\(\int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [630]

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

-15/2*a*b/f/(d*sec(f*x+e))^(1/3)-3/8*(2*a^2-3*b^2)*d*hypergeom([1/2, 2/3],[5/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*
sec(f*x+e))^(4/3)/(sin(f*x+e)^2)^(1/2)+3/2*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(1/3)

Rubi [A] (verified)

Time = 0.18 (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} \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right )}{8 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]

[In]

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]

[Out]

(-15*a*b)/(2*f*(d*Sec[e + f*x])^(1/3)) - (3*(2*a^2 - 3*b^2)*d*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[e + f*x]^2]
*Sin[e + f*x])/(8*f*(d*Sec[e + f*x])^(4/3)*Sqrt[Sin[e + f*x]^2]) + (3*b*(a + b*Tan[e + f*x]))/(2*f*(d*Sec[e +
f*x])^(1/3))

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 (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3}{2} \int \frac {\frac {2 a^2}{3}-b^2+\frac {5}{3} a b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt [3]{d \sec (e+f x)}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (\left (2 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 \sqrt [3]{\frac {\cos (e+f x)}{d}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{8 d f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{f \sqrt [3]{d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \]

[In]

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]

[Out]

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

Maple [F]

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \]

[In]

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

[Out]

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