\(\int \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [70]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45 \[ \int \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {2} \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )-\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right )\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

Integrate[Csc[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(Sqrt[2]*(Sqrt[b]*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]] - Sqrt[a 
 + b]*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]])*Cos[e + f*x]*Sq 
rt[a + b*Sec[e + f*x]^2])/(f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4622, 25, 301, 224, 219, 291, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\sin (e+f x)}dx\)

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 301

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 301

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[b/ 
d   Int[(a + b*x^2)^(p - 1), x], x] - Simp[(b*c - a*d)/d   Int[(a + b*x^2)^ 
(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] 
&& GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4] || (EqQ[p, 2/3] && E 
qQ[b*c + 3*a*d, 0]))
 

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(553\) vs. \(2(70)=140\).

Time = 3.98 (sec) , antiderivative size = 554, normalized size of antiderivative = 6.76

method	result	size
default	\(-\frac {\sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (2 \sqrt {b}\, \ln \left (\frac {4 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 b}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right ) \sqrt {a +b}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right ) a +2 b}{\sqrt {a +b}\, \left (1+\cos \left (f x +e \right )\right )}\right ) a -b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right ) a +2 b}{\sqrt {a +b}\, \left (1+\cos \left (f x +e \right )\right )}\right )-\ln \left (-\frac {4 \left (\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\cos \left (f x +e \right ) a +b \right )}{-1+\cos \left (f x +e \right )}\right ) a -\ln \left (-\frac {4 \left (\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\cos \left (f x +e \right ) a +b \right )}{-1+\cos \left (f x +e \right )}\right ) b \right )}{4 f \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(554\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 536, normalized size of antiderivative = 6.54 \[ \int \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {\sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {2 \, \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) + \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) - \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, f}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) - \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right )}{f}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*csc(e + f*x), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*csc(f*x + e), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (70) = 140\).

Time = 0.54 (sec) , antiderivative size = 405, normalized size of antiderivative = 4.94 \[ \int \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {{\left (\frac {4 \, b \arctan \left (-\frac {\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} - \sqrt {a + b}}{2 \, \sqrt {-b}}\right )}{\sqrt {-b}} + \sqrt {a + b} \log \left ({\left | -\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} + \sqrt {a + b} \right |}\right ) - \sqrt {a + b} \log \left ({\left | -\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} - \sqrt {a + b} \right |}\right ) - \sqrt {a + b} \log \left ({\left | {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{2 \, f} \] Input:

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

Output:

-1/2*(4*b*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2 
*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 
2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b))/sqrt(-b))/sqrt(-b) + sq 
rt(a + b)*log(abs(-sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a*tan(1/2*f*x 
 + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b* 
tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))) - sqrt(a + b)*log(abs(-sqr 
t(a + b)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/ 
2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 
 + a + b) - sqrt(a + b))) - sqrt(a + b)*log(abs((sqrt(a + b)*tan(1/2*f*x + 
 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a 
*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*(a + b) - s 
qrt(a + b)*(a - b))))*sgn(cos(f*x + e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x),x)