\(\int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx\) [467]

Optimal result

Integrand size = 23, antiderivative size = 238 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\frac {\sqrt {b} \arctan \left (1-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{\sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\sqrt {b} \arctan \left (1+\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{\sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\left (\sqrt {b}+\sqrt {b} \cot (e+f x)\right ) \sqrt {\sin (e+f x)}}\right )}{\sqrt {2} f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\frac {b \left (\arctan \left (\frac {-1+\sqrt {\tan ^2(e+f x)}}{\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}{1+\sqrt {\tan ^2(e+f x)}}\right )\right ) \tan ^2(e+f x)^{3/4}}{\sqrt {2} f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \] Input:

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]),x]
 

Output:

(b*(ArcTan[(-1 + Sqrt[Tan[e + f*x]^2])/(Sqrt[2]*(Tan[e + f*x]^2)^(1/4))] + 
 ArcTanh[(Sqrt[2]*(Tan[e + f*x]^2)^(1/4))/(1 + Sqrt[Tan[e + f*x]^2])])*(Ta 
n[e + f*x]^2)^(3/4))/(Sqrt[2]*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3065, 3042, 3055, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {1}{\sqrt {\sin (e+f x)} \sqrt {b \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\sin (e+f x)} \sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3065

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3055

\(\displaystyle -\frac {2 b \int \frac {b \cot (e+f x)}{\cot ^2(e+f x) b^2+b^2}d\frac {\sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}}{f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {2 b \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \cos (e+f x)}}{\sqrt {b} \sqrt {\sin (e+f x)}}\right )}{\sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\log \left (b \cot (e+f x)-\frac {\sqrt {2} \sqrt {b} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}+b\right )}{2 \sqrt {2} \sqrt {b}}-\frac {\log \left (b \cot (e+f x)+\frac {\sqrt {2} \sqrt {b} \sqrt {b \cos (e+f x)}}{\sqrt {\sin (e+f x)}}+b\right )}{2 \sqrt {2} \sqrt {b}}\right )\right )}{f \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}\)

Input:

Int[1/(Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]),x]
 

Output:

(-2*b*((-(ArcTan[1 - (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sqrt[b]*Sqrt[Sin[e + 
f*x]])]/(Sqrt[2]*Sqrt[b])) + ArcTan[1 + (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sq 
rt[b]*Sqrt[Sin[e + f*x]])]/(Sqrt[2]*Sqrt[b]))/2 + (Log[b + b*Cot[e + f*x] 
- (Sqrt[2]*Sqrt[b]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[e + f*x]]]/(2*Sqrt[2]*Sq 
rt[b]) - Log[b + b*Cot[e + f*x] + (Sqrt[2]*Sqrt[b]*Sqrt[b*Cos[e + f*x]])/S 
qrt[Sin[e + f*x]]]/(2*Sqrt[2]*Sqrt[b]))/2))/(f*Sqrt[b*Cos[e + f*x]]*Sqrt[b 
*Sec[e + f*x]])
 

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 217

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

rule 826

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

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

rule 3055

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> With[{k = Denominator[m]}, Simp[(-k)*a*(b/f)   Subst[Int[x 
^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Sin[ 
e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 
0] && LtQ[m, 1]
 

rule 3065

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n   Int[(a*Sin[e 
+ f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int 
egerQ[m - 1/2] && IntegerQ[n - 1/2]
 

Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {\sqrt {2}\, \left (1-\cos \left (f x +e \right )\right ) \left (\ln \left (\frac {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )+2 \sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+2-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )}{1-\cos \left (f x +e \right )}\right )-2 \arctan \left (\frac {-\sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{-1+\cos \left (f x +e \right )}\right )-\ln \left (-\frac {-\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )+2 \sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )-2+2 \cos \left (f x +e \right )+\sin \left (f x +e \right )}{1-\cos \left (f x +e \right )}\right )+2 \arctan \left (\frac {\sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{-1+\cos \left (f x +e \right )}\right )\right )}{4 f \sin \left (f x +e \right )^{\frac {3}{2}} \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}}\)	\(347\)

