\(\int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx\) [286]

Optimal result

Integrand size = 25, antiderivative size = 327 \[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=-\frac {3 a \sec ^{\frac {5}{3}}(c+d x) \sin (c+d x)}{2 d \sqrt [3]{a (1+\sec (c+d x))}}+\frac {9 \sec ^{\frac {2}{3}}(c+d x) (a (1+\sec (c+d x)))^{2/3} \sin (c+d x)}{4 d}-\frac {9 (a (1+\sec (c+d x)))^{2/3} \tan (c+d x)}{4 d \sqrt [3]{\frac {1}{1+\cos (c+d x)}} (1+\sec (c+d x))^{7/3}}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},\tan ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{2/3} \tan (c+d x)}{8 d \sqrt [3]{\frac {1}{1+\cos (c+d x)}} (1+\sec (c+d x))^{4/3}}-\frac {5 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},\tan ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{2/3} \tan ^3(c+d x)}{8 d \sqrt [3]{\frac {1}{1+\cos (c+d x)}} (1+\sec (c+d x))^{10/3}} \]

[Out]

-3/2*a*sec(d*x+c)^(5/3)*sin(d*x+c)/d/(a*(1+sec(d*x+c)))^(1/3)+9/4*sec(d*x+c)^(2/3)*(a*(1+sec(d*x+c)))^(2/3)*si
n(d*x+c)/d-9/4*(a*(1+sec(d*x+c)))^(2/3)*tan(d*x+c)/d/(1/(1+cos(d*x+c)))^(1/3)/(1+sec(d*x+c))^(7/3)+1/8*hyperge
om([1/4, 1/3],[5/4],tan(1/2*d*x+1/2*c)^4)*(cos(d*x+c)*sec(1/2*d*x+1/2*c)^4)^(1/3)*(a*(1+sec(d*x+c)))^(2/3)*tan
(d*x+c)/d/(1/(1+cos(d*x+c)))^(1/3)/(1+sec(d*x+c))^(4/3)-5/8*hypergeom([1/3, 3/4],[7/4],tan(1/2*d*x+1/2*c)^4)*(
cos(d*x+c)*sec(1/2*d*x+1/2*c)^4)^(1/3)*(a*(1+sec(d*x+c)))^(2/3)*tan(d*x+c)^3/d/(1/(1+cos(d*x+c)))^(1/3)/(1+sec
(d*x+c))^(10/3)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3913, 3910, 138} \[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\frac {2 \sqrt [6]{2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{3},-\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}} \]

[In]

Int[Sec[c + d*x]^(5/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

(2*2^(1/6)*AppellF1[1/2, -2/3, -1/6, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(2/3)*T
an[c + d*x])/(d*(1 + Sec[c + d*x])^(7/6))

Rule 138

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

Rule 3910

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

Rule 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+a \sec (c+d x))^{2/3} \int \sec ^{\frac {5}{3}}(c+d x) (1+\sec (c+d x))^{2/3} \, dx}{(1+\sec (c+d x))^{2/3}} \\ & = \frac {\left ((a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{2/3} \sqrt [6]{2-x}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {2 \sqrt [6]{2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{3},-\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{d (1+\sec (c+d x))^{7/6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.27 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.84 \[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\frac {(a (1+\sec (c+d x)))^{2/3} \left (-3 \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt [3]{1+\sec (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {3}{2} (c+d x)\right )\right )+\sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},\tan ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )} \sqrt [3]{\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right )-5 \sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},\tan ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right )} \sqrt [3]{\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \tan ^3\left (\frac {1}{2} (c+d x)\right )\right )}{8 d \sqrt [3]{\frac {1}{1+\cos (c+d x)}} (1+\sec (c+d x))^{2/3}} \]

[In]

Integrate[Sec[c + d*x]^(5/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

((a*(1 + Sec[c + d*x]))^(2/3)*(-3*Sec[(c + d*x)/2]^3*Sec[c + d*x]*(1 + Sec[c + d*x])^(1/3)*(Sin[(c + d*x)/2] -
2*Sin[(3*(c + d*x))/2]) + 2^(1/3)*Hypergeometric2F1[1/4, 1/3, 5/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Sec[(c +
d*x)/2]^4)^(1/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)*Tan[(c + d*x)/2] - 5*2^(1/3)*Hypergeometric2F1[1/3,
3/4, 7/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Sec[(c + d*x)/2]^4)^(1/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)*
Tan[(c + d*x)/2]^3))/(8*d*((1 + Cos[c + d*x])^(-1))^(1/3)*(1 + Sec[c + d*x])^(2/3))

Maple [F]

\[\int \sec \left (d x +c \right )^{\frac {5}{3}} \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]

[In]

int(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x)

[Out]

int(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x)

Fricas [F]

\[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{\frac {5}{3}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**(5/3)*(a+a*sec(d*x+c))**(2/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{\frac {5}{3}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)

Giac [F]

\[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{\frac {5}{3}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/3} \,d x \]

[In]

int((a + a/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^(5/3),x)

[Out]

int((a + a/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^(5/3), x)