3.2.56 \(\int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\) [156]

3.2.56.1 Optimal result

Integrand size = 21, antiderivative size = 276 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=-\frac {3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{\sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

output

-1/2*3^(3/4)*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec 
(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1 
/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))*EllipticF((1-(2^(1/3)-(1+ 
sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)) 
)^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(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+sec(d*x+ 
c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)*2^(2/3)/d/(1-sec(d*x+c))/(a+a*s 
ec(d*x+c))^(1/3)/(-(1+sec(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^ 
(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)
 

3.2.56.2 Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.24 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {\sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \tan (c+d x)}{d \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a (1+\sec (c+d x))}} \]

input

Integrate[Sec[c + d*x]/(a + a*Sec[c + d*x])^(1/3),x]
 

output

(2^(1/6)*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sec[c + d*x])/2]*Tan[c + d* 
x])/(d*(1 + Sec[c + d*x])^(1/6)*(a*(1 + Sec[c + d*x]))^(1/3))
 

3.2.56.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4315, 3042, 4314, 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.

\(\displaystyle \int \frac {\sec (c+d x)}{\sqrt [3]{a \sec (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4315

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)+1} \int \frac {\sec (c+d x)}{\sqrt [3]{\sec (c+d x)+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4314

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 766

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

input

Int[Sec[c + d*x]/(a + a*Sec[c + d*x])^(1/3),x]
 

output

-((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] 
*(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]*Tan[c + d*x])/(2^(1/3)*d*(1 - Sec[c + d*x])*(a + a*Se 
c[c + d*x])^(1/3)*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)]))
 

3.2.56.3.1 Defintions of rubi rules used

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]
 

3.2.56.4 Maple [F]

\[\int \frac {\sec \left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]

input

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

output

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

3.2.56.5 Fricas [F]

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

input

integrate(sec(d*x+c)/(a+a*sec(d*x+c))^(1/3),x, algorithm="fricas")
 

output

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

3.2.56.6 Sympy [F]

\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

input

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

output

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

3.2.56.7 Maxima [F]

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

input

integrate(sec(d*x+c)/(a+a*sec(d*x+c))^(1/3),x, algorithm="maxima")
 

output

integrate(sec(d*x + c)/(a*sec(d*x + c) + a)^(1/3), x)
 

3.2.56.8 Giac [F]

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

input

integrate(sec(d*x+c)/(a+a*sec(d*x+c))^(1/3),x, algorithm="giac")
 

output

integrate(sec(d*x + c)/(a*sec(d*x + c) + a)^(1/3), x)
 

3.2.56.9 Mupad [F(-1)]

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

input

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

output

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