∫ 1_____________ (d sec(e+fx))5∕2∘ _______

Optimal. Leaf size=72 \[ \frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}+\frac {8 \sqrt {b \tan (e+f x)}}{5 b d^2 f \sqrt {d \sec (e+f x)}} \]

2/5*(b*tan(f*x+e))^(1/2)/b/f/(d*sec(f*x+e))^(5/2)+8/5*(b*tan(f*x+e))^(1/2)/b/d^2/f/(d*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 72, 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 = {2692, 2685} \begin {gather*} \frac {8 \sqrt {b \tan (e+f x)}}{5 b d^2 f \sqrt {d \sec (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}} \end {gather*}

(2*Sqrt[b*Tan[e + f*x]])/(5*b*f*(d*Sec[e + f*x])^(5/2)) + (8*Sqrt[b*Tan[e + f*x]])/(5*b*d^2*f*Sqrt[d*Sec[e + f

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

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

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

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Mathematica [A]
time = 1.27, size = 112, normalized size = 1.56 \begin {gather*} \frac {9 \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {\frac {1}{1+\cos (e+f x)}} \cos (2 (e+f x)) \tan (e+f x)}{5 d^2 f \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

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

*Tan[e + f*x])/(5*d^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	\(\frac {2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )+4\right )}{5 f \cos \left (f x +e \right )^{3} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(60\)

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

2/5/f*sin(f*x+e)*(cos(f*x+e)^2+4)/cos(f*x+e)^3/(d/cos(f*x+e))^(5/2)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)

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

integrate(1/((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e))), x)

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Fricas [A]
time = 0.37, size = 63, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\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 )}}}{5 \, b d^{3} f} \end {gather*}

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

2/5*(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*d^3*f)

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Sympy [A]
time = 72.09, size = 88, normalized size = 1.22 \begin {gather*} \begin {cases} \frac {8 \tan ^{3}{\left (e + f x \right )}}{5 f \sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} + \frac {2 \tan {\left (e + f x \right )}}{f \sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x}{\sqrt {b \tan {\left (e \right )}} \left (d \sec {\left (e \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

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

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

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

integrate(1/((d*sec(f*x + e))^(5/2)*sqrt(b*tan(f*x + e))), x)

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Mupad [B]
time = 3.02, size = 64, normalized size = 0.89 \begin {gather*} \frac {\left (17\,\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{10\,d^3\,f\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

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

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3.4.21 \(\int \frac {1}{(d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx\) [321]