3.4.19 \(\int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx\) [319]

Optimal. Leaf size=32 \[ \frac {2 \sqrt {b \tan (e+f x)}}{b f \sqrt {d \sec (e+f x)}} \]

[Out]

2*(b*tan(f*x+e))^(1/2)/b/f/(d*sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2685} \begin {gather*} \frac {2 \sqrt {b \tan (e+f x)}}{b f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]])

Rule 2685

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-(a*Sec[e
+ f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx &=\frac {2 \sqrt {b \tan (e+f x)}}{b f \sqrt {d \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.44, size = 32, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {b \tan (e+f x)}}{b f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(b*f*Sqrt[d*Sec[e + f*x]])

________________________________________________________________________________________

Maple [A]
time = 0.34, size = 50, normalized size = 1.56

method	result	size
default	\(\frac {2 \sin \left (f x +e \right )}{f \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )}\)	\(50\)
risch	\(-\frac {i \sqrt {2}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{\sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 51, normalized size = 1.59 \begin {gather*} \frac {2 \, \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)/(b*d*f)

________________________________________________________________________________________

Sympy [A]
time = 10.99, size = 51, normalized size = 1.59 \begin {gather*} \begin {cases} \frac {2 \tan {\left (e + f x \right )}}{f \sqrt {b \tan {\left (e + f x \right )}} \sqrt {d \sec {\left (e + f x \right )}}} & \text {for}\: f \neq 0 \\\frac {x}{\sqrt {b \tan {\left (e \right )}} \sqrt {d \sec {\left (e \right )}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*tan(e + f*x)/(f*sqrt(b*tan(e + f*x))*sqrt(d*sec(e + f*x))), Ne(f, 0)), (x/(sqrt(b*tan(e))*sqrt(d*
sec(e))), True))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 2.90, size = 52, normalized size = 1.62 \begin {gather*} \frac {2\,\sin \left (e+f\,x\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{d\,f\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________