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

Optimal. Leaf size=109 \[ \frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{5 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{12 f \sqrt{d \tan (e+f x)}} \]

[Out]

(5*EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(12*f*Sqrt[d*Tan[e + f*x]]) + (5*Cos[e +
f*x]*Sqrt[d*Tan[e + f*x]])/(6*d*f) + (Cos[e + f*x]^3*Sqrt[d*Tan[e + f*x]])/(3*d*f)

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Rubi [A]  time = 0.127342, antiderivative size = 109, 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, 2614, 2573, 2641} \[ \frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{5 \sqrt{\sin (2 e+2 f x)} \sec (e+f x) F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{12 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*EllipticF[e - Pi/4 + f*x, 2]*Sec[e + f*x]*Sqrt[Sin[2*e + 2*f*x]])/(12*f*Sqrt[d*Tan[e + f*x]]) + (5*Cos[e +
f*x]*Sqrt[d*Tan[e + f*x]])/(6*d*f) + (Cos[e + f*x]^3*Sqrt[d*Tan[e + f*x]])/(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 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(e+f x)}{\sqrt{d \tan (e+f x)}} \, dx &=\frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{5}{6} \int \frac{\cos (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{5}{12} \int \frac{\sec (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{\left (5 \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)} \sqrt{\sin (e+f x)}} \, dx}{12 \sqrt{\cos (e+f x)} \sqrt{d \tan (e+f x)}}\\ &=\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}+\frac{\left (5 \sec (e+f x) \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{12 \sqrt{d \tan (e+f x)}}\\ &=\frac{5 F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sec (e+f x) \sqrt{\sin (2 e+2 f x)}}{12 f \sqrt{d \tan (e+f x)}}+\frac{5 \cos (e+f x) \sqrt{d \tan (e+f x)}}{6 d f}+\frac{\cos ^3(e+f x) \sqrt{d \tan (e+f x)}}{3 d f}\\ \end{align*}

Mathematica [C]  time = 0.970626, size = 94, normalized size = 0.86 \[ \frac{11 \sin (e+f x)+\sin (3 (e+f x))-10 \sqrt [4]{-1} \cos (e+f x) \sqrt{\tan (e+f x)} \sqrt{\sec ^2(e+f x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (e+f x)}\right )\right |-1\right )}{12 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sin[e + f*x] + Sin[3*(e + f*x)] - 10*(-1)^(1/4)*Cos[e + f*x]*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[e + f
*x]]], -1]*Sqrt[Sec[e + f*x]^2]*Sqrt[Tan[e + f*x]])/(12*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]  time = 0.174, size = 224, normalized size = 2.1 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( fx+e \right ) -1 \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{12\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) } \left ( 5\,\sin \left ( fx+e \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) -1}{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{\cos \left ( fx+e \right ) -1+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},1/2\,\sqrt{2} \right ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sqrt{2}+2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sqrt{2}-5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{2}+5\,\cos \left ( fx+e \right ) \sqrt{2} \right ){\frac{1}{\sqrt{{\frac{d\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \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)*(5*sin(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))-2*cos(f*x+e)^4*2^(1/2)+2*cos(f*x+e)^3*2^(1/2)-5*cos(f*x+e)^2*2^(1/2)+5*cos(f*x+e)*2^(1/2))*(c
os(f*x+e)+1)^2/sin(f*x+e)^3/cos(f*x+e)/(d*sin(f*x+e)/cos(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )^{3}}{\sqrt{d \tan \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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (f x + e\right )} \cos \left (f x + e\right )^{3}}{d \tan \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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*cos(f*x + e)^3/(d*tan(f*x + e)), 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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )^{3}}{\sqrt{d \tan \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)^3/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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