\(\int \csc ^3(e+f x) \sqrt {b \sec (e+f x)} \, dx\) [376]

Optimal result

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

3/4*b^(1/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-3/4*b^(1/2)*arctanh((b* 
sec(f*x+e))^(1/2)/b^(1/2))/f-1/2*cot(f*x+e)^2*(b*sec(f*x+e))^(3/2)/b/f
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \csc ^3(e+f x) \sqrt {b \sec (e+f x)} \, dx=-\frac {\left (-6 \arctan \left (\sqrt {\sec (e+f x)}\right )-3 \log \left (1-\sqrt {\sec (e+f x)}\right )+3 \log \left (1+\sqrt {\sec (e+f x)}\right )+\frac {4 \csc ^2(e+f x)}{\sqrt {\sec (e+f x)}}\right ) \sqrt {b \sec (e+f x)}}{8 f \sqrt {\sec (e+f x)}} \] Input:

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

Output:

-1/8*((-6*ArcTan[Sqrt[Sec[e + f*x]]] - 3*Log[1 - Sqrt[Sec[e + f*x]]] + 3*L 
og[1 + Sqrt[Sec[e + f*x]]] + (4*Csc[e + f*x]^2)/Sqrt[Sec[e + f*x]])*Sqrt[b 
*Sec[e + f*x]])/(f*Sqrt[Sec[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (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, 827, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3102

\(\displaystyle \frac {\int \frac {b^4 (b \sec (e+f x))^{5/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{b^3 f}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(b*((-3*(-1/2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e 
 + f*x]]/(2*Sqrt[b])))/2 + (b*Sec[e + f*x])^(3/2)/(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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   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. \(461\) vs. \(2(73)=146\).

Time = 3.16 (sec) , antiderivative size = 462, normalized size of antiderivative = 4.97

method	result	size
default	\(\frac {\sqrt {-\frac {b \left (\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}+1\right )}{\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (\sqrt {\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1}\, \left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}+4 \ln \left (2 \left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}+2 \sqrt {\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1}\right ) \left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}-\ln \left (\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}+\sqrt {\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1}\right ) \left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}-\left (\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1\right )^{\frac {3}{2}}+\left (-\cos \left (f x +e \right )+1\right )^{2} \sqrt {\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1}\, \csc \left (f x +e \right )^{2}-3 \arctan \left (\frac {1}{\sqrt {\left (-\cos \left (f x +e \right )+1\right )^{4} \csc \left (f x +e \right )^{4}-1}}\right ) \left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )^{2}}{8 f \sqrt {\left (\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (\left (-\cos \left (f x +e \right )+1\right )^{2} \csc \left (f x +e \right )^{2}-1\right )}\, \left (-\cos \left (f x +e \right )+1\right )^{2}}\)	\(462\)

Input:

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

Output:

1/8/f*(-b*((-cos(f*x+e)+1)^2*csc(f*x+e)^2+1)/((-cos(f*x+e)+1)^2*csc(f*x+e) 
^2-1))^(1/2)*((-cos(f*x+e)+1)^2*csc(f*x+e)^2-1)*(((-cos(f*x+e)+1)^4*csc(f* 
x+e)^4-1)^(1/2)*(-cos(f*x+e)+1)^4*csc(f*x+e)^4+4*ln(2*(-cos(f*x+e)+1)^2*cs 
c(f*x+e)^2+2*((-cos(f*x+e)+1)^4*csc(f*x+e)^4-1)^(1/2))*(-cos(f*x+e)+1)^2*c 
sc(f*x+e)^2-ln((-cos(f*x+e)+1)^2*csc(f*x+e)^2+((-cos(f*x+e)+1)^4*csc(f*x+e 
)^4-1)^(1/2))*(-cos(f*x+e)+1)^2*csc(f*x+e)^2-((-cos(f*x+e)+1)^4*csc(f*x+e) 
^4-1)^(3/2)+(-cos(f*x+e)+1)^2*((-cos(f*x+e)+1)^4*csc(f*x+e)^4-1)^(1/2)*csc 
(f*x+e)^2-3*arctan(1/((-cos(f*x+e)+1)^4*csc(f*x+e)^4-1)^(1/2))*(-cos(f*x+e 
)+1)^2*csc(f*x+e)^2)/(((-cos(f*x+e)+1)^2*csc(f*x+e)^2+1)*((-cos(f*x+e)+1)^ 
2*csc(f*x+e)^2-1))^(1/2)/(-cos(f*x+e)+1)^2*sin(f*x+e)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (73) = 146\).

Time = 0.14 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.01 \[ \int \csc ^3(e+f x) \sqrt {b \sec (e+f x)} \, dx=\left [\frac {6 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 3 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}, \frac {6 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 3 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \] Input:

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

Output:

[1/16*(6*(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)) + 3*(cos(f*x + e)^2 - 1)*sqrt(-b)*l 
og((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/c 
os(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1) 
) + 8*sqrt(b/cos(f*x + e))*cos(f*x + e))/(f*cos(f*x + e)^2 - f), 1/16*(6*( 
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)) + 3*(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)) + 8*sqrt(b/c 
os(f*x + e))*cos(f*x + e))/(f*cos(f*x + e)^2 - f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \csc ^3(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b {\left (\frac {4 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} + \frac {6 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {3 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{\sqrt {b}}\right )}}{8 \, f} \] Input:

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

Output:

1/8*b*(4*(b/cos(f*x + e))^(3/2)/(b^2 - b^2/cos(f*x + e)^2) + 6*arctan(sqrt 
(b/cos(f*x + e))/sqrt(b))/sqrt(b) + 3*log(-(sqrt(b) - sqrt(b/cos(f*x + e)) 
)/(sqrt(b) + sqrt(b/cos(f*x + e))))/sqrt(b))/f
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \csc ^3(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b^{4} {\left (\frac {2 \, \sqrt {b \cos \left (f x + e\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b^{2}} + \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{4 \, f} \] Input:

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

Output:

1/4*b^4*(2*sqrt(b*cos(f*x + e))/((b^2*cos(f*x + e)^2 - b^2)*b^2) + 3*arcta 
n(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^3) - 3*arctan(sqrt(b*cos(f*x 
+ e))/sqrt(b))/b^(7/2))*sgn(cos(f*x + e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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