3.624 \(\int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=78 \[ \frac{3 a d \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{2 f \sqrt{\sin ^2(e+f x)}}+\frac{3 b (d \sec (e+f x))^{5/3}}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.0642202, antiderivative size = 78, 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} \[ \frac{3 a d \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{2 f \sqrt{\sin ^2(e+f x)}}+\frac{3 b (d \sec (e+f x))^{5/3}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x]),x]

[Out]

(3*b*(d*Sec[e + f*x])^(5/3))/(5*f) + (3*a*d*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[e + f*x]^2]*(d*Sec[e + f*x])
^(2/3)*Sin[e + f*x])/(2*f*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 (d \sec (e+f x))^{5/3} (a+b \tan (e+f x)) \, dx &=\frac{3 b (d \sec (e+f x))^{5/3}}{5 f}+a \int (d \sec (e+f x))^{5/3} \, dx\\ &=\frac{3 b (d \sec (e+f x))^{5/3}}{5 f}+\left (a \left (\frac{\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (e+f x)}{d}\right )^{5/3}} \, dx\\ &=\frac{3 b (d \sec (e+f x))^{5/3}}{5 f}+\frac{3 a d \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{2 f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.506981, size = 126, normalized size = 1.62 \[ \frac{d (d \sec (e+f x))^{2/3} (a+b \tan (e+f x)) \left (3 \cos ^2(e+f x)^{2/3} (5 a \sin (2 (e+f x))+4 b)-10 a \sin (e+f x) \cos ^3(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{3}{2},\sin ^2(e+f x)\right )\right )}{20 f \cos ^2(e+f x)^{2/3} (a \cos (e+f x)+b \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x]),x]

[Out]

(d*(d*Sec[e + f*x])^(2/3)*(-10*a*Cos[e + f*x]^3*Hypergeometric2F1[1/3, 1/2, 3/2, Sin[e + f*x]^2]*Sin[e + f*x]
+ 3*(Cos[e + f*x]^2)^(2/3)*(4*b + 5*a*Sin[2*(e + f*x)]))*(a + b*Tan[e + f*x]))/(20*f*(Cos[e + f*x]^2)^(2/3)*(a
*Cos[e + f*x] + b*Sin[e + f*x]))

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Maple [F]  time = 0.1, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{5}{3}}} \left ( a+b\tan \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x)

[Out]

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{5}{3}}{\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*d*sec(f*x + e)*tan(f*x + e) + a*d*sec(f*x + e))*(d*sec(f*x + e))^(2/3), 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((d*sec(f*x+e))**(5/3)*(a+b*tan(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{5}{3}}{\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a), x)