3.626 \(\int \frac{a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{3 a d \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(e+f x)\right )}{4 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac{3 b}{f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

(-3*b)/(f*(d*Sec[e + f*x])^(1/3)) - (3*a*d*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[e + f*x]^2]*Sin[e + f*x])/(4*f
*(d*Sec[e + f*x])^(4/3)*Sqrt[Sin[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0598214, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3486, 3772, 2643} \[ -\frac{3 a d \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(e+f x)\right )}{4 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac{3 b}{f \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(1/3),x]

[Out]

(-3*b)/(f*(d*Sec[e + f*x])^(1/3)) - (3*a*d*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[e + f*x]^2]*Sin[e + f*x])/(4*f
*(d*Sec[e + f*x])^(4/3)*Sqrt[Sin[e + f*x]^2])

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx &=-\frac{3 b}{f \sqrt [3]{d \sec (e+f x)}}+a \int \frac{1}{\sqrt [3]{d \sec (e+f x)}} \, dx\\ &=-\frac{3 b}{f \sqrt [3]{d \sec (e+f x)}}+\left (a \left (\frac{\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \sqrt [3]{\frac{\cos (e+f x)}{d}} \, dx\\ &=-\frac{3 b}{f \sqrt [3]{d \sec (e+f x)}}-\frac{3 a \cos ^2(e+f x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{4 d f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.929204, size = 119, normalized size = 1.57 \[ -\frac{3 (a \cot (e+f x)+b) \left (a \sqrt{\sin ^2(e+f x)} \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\sec ^2(e+f x)\right )+b \sin (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)} \left (a \sqrt{\sin ^2(e+f x)} \cot (e+f x)+b \sin (e+f x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(1/3),x]

[Out]

(-3*(b + a*Cot[e + f*x])*(b*Sin[e + f*x] + a*Cot[e + f*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[e + f*x]^2]*Sq
rt[Sin[e + f*x]^2]*Sqrt[-Tan[e + f*x]^2]))/(f*(d*Sec[e + f*x])^(1/3)*(b*Sin[e + f*x] + a*Cot[e + f*x]*Sqrt[Sin
[e + f*x]^2]))

________________________________________________________________________________________

Maple [F]  time = 0.138, size = 0, normalized size = 0. \begin{align*} \int{(a+b\tan \left ( fx+e \right ) ){\frac{1}{\sqrt [3]{d\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)

[Out]

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(1/3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d \sec \left (f x + e\right )\right )^{\frac{2}{3}}{\left (b \tan \left (f x + e\right ) + a\right )}}{d \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

integral((d*sec(f*x + e))^(2/3)*(b*tan(f*x + e) + a)/(d*sec(f*x + e)), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (e + f x \right )}}{\sqrt [3]{d \sec{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(1/3),x)

[Out]

Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(1/3), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(1/3), x)