∫ (a+a sec(c+dx))4∕3 _____________________________ 3∘______

3.3.87 \(\int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx\) [287]

Optimal. Leaf size=80 \[ \frac {2\ 2^{5/6} a F_1\left (\frac {1}{2};\frac {4}{3},-\frac {5}{6};\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}} \]

[Out]

2*2^(5/6)*a*AppellF1(1/2,4/3,-5/6,3/2,1-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)

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Rubi [A]
time = 0.09, antiderivative size = 80, 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\ 2^{5/6} a \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac {1}{2};\frac {4}{3},-\frac {5}{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[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]

[Out]

(2*2^(5/6)*a*AppellF1[1/2, 4/3, -5/6, 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 \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx &=\frac {\left (a \sqrt [3]{a+a \sec (c+d x)}\right ) \int \frac {(1+\sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx}{\sqrt [3]{1+\sec (c+d x)}}\\ &=\frac {\left (a \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(2-x)^{5/6}}{(1-x)^{4/3} \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\ 2^{5/6} a F_1\left (\frac {1}{2};\frac {4}{3},-\frac {5}{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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 31.33, size = 2325, normalized size = 29.06 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]

[Out]

(-3*(a*(1 + Sec[c + d*x]))^(4/3)*((1 + Sec[c + d*x])^(1/3)/Sec[c + d*x]^(1/3) + Sec[c + d*x]^(2/3)*(1 + Sec[c

+ d*x])^(1/3))*(-8*Tan[(c + d*x)/2] + (AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1

+ I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I

+ Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c +

d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c

+ d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2))/(4*2^(2/3)*d*(Sec[(c + d*x)/2]^2)^(1/3)*(1

+ Sec[c + d*x])^(4/3)*((Tan[(c + d*x)/2]*(-8*Tan[(c + d*x)/2] + (AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-

1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/(-1 + T

an[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - AppellF1[-4/3, -2/3, -2/3, -

1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*

x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2))/(4*2^(2/3)*(Sec

[(c + d*x)/2]^2)^(1/3)) - (3*(-4*Sec[(c + d*x)/2]^2 + (Sec[(c + d*x)/2]^2*(((-4/3 + (4*I)/3)*AppellF1[-1/3, -2

/3, 1/3, 2/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(-1 + Ta

n[(c + d*x)/2])^2 - ((4/3 + (4*I)/3)*AppellF1[-1/3, 1/3, -2/3, 2/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)

/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(-1 + Tan[(c + d*x)/2])^2))/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[

(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) + (AppellF1[-4/3, -2/3, -2/3, -1/

3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/((

(-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3))

- AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*Sec[(c + d

*x)/2]^2*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2])

)^(1/3)*(1 + Tan[(c + d*x)/2]) - (2*AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I

)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*(Sec[(c + d*x)/2]^2/(2*(-1 + Tan[(c + d*x)/2])) - (Sec[(c + d*x)

/2]^2*(-I + Tan[(c + d*x)/2]))/(2*(-1 + Tan[(c + d*x)/2])^2)))/(3*((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)

/2]))^(5/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - (2*AppellF1[-4/3, -2/3, -2/3, -1/3, (-1

- I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*(Sec[(c + d*x)/2]^2/(2*(-1

+ Tan[(c + d*x)/2])) - (Sec[(c + d*x)/2]^2*(I + Tan[(c + d*x)/2]))/(2*(-1 + Tan[(c + d*x)/2])^2)))/(3*((-I + T

an[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(5/3)) - ((-I

+ Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 +

Tan[(c + d*x)/2])^2*(((4/3 + (4*I)/3)*AppellF1[-1/3, -2/3, 1/3, 2/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(

1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2])^2 + ((4/3 - (4*I)/3)*AppellF1[-1/3, 1/3, -2/

3, 2/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)

/2])^2) - (AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((

I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2*(-1/2*(Sec[(c + d*x)/2]^2*(-I + T

an[(c + d*x)/2]))/(1 + Tan[(c + d*x)/2])^2 + Sec[(c + d*x)/2]^2/(2*(1 + Tan[(c + d*x)/2]))))/(3*((-I + Tan[(c

+ d*x)/2])/(1 + Tan[(c + d*x)/2]))^(2/3)) - (AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]),

(1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^

2*(-1/2*(Sec[(c + d*x)/2]^2*(I + Tan[(c + d*x)/2]))/(1 + Tan[(c + d*x)/2])^2 + Sec[(c + d*x)/2]^2/(2*(1 + Tan[

(c + d*x)/2]))))/(3*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(2/3))))/(4*2^(2/3)*(Sec[(c + d*x)/2]^2)^(

1/3))))

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{\frac {4}{3}}}{\sec \left (d x +c \right )^{\frac {1}{3}}}\, dx\]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)

[Out]

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)

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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((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(4/3)/sec(d*x+c)**(1/3),x)

[Out]

Timed out

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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((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{4/3}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(4/3)/(1/cos(c + d*x))^(1/3),x)

[Out]

int((a + a/cos(c + d*x))^(4/3)/(1/cos(c + d*x))^(1/3), x)

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