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

Optimal. Leaf size=107 \[ \frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac{4 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{5 f \sqrt{\sin (2 e+2 f x)}} \]

[Out]

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

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Rubi [A]  time = 0.129166, antiderivative size = 107, 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, 2615, 2572, 2639} \[ \frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac{4 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{5 f \sqrt{\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(5*f*Sqrt[Sin[2*e + 2*f*x]]) + (4*Cos[e +
f*x]*(d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*Sec[e + f*x]*(d*Tan[e + f*x])^(3/2))/(5*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 ^3(e+f x) \sqrt{d \tan (e+f x)} \, dx &=\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2}{5} \int \sec (e+f x) \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac{4}{5} \int \cos (e+f x) \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac{\left (4 \sqrt{\cos (e+f x)} \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \sqrt{\sin (e+f x)} \, dx}{5 \sqrt{\sin (e+f x)}}\\ &=\frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac{\left (4 \cos (e+f x) \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\sin (2 e+2 f x)} \, dx}{5 \sqrt{\sin (2 e+2 f x)}}\\ &=-\frac{4 \cos (e+f x) E\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{d \tan (e+f x)}}{5 f \sqrt{\sin (2 e+2 f x)}}+\frac{4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}\\ \end{align*}

Mathematica [C]  time = 0.42256, size = 102, normalized size = 0.95 \[ \frac{2 \sqrt{d \tan (e+f x)} \left (3 \sqrt{\sec ^2(e+f x)} (2 \sin (e+f x)+\tan (e+f x) \sec (e+f x))-4 \tan (e+f x) \sec (e+f x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(e+f x)\right )\right )}{15 f \sqrt{\sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*Tan[e + f*x]]*(-4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[e + f*x]^2]*Sec[e + f*x]*Tan[e + f*x] + 3*Sq
rt[Sec[e + f*x]^2]*(2*Sin[e + f*x] + Sec[e + f*x]*Tan[e + f*x])))/(15*f*Sqrt[Sec[e + f*x]^2])

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Maple [B]  time = 0.16, size = 551, normalized size = 5.2 \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)^3*(d*tan(f*x+e))^(1/2),x)

[Out]

-1/5/f*2^(1/2)*(cos(f*x+e)-1)^2*(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)^3*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))
^(1/2),1/2*2^(1/2))-4*((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*EllipticE(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^
(1/2))+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*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-4*((co
s(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*EllipticE(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+2*cos(f*x+e)^3*2^(1
/2)-cos(f*x+e)^2*2^(1/2)-2^(1/2))*(cos(f*x+e)+1)^2*(d*sin(f*x+e)/cos(f*x+e))^(1/2)/sin(f*x+e)^5/cos(f*x+e)^2

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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 )^{3}\,{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(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, 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 )^{3}, 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, 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 ^{3}{\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(sqrt(d*tan(e + f*x))*sec(e + f*x)**3, 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 )^{3}\,{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(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)