∫ 1_____________∘ d sec(e + fx) (b tan(e+fx))5∕2 dx [332]

Optimal. Leaf size=69 \[ -\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \]

-8/3*(b*tan(f*x+e))^(1/2)/b^3/f/(d*sec(f*x+e))^(1/2)-2/3/b/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2689, 2685} \begin {gather*} -\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}} \end {gather*}

-2/(3*b*f*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2)) - (8*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*Sqrt[d*Sec[e + f*x]

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]

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*(n + 1))), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(

b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

\begin {align*} \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx &=-\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {4 \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.91, size = 110, normalized size = 1.59 \begin {gather*} -\frac {2 \left (\sqrt {\frac {1}{1+\cos (e+f x)}} \csc (e+f x) \sec (e+f x)+3 \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )\right )}{3 b^2 f \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

(-2*(Sqrt[(1 + Cos[e + f*x])^(-1)]*Csc[e + f*x]*Sec[e + f*x] + 3*Sqrt[Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]*Tan

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

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method	result	size

default	\(\frac {2 \sin \left (f x +e \right ) \left (3 \left (\cos ^{2}\left (f x +e \right )\right )-4\right )}{3 f \cos \left (f x +e \right )^{3} \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(62\)

int(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

2/3/f*sin(f*x+e)*(3*cos(f*x+e)^2-4)/cos(f*x+e)^3/(d/cos(f*x+e))^(1/2)/(b*sin(f*x+e)/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*}

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

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Fricas [A]
time = 0.43, size = 81, normalized size = 1.17 \begin {gather*} -\frac {2 \, {\left (3 \, \cos \left (f x + e\right )^{3} - 4 \, \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 )}}}{3 \, {\left (b^{3} d f \cos \left (f x + e\right )^{2} - b^{3} d f\right )}} \end {gather*}

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

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

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Sympy [A]
time = 72.88, size = 92, normalized size = 1.33 \begin {gather*} \begin {cases} - \frac {8 \tan ^{3}{\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} - \frac {2 \tan {\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} & \text {for}\: f \neq 0 \\\frac {x}{\left (b \tan {\left (e \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e \right )}}} & \text {otherwise} \end {cases} \end {gather*}

Piecewise((-8*tan(e + f*x)**3/(3*f*(b*tan(e + f*x))**(5/2)*sqrt(d*sec(e + f*x))) - 2*tan(e + f*x)/(3*f*(b*tan(

e + f*x))**(5/2)*sqrt(d*sec(e + f*x))), Ne(f, 0)), (x/((b*tan(e))**(5/2)*sqrt(d*sec(e))), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")

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Mupad [B]
time = 3.56, size = 81, normalized size = 1.17 \begin {gather*} \frac {\left (\frac {13\,\sin \left (e+f\,x\right )}{3}-\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{b^2\,d\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

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

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3.4.32 \(\int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx\) [332]