\(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx\) [627]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {3 b}{5 f (d \sec (e+f x))^{5/3}}-\frac {3 a d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{8/3} \sqrt {\sin ^2(e+f x)}} \]

[Out]

-3/5*b/f/(d*sec(f*x+e))^(5/3)-3/8*a*d*hypergeom([1/2, 4/3],[7/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*sec(f*x+e))^(8/
3)/(sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 78, 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} \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {3 a d \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(e+f x)\right )}{8 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}-\frac {3 b}{5 f (d \sec (e+f x))^{5/3}} \]

[In]

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

[Out]

(-3*b)/(5*f*(d*Sec[e + f*x])^(5/3)) - (3*a*d*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[e + f*x]^2]*Sin[e + f*x])/(8
*f*(d*Sec[e + f*x])^(8/3)*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}{5 f (d \sec (e+f x))^{5/3}}+a \int \frac {1}{(d \sec (e+f x))^{5/3}} \, dx \\ & = -\frac {3 b}{5 f (d \sec (e+f x))^{5/3}}+\left (a \sqrt [3]{\frac {\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)}\right ) \int \left (\frac {\cos (e+f x)}{d}\right )^{5/3} \, dx \\ & = -\frac {3 b}{5 f (d \sec (e+f x))^{5/3}}-\frac {3 a \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \sin (e+f x)}{8 d^2 f \sqrt {\sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {3 \left (b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{5 f (d \sec (e+f x))^{5/3}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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