3.8.14 \(\int \frac {\sqrt [3]{\sec (c+d x)}}{a+b \sec (c+d x)} \, dx\) [714]

3.8.14.1 Optimal result

Integrand size = 23, antiderivative size = 174 \[ \int \frac {\sqrt [3]{\sec (c+d x)}}{a+b \sec (c+d x)} \, dx=\frac {a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)}-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d} \]

output

a*AppellF1(1/2,-1/3,1,3/2,sin(d*x+c)^2,a^2*sin(d*x+c)^2/(a^2-b^2))*sin(d*x 
+c)/(a^2-b^2)/d/(cos(d*x+c)^2)^(1/3)/sec(d*x+c)^(2/3)-b*AppellF1(1/2,1/6,1 
,3/2,sin(d*x+c)^2,a^2*sin(d*x+c)^2/(a^2-b^2))*(cos(d*x+c)^2)^(1/6)*sec(d*x 
+c)^(1/3)*sin(d*x+c)/(a^2-b^2)/d
 

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

Leaf count is larger than twice the leaf count of optimal. \(4544\) vs. \(2(174)=348\).

Time = 38.04 (sec) , antiderivative size = 4544, normalized size of antiderivative = 26.11 \[ \int \frac {\sqrt [3]{\sec (c+d x)}}{a+b \sec (c+d x)} \, dx=\text {Result too large to show} \]

input

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

output

(9*(a^2 - b^2)*Sec[c + d*x]^(4/3)*Sin[c + d*x]*((b*AppellF1[1/2, 1/3, 1, 3 
/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x]^2 
])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + 
 d*x]^2)/(a^2 - b^2)] + 2*(3*b^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c + d*x]^ 
2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 4/3, 1, 
5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) + 
 (a*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 
- b^2)])/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, (b^2* 
Tan[c + d*x]^2)/(a^2 - b^2)] + (-6*b^2*AppellF1[3/2, 5/6, 2, 5/2, -Tan[c + 
 d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 5*(a^2 - b^2)*AppellF1[3/2, 1 
1/6, 1, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d 
*x]^2)))/(d*(Sec[c + d*x]^2)^(5/6)*(a + b*Sec[c + d*x])*(-a^2 + b^2*Sec[c 
+ d*x]^2)*((9*(a^2 - b^2)*(Sec[c + d*x]^2)^(1/6)*((b*AppellF1[1/2, 1/3, 1, 
 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sec[c + d*x] 
^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/3, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c 
 + d*x]^2)/(a^2 - b^2)] + 2*(3*b^2*AppellF1[3/2, 1/3, 2, 5/2, -Tan[c + d*x 
]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[3/2, 4/3, 1 
, 5/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2) 
 + (a*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^ 
2 - b^2)])/(-9*(a^2 - b^2)*AppellF1[1/2, 5/6, 1, 3/2, -Tan[c + d*x]^2, ...
 

3.8.14.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4356, 3042, 3302, 3042, 3668, 25, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4356

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3302

\(\displaystyle \sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)} \left (b \int \frac {\cos ^{\frac {2}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)}dx-a \int \frac {\cos ^{\frac {5}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)} \left (b \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{2/3}}{b^2-a^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}dx-a \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/3}}{b^2-a^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )\)

\(\Big \downarrow \) 3668

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 333

\(\displaystyle \sqrt [3]{\cos (c+d x)} \sqrt [3]{\sec (c+d x)} \left (\frac {a \sin (c+d x) \cos ^{\frac {2}{3}}(c+d x) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos ^2(c+d x)}}-\frac {b \sin (c+d x) \sqrt [6]{\cos ^2(c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [3]{\cos (c+d x)}}\right )\)

input

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

output

Cos[c + d*x]^(1/3)*Sec[c + d*x]^(1/3)*((a*AppellF1[1/2, -1/3, 1, 3/2, Sin[ 
c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Cos[c + d*x]^(2/3)*Sin[c + d 
*x])/((a^2 - b^2)*d*(Cos[c + d*x]^2)^(1/3)) - (b*AppellF1[1/2, 1/6, 1, 3/2 
, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*(Cos[c + d*x]^2)^(1/6) 
*Sin[c + d*x])/((a^2 - b^2)*d*Cos[c + d*x]^(1/3)))
 

3.8.14.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 3042

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

rule 3302

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[a   Int[(d*Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x] 
^2), x], x] - Simp[b/d   Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e + f* 
x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
 

rule 3668

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( 
-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) 
/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]))   Subst[Int[(1 - ff^2*x^2)^((m - 
 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, 
 d, e, f, m, p}, x] &&  !IntegerQ[m]
 

rule 4356

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

3.8.14.4 Maple [F]

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

input

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

output

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

3.8.14.5 Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{\sec (c+d x)}}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.8.14.6 Sympy [F]

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

input

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

output

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

3.8.14.7 Maxima [F]

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

input

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

output

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

3.8.14.8 Giac [F]

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

input

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

output

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

3.8.14.9 Mupad [F(-1)]

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

input

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

output

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