3.4.13 \(\int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{7/2}} \, dx\) [313]

Optimal. Leaf size=34 \[ \frac {2 (b \tan (e+f x))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}} \]

[Out]

2/7*(b*tan(f*x+e))^(7/2)/b/f/(d*sec(f*x+e))^(7/2)

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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))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*(b*Tan[e + f*x])^(7/2))/(7*b*f*(d*Sec[e + f*x])^(7/2))

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))^{5/2}}{(d \sec (e+f x))^{7/2}} \, dx &=\frac {2 (b \tan (e+f x))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 45, normalized size = 1.32 \begin {gather*} \frac {2 b^2 \sin ^3(e+f x) \sqrt {b \tan (e+f x)}}{7 d^3 f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*b^2*Sin[e + f*x]^3*Sqrt[b*Tan[e + f*x]])/(7*d^3*f*Sqrt[d*Sec[e + f*x]])

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Maple [A]
time = 0.34, size = 50, normalized size = 1.47

method	result	size
default	\(\frac {2 \sin \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{7 f \cos \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/7/f*sin(f*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(5/2)/cos(f*x+e)/(d/cos(f*x+e))^(7/2)

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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))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(7/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (30) = 60\).
time = 0.38, size = 74, normalized size = 2.18 \begin {gather*} -\frac {2 \, {\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \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 )}} \sin \left (f x + e\right )}{7 \, d^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-2/7*(b^2*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x +
e)/(d^4*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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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))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(7/2), x)

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Mupad [B]
time = 3.21, size = 72, normalized size = 2.12 \begin {gather*} \frac {b^2\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (2\,\sin \left (2\,e+2\,f\,x\right )-\sin \left (4\,e+4\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{28\,d^4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(5/2)/(d/cos(e + f*x))^(7/2),x)

[Out]

(b^2*(d/cos(e + f*x))^(1/2)*(2*sin(2*e + 2*f*x) - sin(4*e + 4*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) +
1))^(1/2))/(28*d^4*f)

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