3.232 \(\int \sec (e+f x) \sqrt{d \tan (e+f x)} \, dx\)

Optimal. Leaf size=75 \[ \frac{2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-\frac{2 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{f \sqrt{\sin (2 e+2 f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0899136, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2613, 2615, 2572, 2639} \[ \frac{2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-\frac{2 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{f \sqrt{\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[Sin[2*e + 2*f*x]]) + (2*Cos[e + f*
x]*(d*Tan[e + f*x])^(3/2))/(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 2615

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

Rule 2572

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

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 0.266433, size = 61, normalized size = 0.81 \[ -\frac{2 \sin (e+f x) \sqrt{d \tan (e+f x)} \left (2 \sqrt{\sec ^2(e+f x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(e+f x)\right )-3\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-3 + 2*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[e + f*x]^2]*Sqrt[Sec[e + f*x]^2])*Sin[e + f*x]*Sqrt[d*Tan[e
+ f*x]])/(3*f)

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Maple [B]  time = 0.15, size = 505, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*2^(1/2)*(d*sin(f*x+e)/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(2*EllipticE(((1-cos(f*x+e)+sin(
f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*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)-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)*((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)+2*EllipticE(((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)*((cos(f*x+e)-1+sin(f*x+e
))/sin(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)*((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))/sin(f*x+e)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (f x + e\right )} \sec \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)*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \tan \left (f x + e\right )} \sec \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)*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (f x + e\right )} \sec \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)*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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