3.3.54 \(\int \frac {\sec ^3(e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\) [254]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2693, 2694, 2653, 2720} \begin {gather*} \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)}} \end {gather*}

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 2653

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 2693

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 2694

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 2720

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.29, size = 68, normalized size = 0.86 \begin {gather*} \frac {2 \left (\sec ^2(e+f x)+2 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x)}\right ) \sin (e+f x)}{3 f \sqrt {d \tan (e+f x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(195\) vs. \(2(94)=188\).
time = 0.60, size = 196, normalized size = 2.48

method	result	size
default	\(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (2 \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right ) \cos \left (f x +e \right )-\cos \left (f x +e \right ) \sqrt {2}+\sqrt {2}\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {2}}{3 f \sin \left (f x +e \right )^{3} \cos \left (f x +e \right )^{2} \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(196\)

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,method=_RETURNVERBOSE)

[Out]

-1/3/f*(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))*((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)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2
)*sin(f*x+e)*cos(f*x+e)-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)/c
os(f*x+e))^(1/2)*2^(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*}

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 [C] Result contains complex when optimal does not.
time = 0.10, size = 104, normalized size = 1.32 \begin {gather*} -\frac {2 \, {\left (\sqrt {i \, d} \cos \left (f x + e\right ) {\rm ellipticF}\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ), -1\right ) + \sqrt {-i \, d} \cos \left (f x + e\right ) {\rm ellipticF}\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ), -1\right ) - \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}\right )}}{3 \, d f \cos \left (f x + e\right )} \end {gather*}

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]

-2/3*(sqrt(I*d)*cos(f*x + e)*ellipticF(cos(f*x + e) + I*sin(f*x + e), -1) + sqrt(-I*d)*cos(f*x + e)*ellipticF(
cos(f*x + e) - I*sin(f*x + e), -1) - sqrt(d*sin(f*x + e)/cos(f*x + e)))/(d*f*cos(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (e+f\,x\right )}^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^3*(d*tan(e + f*x))^(1/2)),x)

[Out]

int(1/(cos(e + f*x)^3*(d*tan(e + f*x))^(1/2)), x)

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