3.233 \(\int \cos (e+f x) \sqrt{d \tan (e+f x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{\cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{f \sqrt{\sin (2 e+2 f x)}} \]

[Out]

(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]])

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Rubi [A]  time = 0.0649298, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2615, 2572, 2639} \[ \frac{\cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{f \sqrt{\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]])

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 \cos (e+f x) \sqrt{d \tan (e+f x)} \, dx &=\frac{\left (\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{\left (\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{\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)}}\\ \end{align*}

Mathematica [C]  time = 0.094254, size = 57, normalized size = 1.21 \[ \frac{2 \sin (e+f x) \sqrt{\sec ^2(e+f x)} \sqrt{d \tan (e+f x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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 = 0.157, size = 523, normalized size = 11.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*2^(1/2)*(cos(f*x+e)-1)^2*(2*cos(f*x+e)*EllipticE((-(cos(f*x+e)-1-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)*((cos(f*x+e)-1+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)*EllipticF((-(cos(f*x+e)-1-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)*((cos(f*x+e)-1+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((-(cos(f*x+e)-1-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)*
((cos(f*x+e)-1+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((-(cos(f*
x+e)-1-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)*((cos(f*x+e)-1+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*2^(1/2)-cos(f*x+e)*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*cos(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 )} \cos \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*cos(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 )}} \cos{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*tan(e + f*x))*cos(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 )} \cos \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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