\(\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\) [282]
Optimal result
Integrand size = 27, antiderivative size = 76 \[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (-\frac {1}{3},\frac {1}{2},1,\frac {2}{3},\sec (c+d x),-\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}
\]
[Out]
3*AppellF1(-1/3,1,1/2,2/3,-sec(d*x+c),sec(d*x+c))*tan(d*x+c)/d/(e*sec(d*x+c))^(1/3)/(1-sec(d*x+c))^(1/2)/(a+a*
sec(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3913, 3912, 129, 524}
\[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {3 \tan (c+d x) \operatorname {AppellF1}\left (-\frac {1}{3},\frac {1}{2},1,\frac {2}{3},\sec (c+d x),-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a} \sqrt [3]{e \sec (c+d x)}}
\]
[In]
Int[1/((e*Sec[c + d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]]),x]
[Out]
(3*AppellF1[-1/3, 1/2, 1, 2/3, Sec[c + d*x], -Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(e*Sec[c +
d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]])
Rule 129
Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]
Rule 524
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 3912
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[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] && !In
tegerQ[m] && GtQ[a, 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 {\sqrt {1+\sec (c+d x)} \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx}{\sqrt {a+a \sec (c+d x)}} \\ & = -\frac {(e \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (e x)^{4/3} (1+x)} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {(3 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x^3}{e}} \left (1+\frac {x^3}{e}\right )} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = \frac {3 \operatorname {AppellF1}\left (-\frac {1}{3},\frac {1}{2},1,\frac {2}{3},\sec (c+d x),-\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \\
\end{align*}
Mathematica [B] (warning: unable to verify)
Leaf count is larger than twice the leaf count of optimal. \(3346\) vs. \(2(76)=152\).
Time = 19.02 (sec) , antiderivative size = 3346, normalized size of antiderivative = 44.03
\[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\text {Result too large to show}
\]
[In]
Integrate[1/((e*Sec[c + d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]]),x]
[Out]
-(((Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1
/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6)
+ (3*(1 + (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)
*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2
])*Tan[(c + d*x)/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3)))/(d*(e*Sec[c + d*x])^(1/3)*Sqrt[a*(1 + Sec[c + d*x])]*(-
(Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*Tan[(c + d*x)/2]^2*((2*AppellF1[3/2, 1/6, 1/3, 5/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6) + (3*(1
+ (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(9*App
ellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[
(c + d*x)/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3))) - (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*
(-1 + Tan[(c + d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d
*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6) + (3*(1 + (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2
, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6,
1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3)))/2 - (
Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1/3,
5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(Cos[c + d*x]*Sec[(c + d*x)
/2]^2)^(5/6) + (2*Tan[(c + d*x)/2]^2*(-1/5*(AppellF1[5/2, 1/6, 4/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) + (AppellF1[5/2, 7/6, 1/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/10))/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6) - (5*AppellF1[3/2, 1/6,
1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Co
s[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(11/6)) - (Tan[(c + d*x)
/2]*(1 + (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*
(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Ta
n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
)*Tan[(c + d*x)/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3) + (3*((-3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/((-1 + Tan[(c + d*x)/2]^2)^2*(9*AppellF1[1/2, 1/6,
1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))
+ (3*(-1/9*(AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c +
d*x)/2]) + (AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c +
d*x)/2])/18))/((-1 + Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2
]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)) - (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2]*((-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + App
ellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] + 9*(-
1/9*(AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]
) + (AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]
)/18) + Tan[(c + d*x)/2]^2*(-1/5*(AppellF1[5/2, 7/6, 4/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c
+ d*x)/2]^2*Tan[(c + d*x)/2]) + (7*AppellF1[5/2, 13/6, 1/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec
[(c + d*x)/2]^2*Tan[(c + d*x)/2])/10 - 2*((-4*AppellF1[5/2, 1/6, 7/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (AppellF1[5/2, 7/6, 4/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x
)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/10))))/((-1 + Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2, 1/6, 1/3, 3/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x
)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)^2)))/(Sec
[(c + d*x)/2]^2)^(1/3)) - (Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1/3, 5/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6) + (3*(1 + (3*App
ellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2
, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^
2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)
/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3))*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*
Sec[c + d*x]*Tan[c + d*x]))/(6*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(5/6)))))
Maple [F]
\[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}} \sqrt {a +a \sec \left (d x +c \right )}}d x\]
[In]
int(1/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x)
[Out]
int(1/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x)
Fricas [F(-1)]
Timed out. \[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out}
\]
[In]
integrate(1/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
Timed out
Sympy [F]
\[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sqrt [3]{e \sec {\left (c + d x \right )}}}\, dx
\]
[In]
integrate(1/(e*sec(d*x+c))**(1/3)/(a+a*sec(d*x+c))**(1/2),x)
[Out]
Integral(1/(sqrt(a*(sec(c + d*x) + 1))*(e*sec(c + d*x))**(1/3)), x)
Maxima [F]
\[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x }
\]
[In]
integrate(1/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3)), x)
Giac [F]
\[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x }
\]
[In]
integrate(1/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x
\]
[In]
int(1/((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3)),x)
[Out]
int(1/((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3)), x)