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

Optimal. Leaf size=81 \[ \frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}+\frac{\cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{2 f \sqrt{\sin (2 e+2 f x)}} \]

[Out]

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

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Rubi [A]  time = 0.101796, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2612, 2615, 2572, 2639} \[ \frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}+\frac{\cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{2 f \sqrt{\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^3*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]])/(2*f*Sqrt[Sin[2*e + 2*f*x]]) + (Cos[e + f*x]^
3*(d*Tan[e + f*x])^(3/2))/(3*d*f)

Rule 2612

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[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 \cos ^3(e+f x) \sqrt{d \tan (e+f x)} \, dx &=\frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}+\frac{1}{2} \int \cos (e+f x) \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}+\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}{2 \sqrt{\sin (e+f x)}}\\ &=\frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}+\frac{\left (\cos (e+f x) \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\sin (2 e+2 f x)} \, dx}{2 \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)}}{2 f \sqrt{\sin (2 e+2 f x)}}+\frac{\cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{3 d f}\\ \end{align*}

Mathematica [C]  time = 0.4308, size = 94, normalized size = 1.16 \[ \frac{\sqrt{d \tan (e+f x)} \left (4 \tan (e+f x) \sec (e+f x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(e+f x)\right )+(\sin (e+f x)+\sin (3 (e+f x))) \sqrt{\sec ^2(e+f x)}\right )}{12 f \sqrt{\sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*Tan[e + f*x]]*(Sqrt[Sec[e + f*x]^2]*(Sin[e + f*x] + Sin[3*(e + f*x)]) + 4*Hypergeometric2F1[3/4, 3/2,
7/4, -Tan[e + f*x]^2]*Sec[e + f*x]*Tan[e + f*x]))/(12*f*Sqrt[Sec[e + f*x]^2])

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Maple [B]  time = 0.192, size = 536, normalized size = 6.6 \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)^3*(d*tan(f*x+e))^(1/2),x)

[Out]

-1/12/f*2^(1/2)*(cos(f*x+e)-1)^2*(2*cos(f*x+e)^4*2^(1/2)-3*cos(f*x+e)*EllipticF((-(cos(f*x+e)-1-sin(f*x+e))/si
n(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)+6*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)-3*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)+6*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)^2*2^(1
/2)-3*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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*cos(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 )} \cos \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*cos(f*x + e)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError