\(\int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx\) [629]

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx=\frac {21 a b \sqrt [3]{d \sec (e+f x)}}{4 f}-\frac {3 \left (4 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{2/3} \sqrt {\sin ^2(e+f x)}}+\frac {3 b \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f} \]

[Out]

21/4*a*b*(d*sec(f*x+e))^(1/3)/f-3/8*(4*a^2-3*b^2)*d*hypergeom([1/3, 1/2],[4/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*s
ec(f*x+e))^(2/3)/(sin(f*x+e)^2)^(1/2)+3/4*b*(d*sec(f*x+e))^(1/3)*(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} \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx=-\frac {3 d \left (4 a^2-3 b^2\right ) \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right )}{8 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{2/3}}+\frac {21 a b \sqrt [3]{d \sec (e+f x)}}{4 f}+\frac {3 b \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f} \]

[In]

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

[Out]

(21*a*b*(d*Sec[e + f*x])^(1/3))/(4*f) - (3*(4*a^2 - 3*b^2)*d*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[e + f*x]^2]*
Sin[e + f*x])/(8*f*(d*Sec[e + f*x])^(2/3)*Sqrt[Sin[e + f*x]^2]) + (3*b*(d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f
*x]))/(4*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 \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f}+\frac {3}{4} \int \sqrt [3]{d \sec (e+f x)} \left (\frac {4 a^2}{3}-b^2+\frac {7}{3} a b \tan (e+f x)\right ) \, dx \\ & = \frac {21 a b \sqrt [3]{d \sec (e+f x)}}{4 f}+\frac {3 b \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f}+\frac {1}{4} \left (4 a^2-3 b^2\right ) \int \sqrt [3]{d \sec (e+f x)} \, dx \\ & = \frac {21 a b \sqrt [3]{d \sec (e+f x)}}{4 f}+\frac {3 b \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f}+\frac {1}{4} \left (\left (4 a^2-3 b^2\right ) \sqrt [3]{\frac {\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)}\right ) \int \frac {1}{\sqrt [3]{\frac {\cos (e+f x)}{d}}} \, dx \\ & = \frac {21 a b \sqrt [3]{d \sec (e+f x)}}{4 f}-\frac {3 \left (4 a^2-3 b^2\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \sin (e+f x)}{8 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89 \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx=\frac {3 \sqrt [3]{d \sec (e+f x)} \left (\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{6},\frac {7}{6},\sec ^2(e+f x)\right ) \tan (e+f x)}{\sqrt {-\tan ^2(e+f x)}}+a \left (2 b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )\right )}{f} \]

[In]

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

[Out]

(3*(d*Sec[e + f*x])^(1/3)*((b^2*Hypergeometric2F1[-1/2, 1/6, 7/6, Sec[e + f*x]^2]*Tan[e + f*x])/Sqrt[-Tan[e +
f*x]^2] + a*(2*b + a*Cot[e + f*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[e + f*x]^2]*Sqrt[-Tan[e + f*x]^2])))/f

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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