3.4.4 \(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx\) [304]

Optimal. Leaf size=34 \[ \frac {2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

[Out]

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

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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 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(34)=68\).
time = 1.68, size = 141, normalized size = 4.15 \begin {gather*} -\frac {b \sec ^{\frac {3}{2}}(e+f x) \left (\sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {\sec (e+f x)}+\sqrt {\frac {1}{1+\cos (e+f x)}} \cos (3 (e+f x)) \sec ^{\frac {3}{2}}(e+f x)-\sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {1+\sec (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{10 f \sqrt {\frac {1}{1+\cos (e+f x)}} (d \sec (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*(b*Sec[e + f*x]^(3/2)*(Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[Sec[e + f*x]] + Sqrt[(1 + Cos[e + f*x])^(-1)]*
Cos[3*(e + f*x)]*Sec[e + f*x]^(3/2) - Sec[(e + f*x)/2]^2*Sqrt[1 + Sec[e + f*x]])*Sqrt[b*Tan[e + f*x]])/(f*Sqrt
[(1 + Cos[e + f*x])^(-1)]*(d*Sec[e + f*x])^(5/2))

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Maple [A]
time = 0.33, size = 50, normalized size = 1.47

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(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. 63 vs. \(2 (30) = 60\).
time = 0.41, size = 63, normalized size = 1.85 \begin {gather*} -\frac {2 \, {\left (b \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 62.86, size = 53, normalized size = 1.56 \begin {gather*} \begin {cases} \frac {2 \left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}}{5 f \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x \left (b \tan {\left (e \right )}\right )^{\frac {3}{2}}}{\left (d \sec {\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((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(5/2),x)

[Out]

Piecewise((2*(b*tan(e + f*x))**(3/2)*tan(e + f*x)/(5*f*(d*sec(e + f*x))**(5/2)), Ne(f, 0)), (x*(b*tan(e))**(3/
2)/(d*sec(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((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.16, size = 65, normalized size = 1.91 \begin {gather*} \frac {b\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (\cos \left (e+f\,x\right )-\cos \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{10\,d^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*(d/cos(e + f*x))^(1/2)*(cos(e + f*x) - cos(3*e + 3*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2
))/(10*d^3*f)

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