\(\int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx\) [145]

Optimal result

Integrand size = 21, antiderivative size = 326 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{2 d (1+\sec (c+d x))}-\frac {3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) (a+a \sec (c+d x))^{2/3} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{2 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \] Output:

3/2*(a+a*sec(d*x+c))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))-1/4*3^(3/4)*Inverse 
JacobiAM(arccos((2^(1/3)-(1-3^(1/2))*(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+3^( 
1/2))*(1+sec(d*x+c))^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))*(a+a*sec(d*x+c))^(2/ 
3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+( 
1+sec(d*x+c))^(2/3))/(2^(1/3)-(1+3^(1/2))*(1+sec(d*x+c))^(1/3))^2)^(1/2)*t 
an(d*x+c)*2^(2/3)/d/(1-sec(d*x+c))/(1+sec(d*x+c))/(-(1+sec(d*x+c))^(1/3)*( 
2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+3^(1/2))*(1+sec(d*x+c))^(1/3))^2 
)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.20 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\frac {2 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) (a (1+\sec (c+d x)))^{2/3} \tan (c+d x)}{d (1+\sec (c+d x))^{7/6}} \] Input:

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

Output:

(2*2^(1/6)*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - Sec[c + d*x])/2]*(a*(1 + 
 Sec[c + d*x]))^(2/3)*Tan[c + d*x])/(d*(1 + Sec[c + d*x])^(7/6))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4315, 3042, 4314, 60, 73, 766}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{2/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{2/3}dx\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {(a \sec (c+d x)+a)^{2/3} \int \sec (c+d x) (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a \sec (c+d x)+a)^{2/3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}\)

\(\Big \downarrow \) 4314

\(\displaystyle -\frac {\tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 60

\(\displaystyle -\frac {\tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (3 \int \frac {1}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 766

\(\displaystyle -\frac {\tan (c+d x) (a \sec (c+d x)+a)^{2/3} \left (\frac {3^{3/4} \sqrt [6]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [3]{2} \sqrt {1-\sec (c+d x)} \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}}-\frac {3}{2} \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}\right )}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

Input:

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

Output:

-(((a + a*Sec[c + d*x])^(2/3)*((-3*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x 
])^(1/6))/2 + (3^(3/4)*EllipticF[ArcCos[(2^(1/3) - (1 - Sqrt[3])*(1 + Sec[ 
c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + 
 Sqrt[3])/4]*(1 + Sec[c + d*x])^(1/6)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)) 
*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/ 
3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2])/(2*2^(1/3)*Sqrt 
[1 - Sec[c + d*x]]*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c 
+ d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2)]))*T 
an[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])^(7/6)))
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4314

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x 
]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 
)/Sqrt[a - b*x]), x], x, 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]
 

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \sec {\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral((a*(sec(c + d*x) + 1))**(2/3)*sec(c + d*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^{2/3} \, dx=a^{\frac {2}{3}} \left (\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}} \sec \left (d x +c \right )d x \right ) \] Input:

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

Output:

a**(2/3)*int((sec(c + d*x) + 1)**(2/3)*sec(c + d*x),x)