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

Optimal. Leaf size=78 \[ -\frac{3 a d \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{4}{3},\frac{7}{3},\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}-\frac{3 b}{5 f (d \sec (e+f x))^{5/3}} \]

[Out]

(-3*b)/(5*f*(d*Sec[e + f*x])^(5/3)) - (3*a*d*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[e + f*x]^2]*Sin[e + f*x])/(8
*f*(d*Sec[e + f*x])^(8/3)*Sqrt[Sin[e + f*x]^2])

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Rubi [A]  time = 0.0639063, 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} \[ -\frac{3 a d \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{4}{3},\frac{7}{3},\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}-\frac{3 b}{5 f (d \sec (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b)/(5*f*(d*Sec[e + f*x])^(5/3)) - (3*a*d*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[e + f*x]^2]*Sin[e + f*x])/(8
*f*(d*Sec[e + f*x])^(8/3)*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)}{(d \sec (e+f x))^{5/3}} \, dx &=-\frac{3 b}{5 f (d \sec (e+f x))^{5/3}}+a \int \frac{1}{(d \sec (e+f x))^{5/3}} \, dx\\ &=-\frac{3 b}{5 f (d \sec (e+f x))^{5/3}}+\left (a \sqrt [3]{\frac{\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)}\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{5/3} \, dx\\ &=-\frac{3 b}{5 f (d \sec (e+f x))^{5/3}}-\frac{3 a \cos ^3(e+f x) \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \sin (e+f x)}{8 d^2 f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.433942, size = 94, normalized size = 1.21 \[ \frac{2 a \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{3}{2},\sin ^2(e+f x)\right )+3 \sqrt [3]{\cos ^2(e+f x)} (a \sin (e+f x)-b \cos (e+f x))}{5 d f \sqrt [3]{\cos ^2(e+f x)} (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.094, size = 0, normalized size = 0. \begin{align*} \int{(a+b\tan \left ( fx+e \right ) ) \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{5}{3}}}}\, 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))^(5/3),x)

[Out]

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

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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{5}{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))^(5/3),x, algorithm="maxima")

[Out]

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

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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{1}{3}}{\left (b \tan \left (f x + e\right ) + a\right )}}{d^{2} \sec \left (f x + e\right )^{2}}, 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))^(5/3),x, algorithm="fricas")

[Out]

integral((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a)/(d^2*sec(f*x + e)^2), x)

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

[Out]

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

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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{5}{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))^(5/3),x, algorithm="giac")

[Out]

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