\(\int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx\) [452]

Optimal result

Integrand size = 25, antiderivative size = 275 \[ \int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx=-\frac {\sqrt {a} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{\sqrt {2} \sqrt {b} f}+\frac {\sqrt {a} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {a} \sqrt {b \cos (e+f x)}}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{\sqrt {2} \sqrt {b} f}-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {a \sin (e+f x)}}{\sqrt {b \cos (e+f x)} \left (\sqrt {a}+\sqrt {a} \tan (e+f x)\right )}\right ) \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}{\sqrt {2} \sqrt {b} f} \] Output:

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

Mathematica [A] (verified)

Time = 1.52 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.44 \[ \int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx=\frac {\left (\arctan \left (\frac {-1+\sqrt {\tan ^2(e+f x)}}{\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}\right )-\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{\tan ^2(e+f x)}}{1+\sqrt {\tan ^2(e+f x)}}\right )\right ) \cot (e+f x) \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \sqrt [4]{\tan ^2(e+f x)}}{\sqrt {2} f} \] Input:

Integrate[Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]],x]
 

Output:

((ArcTan[(-1 + Sqrt[Tan[e + f*x]^2])/(Sqrt[2]*(Tan[e + f*x]^2)^(1/4))] - A 
rcTanh[(Sqrt[2]*(Tan[e + f*x]^2)^(1/4))/(1 + Sqrt[Tan[e + f*x]^2])])*Cot[e 
 + f*x]*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]]*(Tan[e + f*x]^2)^(1/4))/ 
(Sqrt[2]*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3065, 3042, 3054, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3065

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3054

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]],x]
 

Output:

(2*a*b*Sqrt[b*Cos[e + f*x]]*((-(ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + 
 f*x]])/(Sqrt[a]*Sqrt[b*Cos[e + f*x]])]/(Sqrt[2]*Sqrt[a]*Sqrt[b])) + ArcTa 
n[1 + (Sqrt[2]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/(Sqrt[a]*Sqrt[b*Cos[e + f*x]] 
)]/(Sqrt[2]*Sqrt[a]*Sqrt[b]))/(2*b) - (-1/2*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[ 
b]*Sqrt[a*Sin[e + f*x]])/Sqrt[b*Cos[e + f*x]] + a*Tan[e + f*x]]/(Sqrt[2]*S 
qrt[a]*Sqrt[b]) + Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[b]*Sqrt[a*Sin[e + f*x]])/S 
qrt[b*Cos[e + f*x]] + a*Tan[e + f*x]]/(2*Sqrt[2]*Sqrt[a]*Sqrt[b]))/(2*b))* 
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 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 3054

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), 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*Sin[e + f*x])^(1/k)/(b*Cos[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 = 3.52 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {a \sin \left (f x +e \right )}\, \sqrt {b \sec \left (f x +e \right )}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (\ln \left (\frac {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )+2 \sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+2-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )}{1-\cos \left (f x +e \right )}\right )+2 \arctan \left (\frac {-\sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{-1+\cos \left (f x +e \right )}\right )-\ln \left (-\frac {-\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )+2 \sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )-2+2 \cos \left (f x +e \right )+\sin \left (f x +e \right )}{1-\cos \left (f x +e \right )}\right )-2 \arctan \left (\frac {\sqrt {-\frac {2 \sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{-1+\cos \left (f x +e \right )}\right )\right )}{8 f \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(362\)

Input:

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

Output:

1/8/f*2^(1/2)*(a*sin(f*x+e))^(1/2)*(b*sec(f*x+e))^(1/2)*((1-cos(f*x+e))^2* 
csc(f*x+e)^2-1)*(ln(1/(1-cos(f*x+e))*((1-cos(f*x+e))^2*csc(f*x+e)+2*(-2*si 
n(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)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (214) = 428\).

Time = 0.16 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.69 \[ \int \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)} \, dx=-\frac {2 \, \sqrt {2} \sqrt {a b} \arctan \left (-\frac {\sqrt {2} \sqrt {a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a b \cos \left (f x + e\right ) - a b \sin \left (f x + e\right )}\right ) + \sqrt {2} \sqrt {a b} \arctan \left (\frac {2 \, a b \cos \left (f x + e\right )^{2} - 2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a b + \sqrt {2} \sqrt {a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a b\right )}}\right ) + \sqrt {2} \sqrt {a b} \arctan \left (-\frac {2 \, a b \cos \left (f x + e\right )^{2} - 2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a b - \sqrt {2} \sqrt {a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a b\right )}}\right ) + \sqrt {2} \sqrt {a b} \log \left (4 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, \sqrt {2} \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + a b\right ) - \sqrt {2} \sqrt {a b} \log \left (4 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \sqrt {2} \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + a b\right )}{8 \, f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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