Input:

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

Output:

-1/4/f*2^(1/2)/sin(f*x+e)^(3/2)*(1-cos(f*x+e))*(ln(1/(1-cos(f*x+e))*((1-co 
s(f*x+e))^2*csc(f*x+e)+2*(-2*sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2) 
*sin(f*x+e)+2-2*cos(f*x+e)-sin(f*x+e)))-2*arctan((-(-2*sin(f*x+e)*cos(f*x+ 
e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/(-1+cos(f*x+e)))-ln(-1 
/(1-cos(f*x+e))*(-(1-cos(f*x+e))^2*csc(f*x+e)+2*(-2*sin(f*x+e)*cos(f*x+e)/ 
(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-2+2*cos(f*x+e)+sin(f*x+e)))+2*arctan((( 
-2*sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/ 
(-1+cos(f*x+e))))/(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*sec(f 
*x+e))^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (190) = 380\).

Time = 0.13 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\frac {\frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )}}{2 \, \sqrt {b} \sqrt {\sin \left (f x + e\right )}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \arctan \left (\frac {2 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) + \frac {\sqrt {2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{\sqrt {b}} - 2}{2 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 1\right )}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \arctan \left (-\frac {2 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - \frac {\sqrt {2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{\sqrt {b}} - 2}{2 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 1\right )}}\right )}{\sqrt {b}} + \frac {\sqrt {2} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{\sqrt {b}} + 4 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 1\right )}{\sqrt {b}} - \frac {\sqrt {2} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{\sqrt {b}} + 4 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 1\right )}{\sqrt {b}}}{8 \, f} \] Input:

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

Output:

1/8*(2*sqrt(2)*arctan(-1/2*sqrt(2)*sqrt(b/cos(f*x + e))*(cos(f*x + e) - si 
n(f*x + e))/(sqrt(b)*sqrt(sin(f*x + e))))/sqrt(b) - sqrt(2)*arctan(1/2*(2* 
cos(f*x + e)^2 - 2*cos(f*x + e)*sin(f*x + e) + sqrt(2)*sqrt(b/cos(f*x + e) 
)*sqrt(sin(f*x + e))/sqrt(b) - 2)/(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + 
 e) - 1))/sqrt(b) - sqrt(2)*arctan(-1/2*(2*cos(f*x + e)^2 - 2*cos(f*x + e) 
*sin(f*x + e) - sqrt(2)*sqrt(b/cos(f*x + e))*sqrt(sin(f*x + e))/sqrt(b) - 
2)/(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e) - 1))/sqrt(b) + sqrt(2)*log 
(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e))*sqrt(b/cos(f*x + e 
))*sqrt(sin(f*x + e))/sqrt(b) + 4*cos(f*x + e)*sin(f*x + e) + 1)/sqrt(b) - 
 sqrt(2)*log(-2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e)*sin(f*x + e))*sqrt( 
b/cos(f*x + e))*sqrt(sin(f*x + e))/sqrt(b) + 4*cos(f*x + e)*sin(f*x + e) + 
 1)/sqrt(b))/f
 

Sympy [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {b \sec {\left (e + f x \right )}} \sqrt {\sin {\left (e + f x \right )}}}\, dx \] Input:

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

Output:

Integral(1/(sqrt(b*sec(e + f*x))*sqrt(sin(e + f*x))), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sqrt {\sin \left (f x + e\right )}} \,d x } \] Input:

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

Output:

integrate(1/(sqrt(b*sec(f*x + e))*sqrt(sin(f*x + e))), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sqrt {\sin \left (f x + e\right )}} \,d x } \] Input:

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

Output:

integrate(1/(sqrt(b*sec(f*x + e))*sqrt(sin(f*x + e))), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sqrt {\sin (e+f x)}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right ) \sin \left (f x +e \right )}d x \right )}{b} \] Input:

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

Output:

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