3.701 \(\int \frac{\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=219 \[ \frac{\sqrt{2} \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{\sqrt{2} a \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}} \]
[Out]
(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*AppellF
1[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))
________________________________________________________________________________________
Rubi [A] time = 0.228642, antiderivative size = 219, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{3838, 3834, 139, 138} \[ \frac{\sqrt{2} \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac{\sqrt{2} a \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(1/3),x]
[Out]
(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*AppellF
1[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))
Rule 3838
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[a/b, Int[Csc[e +
f*x]*(a + b*Csc[e + f*x])^m, x], x] + Dist[1/b, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; Free
Q[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]
Rule 3834
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[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 139
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((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*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m]
&& !IntegerQ[n] && !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && !GtQ[b/(b*e - a*f), 0]
Rule 138
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !Inte
gerQ[n] && !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] && !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) && !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx &=\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}\\ &=-\frac{\tan (c+d x) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}+\frac{(a \tan (c+d x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=-\frac{\left ((a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac{\left (a \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac{\sqrt{2} F_1\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 F_1\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)}}\\ \end{align*}
Mathematica [B] time = 19.0141, size = 2759, normalized size = 12.6 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(1/3),x]
[Out]
(3*(b + a*Cos[c + d*x])*Tan[c + d*x])/(2*b*d*(a + b*Sec[c + d*x])^(1/3)) - ((b + 3*a*Cos[c + d*x])*(3*(b + a*C
os[c + d*x])^(2/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(2/3) - (3*(a + b*Sec[c + d*x])*Sqrt[(1 - Sqrt[b^(-2)
]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*a*App
ellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-
2)])] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a
+ 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x])))/(5*b*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x
]^(4/3))))/(2*b*d*(a + b*Sec[c + d*x])^(1/3)*((3*(b + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*
x]^2]*Sec[c + d*x]^(1/3)) - (2*a*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(2/3)*Sin[c + d*x])/(b + a*Cos[c + d*x]
)^(1/3) + 2*(b + a*Cos[c + d*x])^(2/3)*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(5/3)*Sin[c + d*x] - (3*Sqrt[b^(-
2)]*Sec[c + d*x]^(2/3)*(a + b*Sec[c + d*x])*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*(-5*a
*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[
b^(-2)])] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])
/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(10*(1 - a*Sqrt[b^(-2)])*(b + a*Cos[c + d*x])^(1/3)
*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]) + (3*Sqrt[b^(-2)]*Sec[
c + d*x]^(2/3)*(a + b*Sec[c + d*x])*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*a*AppellF
1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])
] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/
Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(10*(1 + a*Sqrt[b^(-2)])*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1
- Cos[c + d*x]^2]*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]) - (3*Sec[c + d*x]^(2/3)*Sqrt[(
1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-
2)])]*(-5*a*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(
a + 1/Sqrt[b^(-2)])] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Se
c[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1 - C
os[c + d*x]^2]) + (3*(a + b*Sec[c + d*x])*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1
+ Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*a*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x]
)/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b
*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c
+ d*x])/(5*b*(b + a*Cos[c + d*x])^(1/3)*(1 - Cos[c + d*x]^2)^(3/2)*Sec[c + d*x]^(7/3)) - (a*(a + b*Sec[c + d*x
])*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*
Sqrt[b^(-2)])]*(-5*a*AppellF1[2/3, 1/2, 1/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c
+ d*x])/(a + 1/Sqrt[b^(-2)])] + 2*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])),
(a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*b*(b + a*Cos[c + d*x])^(4/3)
*Sqrt[1 - Cos[c + d*x]^2]*Sec[c + d*x]^(4/3)) + (4*(a + b*Sec[c + d*x])*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])
/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(-5*a*AppellF1[2/3, 1/2, 1
/2, 5/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])] + 2*AppellF
1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])
]*(a + b*Sec[c + d*x]))*Sin[c + d*x])/(5*b*(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])*Sqrt[(1 - Sqrt[b^(-2)]*b*Sec[c + d*x])/(1 + a*Sqrt[b^(-2)])]*Sqrt[(1 + Sqrt[b^(
-2)]*b*Sec[c + d*x])/(1 - a*Sqrt[b^(-2)])]*(2*b*AppellF1[5/3, 1/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sq
rt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c + d*x] - 5*a*((b*AppellF1[5/3, 1/2
, 3/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c +
d*x]*Tan[c + d*x])/(5*(a + 1/Sqrt[b^(-2)])) - (b*AppellF1[5/3, 3/2, 1/2, 8/3, -((a + b*Sec[c + d*x])/(-a + 1/S
qrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c + d*x])/(5*(-a + 1/Sqrt[b^(-2)]))
) + 2*(a + b*Sec[c + d*x])*((5*b*AppellF1[8/3, 1/2, 3/2, 11/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])),
(a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)])]*Sec[c + d*x]*Tan[c + d*x])/(16*(a + 1/Sqrt[b^(-2)])) - (5*b*AppellF
1[8/3, 3/2, 1/2, 11/3, -((a + b*Sec[c + d*x])/(-a + 1/Sqrt[b^(-2)])), (a + b*Sec[c + d*x])/(a + 1/Sqrt[b^(-2)]
)]*Sec[c + d*x]*Tan[c + d*x])/(16*(-a + 1/Sqrt[b^(-2)])))))/(5*b*(b + a*Cos[c + d*x])^(1/3)*Sqrt[1 - Cos[c + d
*x]^2]*Sec[c + d*x]^(4/3))))
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Maple [F] time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{a+b\sec \left ( dx+c \right ) }}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x)
[Out]
int(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="maxima")
[Out]
integrate(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(1/3), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="fricas")
[Out]
integral(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(1/3), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{a + b \sec{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**2/(a+b*sec(d*x+c))**(1/3),x)
[Out]
Integral(sec(c + d*x)**2/(a + b*sec(c + d*x))**(1/3), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(1/3),x, algorithm="giac")
[Out]
integrate(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(1/3), x)