3.3.84 \(\int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx\) [284]
Optimal. Leaf size=78 \[ \frac {2^{5/6} F_1\left (\frac {1}{2};-\frac {1}{3},\frac {1}{6};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}} \]
[Out]
2^(5/6)*AppellF1(1/2,-1/3,1/6,3/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(a+a*sec(d*x+c))^(1/3)*tan(d*x+c)/d/(1+sec(
d*x+c))^(5/6)
________________________________________________________________________________________
Rubi [A]
time = 0.08, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3913, 3910,
138} \begin {gather*} \frac {2^{5/6} \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {1}{2};-\frac {1}{3},\frac {1}{6};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^(4/3)*(a + a*Sec[c + d*x])^(1/3),x]
[Out]
(2^(5/6)*AppellF1[1/2, -1/3, 1/6, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(1/3)*Tan[
c + d*x])/(d*(1 + Sec[c + d*x])^(5/6))
Rule 138
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 3910
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(-(a*(
d/b))^n)*(Cot[e + f*x]/(a^(n - 2)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(a - x)^(n
- 1)*((2*a - x)^(m - 1/2)/Sqrt[x]), x], x, a - b*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] && !IntegerQ[n] && GtQ[a*(d/b), 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*} \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx &=\frac {\sqrt [3]{a+a \sec (c+d x)} \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{1+\sec (c+d x)} \, dx}{\sqrt [3]{1+\sec (c+d x)}}\\ &=\frac {\left (\sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\sqrt [6]{2-x} \sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac {2^{5/6} F_1\left (\frac {1}{2};-\frac {1}{3},\frac {1}{6};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1982\) vs. \(2(78)=156\).
time = 23.25, size = 1982, normalized size = 25.41 \begin {gather*} \text {Too large to display} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^(4/3)*(a + a*Sec[c + d*x])^(1/3),x]
[Out]
(3*Sec[c + d*x]^(1/3)*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(a*(1 + Sec[c + d*x]))^(1/3)*Sin[c + d*x])/(2*d*
(1 + Sec[c + d*x])^(1/3)) + (3*(a*(1 + Sec[c + d*x]))^(1/3)*(-((1 + Sec[c + d*x])^(1/3)/Sec[c + d*x]^(2/3)) +
(Sec[c + d*x]^(1/3)*(1 + Sec[c + d*x])^(1/3))/2)*Tan[(c + d*x)/2]*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
- 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4},
{7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)))/(2^(2/3)*d*(Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*S
ec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(1/3)*((3*(Sec[(c + d*x)/2]^2)^(1/3)*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d
*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2
/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)))/(2*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)
) - (2^(1/3)*Tan[(c + d*x)/2]^2*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2
])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(
c + d*x)/2]^2)))/((Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)) + (3*Tan[(c + d*x)/2]*((
3*((-2*AppellF1[3/2, -1/3, 5/3, 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 - (HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9))/(9
*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d
*x)/2]^2) - (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(-2*(2*AppellF1[3/2, -1/
3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^
4])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] + 9*((-2*AppellF1[3/2, -1/3, 5/3, 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 - (HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4]*S
ec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9) - 2*Tan[(c + d*x)/2]^2*(2*(-(AppellF1[5/2, -1/3, 8/3, 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, 2/3, 5/3, 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) + (3*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-
HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4] + (1 - Tan[(c + d*x)/2]^4)^(-2/3)))/2)))/(9*AppellF1[
1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)^
2))/(2^(2/3)*(Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)) - (Tan[(c + d*x)/2]*(-1 + (3*
AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2))*(-(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]))/(2^(2/3)*(Sec[(c + d*x)/2]^2)^(2/3)
*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(4/3))))
________________________________________________________________________________________
Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (\sec ^{\frac {4}{3}}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(1/3),x)
[Out]
int(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(1/3),x)
________________________________________________________________________________________
Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(1/3),x, algorithm="maxima")
[Out]
integrate((a*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^(4/3), x)
________________________________________________________________________________________
Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(1/3),x, algorithm="fricas")
[Out]
integral((a*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^(4/3), x)
________________________________________________________________________________________
Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**(4/3)*(a+a*sec(d*x+c))**(1/3),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 3004 deep
________________________________________________________________________________________
Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(1/3),x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^(4/3), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a/cos(c + d*x))^(1/3)*(1/cos(c + d*x))^(4/3),x)
[Out]
int((a + a/cos(c + d*x))^(1/3)*(1/cos(c + d*x))^(4/3), x)
________________________________________________________________________________________