3.235 \(\int \cos ^5(e+f x) \sqrt{d \tan (e+f x)} \, dx\)

Optimal. Leaf size=111 \[ \frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{7 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{20 f \sqrt{\sin (2 e+2 f x)}} \]

[Out]

(7*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(20*f*Sqrt[Sin[2*e + 2*f*x]]) + (7*Cos[e +
f*x]^3*(d*Tan[e + f*x])^(3/2))/(30*d*f) + (Cos[e + f*x]^5*(d*Tan[e + f*x])^(3/2))/(5*d*f)

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Rubi [A]  time = 0.139029, antiderivative size = 111, 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 = {2612, 2615, 2572, 2639} \[ \frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{7 \cos (e+f x) E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (e+f x)}}{20 f \sqrt{\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5*Sqrt[d*Tan[e + f*x]],x]

[Out]

(7*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(20*f*Sqrt[Sin[2*e + 2*f*x]]) + (7*Cos[e +
f*x]^3*(d*Tan[e + f*x])^(3/2))/(30*d*f) + (Cos[e + f*x]^5*(d*Tan[e + f*x])^(3/2))/(5*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 ^5(e+f x) \sqrt{d \tan (e+f x)} \, dx &=\frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{7}{10} \int \cos ^3(e+f x) \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{7}{20} \int \cos (e+f x) \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{\left (7 \sqrt{\cos (e+f x)} \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \sqrt{\sin (e+f x)} \, dx}{20 \sqrt{\sin (e+f x)}}\\ &=\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac{\left (7 \cos (e+f x) \sqrt{d \tan (e+f x)}\right ) \int \sqrt{\sin (2 e+2 f x)} \, dx}{20 \sqrt{\sin (2 e+2 f x)}}\\ &=\frac{7 \cos (e+f x) E\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{d \tan (e+f x)}}{20 f \sqrt{\sin (2 e+2 f x)}}+\frac{7 \cos ^3(e+f x) (d \tan (e+f x))^{3/2}}{30 d f}+\frac{\cos ^5(e+f x) (d \tan (e+f x))^{3/2}}{5 d f}\\ \end{align*}

Mathematica [C]  time = 0.749883, size = 86, normalized size = 0.77 \[ \frac{\cos (e+f x) \sqrt{d \tan (e+f x)} \left (28 \tan (e+f x) \sqrt{\sec ^2(e+f x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(e+f x)\right )+20 \sin (2 (e+f x))+3 \sin (4 (e+f x))\right )}{120 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5*Sqrt[d*Tan[e + f*x]],x]

[Out]

(Cos[e + f*x]*Sqrt[d*Tan[e + f*x]]*(20*Sin[2*(e + f*x)] + 3*Sin[4*(e + f*x)] + 28*Hypergeometric2F1[3/4, 3/2,
7/4, -Tan[e + f*x]^2]*Sqrt[Sec[e + f*x]^2]*Tan[e + f*x]))/(120*f)

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Maple [B]  time = 0.171, size = 542, normalized size = 4.9 \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)^5*(d*tan(f*x+e))^(1/2),x)

[Out]

-1/120/f*2^(1/2)*(cos(f*x+e)-1)^2*(12*2^(1/2)*cos(f*x+e)^6+2*cos(f*x+e)^4*2^(1/2)+42*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)-21*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/si
n(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)+42*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+s
in(f*x+e))/sin(f*x+e))^(1/2)-21*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)+7*cos(f*x+e)^2*2^(1/2)-21*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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*cos(f*x + e)^5, 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 )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*cos(f*x + e)^5, 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)**5*(d*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(f*x + e))*cos(f*x + e)^5, x)