\(\int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx\) [284]

Optimal result

Integrand size = 25, antiderivative size = 78 \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\frac {2^{5/6} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}} \] Output:

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*se 
c(d*x+c))^(1/3)*tan(d*x+c)/d/(1+sec(d*x+c))^(5/6)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1982\) vs. \(2(78)=156\).

Time = 20.92 (sec) , antiderivative size = 1982, normalized size of antiderivative = 25.41 \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(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)) + (S 
ec[c + d*x]^(1/3)*(1 + Sec[c + d*x])^(1/3))/2)*Tan[(c + d*x)/2]*(-1 + (3*A 
ppellF1[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*Sec[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}, Ta 
n[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)))/((Sec[(c + d*x)/2]^2)^(2/3)*(C...
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4315, 3042, 4312, 150}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{4/3} \sqrt [3]{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {\sqrt [3]{a \sec (c+d x)+a} \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{\sec (c+d x)+1}dx}{\sqrt [3]{\sec (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{a \sec (c+d x)+a} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{4/3} \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}dx}{\sqrt [3]{\sec (c+d x)+1}}\)

\(\Big \downarrow \) 4312

\(\displaystyle \frac {\tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {\sqrt [3]{\sec (c+d x)}}{\sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}d(1-\sec (c+d x))}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {2^{5/6} \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}}\)

Input:

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

Output:

(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))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> 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] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 3042

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

rule 4312

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-(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 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[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]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{\frac {4}{3}} \left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=a^{\frac {1}{3}} \left (\int \sec \left (d x +c \right )^{\frac {4}{3}} \left (\sec \left (d x +c \right )+1\right )^{\frac {1}{3}}d x \right ) \] Input:

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

Output:

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