3.628 \(\int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=119 \[ \frac{3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]

[Out]

(33*a*b*(d*Sec[e + f*x])^(5/3))/(40*f) + (3*(8*a^2 - 3*b^2)*d*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[e + f*x]^2
]*(d*Sec[e + f*x])^(2/3)*Sin[e + f*x])/(16*f*Sqrt[Sin[e + f*x]^2]) + (3*b*(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e
+ f*x]))/(8*f)

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Rubi [A]  time = 0.152344, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3508, 3486, 3772, 2643} \[ \frac{3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])^2,x]

[Out]

(33*a*b*(d*Sec[e + f*x])^(5/3))/(40*f) + (3*(8*a^2 - 3*b^2)*d*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[e + f*x]^2
]*(d*Sec[e + f*x])^(2/3)*Sin[e + f*x])/(16*f*Sqrt[Sin[e + f*x]^2]) + (3*b*(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e
+ f*x]))/(8*f)

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

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))^2 \, dx &=\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{3}{8} \int (d \sec (e+f x))^{5/3} \left (\frac{8 a^2}{3}-b^2+\frac{11}{3} a b \tan (e+f x)\right ) \, dx\\ &=\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{1}{8} \left (8 a^2-3 b^2\right ) \int (d \sec (e+f x))^{5/3} \, dx\\ &=\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{1}{8} \left (\left (8 a^2-3 b^2\right ) \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{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 \left (8 a^2-3 b^2\right ) 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)}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\\ \end{align*}

Mathematica [A]  time = 1.73674, size = 108, normalized size = 0.91 \[ \frac{(d \sec (e+f x))^{5/3} \left (-5 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x)) \sqrt [3]{\cos ^2(e+f x)} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{3}{2},\sin ^2(e+f x)\right )+15 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x))+12 b (16 a+5 b \tan (e+f x))\right )}{160 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])^2,x]

[Out]

((d*Sec[e + f*x])^(5/3)*(15*(8*a^2 - 3*b^2)*Sin[2*(e + f*x)] - 5*(8*a^2 - 3*b^2)*(Cos[e + f*x]^2)^(1/3)*Hyperg
eometric2F1[1/3, 1/2, 3/2, Sin[e + f*x]^2]*Sin[2*(e + f*x)] + 12*b*(16*a + 5*b*Tan[e + f*x])))/(160*f)

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Maple [F]  time = 0.147, 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 ) ^{2}\, 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))^2,x)

[Out]

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

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Maxima [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))^2,x, algorithm="maxima")

[Out]

Timed out

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

[Out]

integral((b^2*d*sec(f*x + e)*tan(f*x + e)^2 + 2*a*b*d*sec(f*x + e)*tan(f*x + e) + a^2*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))**2,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 )}^{2}\,{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))^2,x, algorithm="giac")

[Out]

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