3.8.1 \(\int \frac {\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\) [701]

3.8.1.1 Optimal result

Integrand size = 23, antiderivative size = 219 \[ \int \frac {\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]

output

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

3.8.1.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1748\) vs. \(2(219)=438\).

Time = 23.82 (sec) , antiderivative size = 1748, normalized size of antiderivative = 7.98 \[ \int \frac {\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx =\text {Too large to display} \]

input

Integrate[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(1/3),x]
 

output

(3*(b + a*Cos[c + d*x])*Tan[c + d*x])/(2*b*d*(a + b*Sec[c + d*x])^(1/3)) - 
 (3*(b + 3*a*Cos[c + d*x])*(a + b*Sec[c + d*x])^(2/3)*(5*(a^2 - b^2) + 3*b 
*AppellF1[5/3, 3/2, 3/2, 8/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + 
 d*x])/(a + b)]*Sec[c + d*x]*Sqrt[(b*(1 + Sec[c + d*x]))/(-a + b)]*Sqrt[(b 
 - b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c + d*x])))/(10*b*(-a^2 + b^2)*d*(b 
 + a*Cos[c + d*x])^(1/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(7/3)*((3*b 
*(5*(a^2 - b^2) + 3*b*AppellF1[5/3, 3/2, 3/2, 8/3, (a + b*Sec[c + d*x])/(a 
 - b), (a + b*Sec[c + d*x])/(a + b)]*Sec[c + d*x]*Sqrt[(b*(1 + Sec[c + d*x 
]))/(-a + b)]*Sqrt[(b - b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c + d*x]))*Sin 
[c + d*x])/(5*(-a^2 + b^2)*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1 - Cos[c + d*x 
]^2]*Sec[c + d*x]^(1/3)) - (3*(a + b*Sec[c + d*x])*(5*(a^2 - b^2) + 3*b*Ap 
pellF1[5/3, 3/2, 3/2, 8/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d* 
x])/(a + b)]*Sec[c + d*x]*Sqrt[(b*(1 + Sec[c + d*x]))/(-a + b)]*Sqrt[(b - 
b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*(-a^2 + b^ 
2)*(b + a*Cos[c + d*x])^(1/3)*(1 - Cos[c + d*x]^2)^(3/2)*Sec[c + d*x]^(10/ 
3)) + (a*(a + b*Sec[c + d*x])*(5*(a^2 - b^2) + 3*b*AppellF1[5/3, 3/2, 3/2, 
 8/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]*Sec[c + 
d*x]*Sqrt[(b*(1 + Sec[c + d*x]))/(-a + b)]*Sqrt[(b - b*Sec[c + d*x])/(a + 
b)]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*(-a^2 + b^2)*(b + a*Cos[c + d*x 
])^(4/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(7/3)) - (7*(a + b*Sec[c...
 

3.8.1.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4325, 3042, 4321, 156, 155}

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.

\(\displaystyle \int \frac {\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4325

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{2/3}dx}{b}-\frac {a \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}}dx}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4321

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} a \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}\)

input

Int[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(1/3),x]
 

output

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

3.8.1.3.1 Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[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] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

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

rule 3042

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

rule 4321

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[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]
 

rule 4325

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

3.8.1.4 Maple [F]

\[\int \frac {\sec \left (d x +c \right )^{2}}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]

input

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

output

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

3.8.1.5 Fricas [F]

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

input

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="fricas")
 

output

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

3.8.1.6 Sympy [F]

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

input

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

output

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

3.8.1.7 Maxima [F]

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

input

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="maxima")
 

output

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

3.8.1.8 Giac [F]

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

input

integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="giac")
 

output

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

3.8.1.9 Mupad [F(-1)]

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

input

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

output

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