3.254 \(\int \frac{\sec ^3(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{2 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{3 f \sqrt{d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0959896, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2613, 2614, 2573, 2641} \[ \frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{2 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{3 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3/Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(3*f*Sqrt[d*Tan[e + f*x]]) + (2*Sec[e + f
*x]*Sqrt[d*Tan[e + f*x]])/(3*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 ^3(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx &=\frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{2}{3} \int \frac{\sec (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{\left (2 \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)} \sqrt{\sin (e+f x)}} \, dx}{3 \sqrt{\cos (e+f x)} \sqrt{d \tan (e+f x)}}\\ &=\frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{\left (2 \sec (e+f x) \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{3 \sqrt{d \tan (e+f x)}}\\ &=\frac{2 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sec (e+f x) \sqrt{\sin (2 e+2 f x)}}{3 f \sqrt{d \tan (e+f x)}}+\frac{2 \sec (e+f x) \sqrt{d \tan (e+f x)}}{3 d f}\\ \end{align*}

Mathematica [C]  time = 0.264323, size = 68, normalized size = 0.86 \[ \frac{2 \sin (e+f x) \left (2 \sqrt{\sec ^2(e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(e+f x)\right )+\sec ^2(e+f x)\right )}{3 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^3/Sqrt[d*Tan[e + f*x]],x]

[Out]

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

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Maple [B]  time = 0.171, size = 194, normalized size = 2.5 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( fx+e \right ) -1 \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,{\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 ) \sin \left ( fx+e \right ) \cos \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 ) }}}-\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)^3/(d*tan(f*x+e))^(1/2),x)

[Out]

-1/3/f*2^(1/2)*(cos(f*x+e)-1)*(2*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*sin(f*x+e
)*cos(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)*2^(1/2)+2^(1/2))*(cos(f*x+e)+1)^2/sin(f*x+e)^3/cos(f*x+e)^2/(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 )^{3}}{\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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^3/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 )^{3}}{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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*sec(f*x + e)^3/(d*tan(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\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)**3/(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sec(e + f*x)**3/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 )^{3}}{\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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^3/sqrt(d*tan(f*x + e)), x)