\(\int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx\) [711]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{(a+b) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]

[Out]

AppellF1(1/2,5/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b*sec(d*x+c))/(a+b))^(2/3)*2^(1/2)*tan
(d*x+c)/(a+b)/d/(a+b*sec(d*x+c))^(2/3)/(1+sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3919, 144, 143} \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\frac {\sqrt {2} \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d (a+b) \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}} \]

[In]

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

[Out]

(Sqrt[2]*AppellF1[1/2, 1/2, 5/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*
x])/(a + b))^(2/3)*Tan[c + d*x])/((a + b)*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(2/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 3919

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{5/3}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = -\frac {\left (\left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{5/3}} \, dx,x,\sec (c+d x)\right )}{(a+b) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ & = \frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{(a+b) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(10363\) vs. \(2(110)=220\).

Time = 49.84 (sec) , antiderivative size = 10363, normalized size of antiderivative = 94.21 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

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

[In]

int(sec(d*x+c)/(a+b*sec(d*x+c))^(5/3),x)

[Out]

int(sec(d*x+c)/(a+b*sec(d*x+c))^(5/3),x)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)/(a+b*sec(d*x+c))**(5/3),x)

[Out]

Integral(sec(c + d*x)/(a + b*sec(c + d*x))**(5/3), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sec(d*x + c)/(b*sec(d*x + c) + a)^(5/3), x)

Giac [F]

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

[In]

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

[Out]

integrate(sec(d*x + c)/(b*sec(d*x + c) + a)^(5/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]

[In]

int(1/(cos(c + d*x)*(a + b/cos(c + d*x))^(5/3)),x)

[Out]

int(1/(cos(c + d*x)*(a + b/cos(c + d*x))^(5/3)), x)