3.4.30 \(\int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{5/2}} \, dx\) [330]

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 = 0.04, 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 = {2685} \begin {gather*} -\frac {2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \end {gather*}

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 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 {(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 = 0.17, size = 34, normalized size = 1.00 \begin {gather*} -\frac {2 (d \sec (e+f x))^{3/2}}{3 b f (b \tan (e+f x))^{3/2}} \end {gather*}

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 = 0.30, size = 50, normalized size = 1.47

method	result	size
default	\(-\frac {2 \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )}{3 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )}\)	\(50\)

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,method=_RETURNVERBOSE)

[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 = 0.00, 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))^(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] Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (30) = 60\).
time = 0.37, size = 66, normalized size = 1.94 \begin {gather*} \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 {gather*}

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 [A]
time = 53.83, size = 54, normalized size = 1.59 \begin {gather*} \begin {cases} - \frac {2 \left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x \left (d \sec {\left (e \right )}\right )^{\frac {3}{2}}}{\left (b \tan {\left (e \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((-2*(d*sec(e + f*x))**(3/2)*tan(e + f*x)/(3*f*(b*tan(e + f*x))**(5/2)), Ne(f, 0)), (x*(d*sec(e))**(3
/2)/(b*tan(e))**(5/2), 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))^(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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Mupad [B]
time = 3.17, size = 55, normalized size = 1.62 \begin {gather*} -\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}}} \end {gather*}

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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