+1)^2)^(1/2))*cos(f*x+e)-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))-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)/(b*sec(f*x+e))^(1/2)*csc(f*x+e)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (73) = 146\).

Time = 0.13 (sec) , antiderivative size = 381, normalized size of antiderivative = 4.10 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\left [\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b f\right )}}, -\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) - 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b f\right )}}\right ] \] Input:

integrate(csc(f*x+e)^3/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(2*(cos(f*x + e)^2 - 1)*sqrt(-b)*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + 
e))*cos(f*x + e)/(b*cos(f*x + e) + b)) + 8*sqrt(b/cos(f*x + e))*cos(f*x + 
e)^2 - (cos(f*x + e)^2 - 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + 
e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b) 
/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)))/(b*f*cos(f*x + e)^2 - b*f), -1/16 
*(2*(cos(f*x + e)^2 - 1)*sqrt(b)*arctan(2*sqrt(b)*sqrt(b/cos(f*x + e))*cos 
(f*x + e)/(b*cos(f*x + e) - b)) - 8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 - 
(cos(f*x + e)^2 - 1)*sqrt(b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + c 
os(f*x + e))*sqrt(b)*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x 
 + e)^2 - 2*cos(f*x + e) + 1)))/(b*f*cos(f*x + e)^2 - b*f)]
 

Sympy [F]

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

integrate(csc(f*x+e)**3/(b*sec(f*x+e))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b {\left (\frac {4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {3}{2}}}\right )}}{8 \, f} \] Input:

integrate(csc(f*x+e)^3/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

1/8*b*(4*sqrt(b/cos(f*x + e))/(b^2 - b^2/cos(f*x + e)^2) - 2*arctan(sqrt(b 
/cos(f*x + e))/sqrt(b))/b^(3/2) + log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(s 
qrt(b) + sqrt(b/cos(f*x + e))))/b^(3/2))/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b^{2} {\left (\frac {2 \, \sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}}\right )}}{4 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \] Input:

integrate(csc(f*x+e)^3/(b*sec(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

1/4*b^2*(2*sqrt(b*cos(f*x + e))*cos(f*x + e)/((b^2*cos(f*x + e)^2 - b^2)*b 
) + arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^2) + arctan(sqrt(b*c 
os(f*x + e))/sqrt(b))/b^(5/2))/(f*sgn(cos(f*x + e)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \] Input:

int(1/(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2)),x)
 

Output:

int(1/(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2)), x)
 

Reduce [F]

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

int(csc(f*x+e)^3/(b*sec(f*x+e))^(1/2),x)
 

Output:

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