3.7.28 \(\int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx\) [628]

Optimal. Leaf size=119 \[ \frac {33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac {3 \left (8 a^2-3 b^2\right ) d \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{16 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]

[Out]

33/40*a*b*(d*sec(f*x+e))^(5/3)/f+3/16*(8*a^2-3*b^2)*d*hypergeom([-1/3, 1/2],[2/3],cos(f*x+e)^2)*(d*sec(f*x+e))
^(2/3)*sin(f*x+e)/f/(sin(f*x+e)^2)^(1/2)+3/8*b*(d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))/f

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Rubi [A]
time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567, 3857, 2722} \begin {gather*} \frac {3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \, _2F_1\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} \end {gather*}

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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3567

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 3589

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[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 3857

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]

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*}

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Mathematica [A]
time = 1.89, size = 108, normalized size = 0.91 \begin {gather*} \frac {(d \sec (e+f x))^{5/3} \left (15 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x))-5 \left (8 a^2-3 b^2\right ) \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right ) \sin (2 (e+f x))+12 b (16 a+5 b \tan (e+f x))\right )}{160 f} \end {gather*}

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 = 0.32, size = 0, normalized size = 0.00 \[\int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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