\(\int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [626]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}-\frac {3 a d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{4 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}} \]

[Out]

-3*b/f/(d*sec(f*x+e))^(1/3)-3/4*a*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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3567, 3857, 2722} \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 a d \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right )}{4 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}} \]

[In]

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

[Out]

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

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 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}{f \sqrt [3]{d \sec (e+f x)}}+a \int \frac {1}{\sqrt [3]{d \sec (e+f x)}} \, dx \\ & = -\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}+\left (a \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 {3 b}{f \sqrt [3]{d \sec (e+f x)}}-\frac {3 a \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)}{4 d f \sqrt {\sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 \left (b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f \sqrt [3]{d \sec (e+f x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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