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]

-2/3*(d*sec(f*x+e))^(3/2)/b/f/(b*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 34, 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 = {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*}

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Mathematica [A]  time = 0.15, size = 34, normalized size = 1.00 \[ -\frac {2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \]

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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fricas [B]  time = 0.53, size = 61, normalized size = 1.79 \[ \frac {2 \, d \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.49, size = 50, normalized size = 1.47 \[ -\frac {2 \sin \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{3 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [B]  time = 3.17, size = 55, normalized size = 1.62 \[ -\frac {2\,d\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{3\,b^2\,f\,\sin \left (e+f\,x\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d*(d/cos(e + f*x))^(1/2))/(3*b^2*f*sin(e + f*x)*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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