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

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac {3 \left (8 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{16 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]

[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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567, 3857, 2722} \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(e+f x)\right )}{16 f \sqrt {\sin ^2(e+f x)}}+\frac {33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]

[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*} \text {integral}& = \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 \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.76 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {3 (d \sec (e+f x))^{5/3} \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{5 f \sqrt {-\tan ^2(e+f x)}} \]

[In]

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

[Out]

(3*(d*Sec[e + f*x])^(5/3)*(b^2*Hypergeometric2F1[-1/2, 5/6, 11/6, Sec[e + f*x]^2]*Tan[e + f*x] + a*(-(a*Hyperg
eometric2F1[1/2, 5/6, 11/6, Sec[e + f*x]^2]*Tan[e + f*x]) + 2*b*Sqrt[-Tan[e + f*x]^2])))/(5*f*Sqrt[-Tan[e + f*
x]^2])

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]

[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)

Fricas [F]

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

[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)

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]

[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)