\(\int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx\) [389]

Optimal result

Integrand size = 19, antiderivative size = 77 \[ \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx=-\frac {b^{3/2} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \] Output:

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx=\frac {\left (-2 \arctan \left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )+4 \sqrt {\sec (e+f x)}\right ) (b \sec (e+f x))^{3/2}}{2 f \sec ^{\frac {3}{2}}(e+f x)} \] Input:

Integrate[Csc[e + f*x]*(b*Sec[e + f*x])^(3/2),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*Sqrt[Sec[e + f*x]])*(b*Sec[e + f*x])^(3/2))/(2*f*Sec 
[e + f*x]^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92, 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, 262, 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.

\(\displaystyle \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc (e+f x) (b \sec (e+f x))^{3/2}dx\)

\(\Big \downarrow \) 3102

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[e + f*x]*(b*Sec[e + f*x])^(3/2),x]
 

Output:

-((b*(2*b^2*(ArcTan[Sqrt[b]*Sec[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Se 
c[e + f*x]]/(2*b^(3/2))) - 2*Sqrt[b*Sec[e + f*x]]))/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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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. \(207\) vs. \(2(63)=126\).

Time = 2.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.70

method	result	size
default	\(\frac {\left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) b \sqrt {b \sec \left (f x +e \right )}}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(208\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (63) = 126\).

Time = 0.12 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.87 \[ \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx=\left [\frac {2 \, \sqrt {-b} b \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + \sqrt {-b} b \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{4 \, f}, -\frac {2 \, b^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) - b^{\frac {3}{2}} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{4 \, f}\right ] \] Input:

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

Output:

[1/4*(2*sqrt(-b)*b*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b* 
cos(f*x + e) + b)) + sqrt(-b)*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*b*sqrt(b/cos(f*x + e)))/f, -1/4*(2* 
b^(3/2)*arctan(2*sqrt(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) 
 - b)) - b^(3/2)*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*b*sqrt(b/cos(f*x + e)))/f]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx=-\frac {{\left (2 \, \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) - \sqrt {b} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) - 4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )} b}{2 \, f} \] Input:

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

Output:

-1/2*(2*sqrt(b)*arctan(sqrt(b/cos(f*x + e))/sqrt(b)) - sqrt(b)*log(-(sqrt( 
b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e)))) - 4*sqrt(b/co 
s(f*x + e)))*b/f
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \csc (e+f x) (b \sec (e+f x))^{3/2} \, dx=\frac {b^{4} {\left (\frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {2}{\sqrt {b \cos \left (f x + e\right )} b^{2}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{f} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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