Integrand size = 25, antiderivative size = 78 \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\frac {2^{5/6} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}} \]
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)
Leaf count is larger than twice the leaf count of optimal. \(1982\) vs. \(2(78)=156\).
Time = 23.47 (sec) , antiderivative size = 1982, normalized size of antiderivative = 25.41 \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx =\text {Too large to display} \]
(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...
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 |
\(\displaystyle \frac {2^{5/6} \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}}\) |
(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))
3.3.84.3.1 Defintions of rubi rules used
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])
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]
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]
\[\int \sec \left (d x +c \right )^{\frac {4}{3}} \left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}d x\]
\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \]
\[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \sec (c+d x)} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]