3.330 \(\int \frac{(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=34 \[ -\frac{2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0568758, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2605} \[ -\frac{2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(3/2)/(b*Tan[e + f*x])^(5/2),x]

[Out]

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

Rule 2605

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{(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{5/2}} \, dx &=-\frac{2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.156271, size = 34, normalized size = 1. \[ -\frac{2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(3/2)/(b*Tan[e + f*x])^(5/2),x]

[Out]

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

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Maple [A]  time = 0.133, size = 50, normalized size = 1.5 \begin{align*} -{\frac{2\,\sin \left ( fx+e \right ) }{3\,f\cos \left ( fx+e \right ) } \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x)

[Out]

-2/3/f*(d/cos(f*x+e))^(3/2)*sin(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e))^(5/2), x)

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Fricas [B]  time = 1.62757, size = 143, normalized size = 4.21 \begin{align*} \frac{2 \, d \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 )}{3 \,{\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/3*d*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)/(b^3*f*cos(f*x + e)^2 - b^3*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(3/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e))^(5/2), x)