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]

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

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Rubi [A]  time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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 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 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*}

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Mathematica [C]  time = 0.54, 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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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (f x + e\right )} \sec \left (f x + e\right )^{5}}{d \tan \left (f x + e\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{5}}{\sqrt {d \tan \left (f x + e\right )}}\,{d x} \]

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)

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maple [A]  time = 0.71, size = 224, normalized size = 2.06 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \sin \left (f x +e \right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (\cos ^{3}\left (f x +e \right )\right )-2 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-\cos \left (f x +e \right ) \sqrt {2}+\sqrt {2}\right ) \left (1+\cos \left (f x +e \right )\right )^{2} \sqrt {2}}{7 f \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )^{4} \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{5}}{\sqrt {d \tan \left (f x + e\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx \]

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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