3.4.24 \(\int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx\) [324]

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

[Out]

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

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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 {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Sec[e + f*x]]/(b*Tan[e + f*x])^(3/2),x]

[Out]

(-2*Sqrt[d*Sec[e + f*x]])/(b*f*Sqrt[b*Tan[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 {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx &=-\frac {2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 32, normalized size = 1.00 \begin {gather*} -\frac {2 \sqrt {d \sec (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Sec[e + f*x]]/(b*Tan[e + f*x])^(3/2),x]

[Out]

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

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Maple [A]
time = 0.33, size = 50, normalized size = 1.56

method	result	size
default	\(-\frac {2 \sin \left (f x +e \right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}}{f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),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))^(3/2)/cos(f*x+e)

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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((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*sec(f*x + e))/(b*tan(f*x + e))^(3/2), x)

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Fricas [A]
time = 0.35, size = 57, normalized size = 1.78 \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^{2} f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),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^2*f*sin(f*x + e))

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Sympy [A]
time = 2.19, size = 53, normalized size = 1.66 \begin {gather*} \begin {cases} - \frac {2 \sqrt {d \sec {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{f \left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}} & \text {for}\: f \neq 0 \\\frac {x \sqrt {d \sec {\left (e \right )}}}{\left (b \tan {\left (e \right )}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(3/2),x)

[Out]

Piecewise((-2*sqrt(d*sec(e + f*x))*tan(e + f*x)/(f*(b*tan(e + f*x))**(3/2)), Ne(f, 0)), (x*sqrt(d*sec(e))/(b*t
an(e))**(3/2), True))

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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((d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*sec(f*x + e))/(b*tan(f*x + e))^(3/2), x)

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Mupad [B]
time = 2.95, size = 46, normalized size = 1.44 \begin {gather*} -\frac {2\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{b\,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((d/cos(e + f*x))^(1/2)/(b*tan(e + f*x))^(3/2),x)

[Out]

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

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