3.253 \(\int \frac{\sec ^5(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=109 \[ \frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{4 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{7 f \sqrt{d \tan (e+f x)}} \]

[Out]

(4*EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(7*f*Sqrt[d*Tan[e + f*x]]) + (4*Sec[e + f
*x]*Sqrt[d*Tan[e + f*x]])/(7*d*f) + (2*Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]])/(7*d*f)

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Rubi [A]  time = 0.136933, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2613, 2614, 2573, 2641} \[ \frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{4 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{7 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^5/Sqrt[d*Tan[e + f*x]],x]

[Out]

(4*EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(7*f*Sqrt[d*Tan[e + f*x]]) + (4*Sec[e + f
*x]*Sqrt[d*Tan[e + f*x]])/(7*d*f) + (2*Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]])/(7*d*f)

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^5(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx &=\frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{6}{7} \int \frac{\sec ^3(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{4}{7} \int \frac{\sec (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{\left (4 \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)} \sqrt{\sin (e+f x)}} \, dx}{7 \sqrt{\cos (e+f x)} \sqrt{d \tan (e+f x)}}\\ &=\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{\left (4 \sec (e+f x) \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{7 \sqrt{d \tan (e+f x)}}\\ &=\frac{4 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sec (e+f x) \sqrt{\sin (2 e+2 f x)}}{7 f \sqrt{d \tan (e+f x)}}+\frac{4 \sec (e+f x) \sqrt{d \tan (e+f x)}}{7 d f}+\frac{2 \sec ^3(e+f x) \sqrt{d \tan (e+f x)}}{7 d f}\\ \end{align*}

Mathematica [C]  time = 0.541142, size = 79, normalized size = 0.72 \[ \frac{2 \sin (e+f x) \left (4 \sqrt{\sec ^2(e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(e+f x)\right )+(\cos (2 (e+f x))+2) \sec ^4(e+f x)\right )}{7 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^5/Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*((2 + Cos[2*(e + f*x)])*Sec[e + f*x]^4 + 4*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[e + f*x]^2]*Sqrt[Sec[e + f
*x]^2])*Sin[e + f*x])/(7*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]  time = 0.196, size = 222, normalized size = 2. \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( fx+e \right ) -1 \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{7\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{4}} \left ( 4\,\sin \left ( fx+e \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) -1}{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{\cos \left ( fx+e \right ) -1+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}} \left ( \cos \left ( fx+e \right ) \right ) ^{3}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},1/2\,\sqrt{2} \right ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sqrt{2}+2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{2}-\cos \left ( fx+e \right ) \sqrt{2}+\sqrt{2} \right ){\frac{1}{\sqrt{{\frac{d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7/f*2^(1/2)*(cos(f*x+e)-1)*(4*sin(f*x+e)*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((1-cos(f*x+e)+sin(f*x+e))/sin(f
*x+e))^(1/2)*((cos(f*x+e)-1+sin(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)^3*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))^(1/2),1/2*2^(1/2))-2*cos(f*x+e)^3*2^(1/2)+2*cos(f*x+e)^2*2^(1/2)-cos(f*x+e)*2^(1/2)+2^(1/2))*(cos(f*
x+e)+1)^2/sin(f*x+e)^3/cos(f*x+e)^4/(d*sin(f*x+e)/cos(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{5}}{\sqrt{d \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (f x + e\right )} \sec \left (f x + e\right )^{5}}{d \tan \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(f*x + e))*sec(f*x + e)^5/(d*tan(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (e + f x \right )}}{\sqrt{d \tan{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(e + f*x)**5/sqrt(d*tan(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{5}}{\sqrt{d \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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