\(\int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \, dx\) [131]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \, dx=\frac {\arctan \left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\text {arctanh}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {a \sin (e+f x)}}{a f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \, dx=\frac {\left (\arctan \left (\sqrt [4]{\cos ^2(e+f x)}\right )-\text {arctanh}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right ) \sin (2 (e+f x))}{2 f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.69, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3081, 27, 3042, 3045, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-2*(-1/2*ArcTan[Sqrt[Cos[e + f*x]]] + ArcTanh[Sqrt[Cos[e + f*x]]]/2)*Sqrt 
[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])
 

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 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 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3081

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ 
n)   Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 
1)]) || IntegersQ[m - 1/2, n - 1/2])
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.48

method	result	size
default	\(-\frac {\left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\right )+\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\cos \left (f x +e \right )+1}{1+\cos \left (f x +e \right )}\right )\right ) \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}{2 f \sqrt {a \sin \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b \tan \left (f x +e \right )}}\)	\(157\)

Input:

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

Output:

-1/2/f*(arctan(1/2/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2))+ln((2*cos(f*x+e)* 
(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2 
)-cos(f*x+e)+1)/(1+cos(f*x+e))))/(a*sin(f*x+e))^(1/2)/(-cos(f*x+e)/(1+cos( 
f*x+e))^2)^(1/2)/(b*tan(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (90) = 180\).

Time = 0.28 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.95 \[ \int \frac {1}{\sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \, dx=\left [\frac {2 \, \sqrt {-a b} \arctan \left (\frac {2 \, \sqrt {-a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt {-a b} \log \left (-\frac {a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )^{2} + 4 \, \sqrt {-a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a b \cos \left (f x + e\right ) + a b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, a b f}, -\frac {2 \, \sqrt {a b} \arctan \left (\frac {2 \, \sqrt {a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) - a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt {a b} \log \left (\frac {4 \, \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - {\left (a b \cos \left (f x + e\right )^{2} + 6 \, a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right )}{4 \, a b f}\right ] \] Input:

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

Output:

[1/4*(2*sqrt(-a*b)*arctan(2*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x 
 + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) + a*b)*sin(f*x + e))) 
- sqrt(-a*b)*log(-(a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e)^2 + 4*sqrt(-a*b 
)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin( 
f*x + e) - 5*a*b*cos(f*x + e) + a*b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 
3*cos(f*x + e) + 1)))/(a*b*f), -1/4*(2*sqrt(a*b)*arctan(2*sqrt(a*b)*sqrt(a 
*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f* 
x + e) - a*b)*sin(f*x + e))) - sqrt(a*b)*log((4*sqrt(a*b)*(cos(f*x + e)^2 
+ cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e)) - ( 
a*b*cos(f*x + e)^2 + 6*a*b*cos(f*x + e) + a*b)*sin(f*x + e))/((cos(f*x + e 
)^2 - 2*cos(f*x + e) + 1)*sin(f*x + e))))/(a*b*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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