Integrand size = 25, antiderivative size = 79 \[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\frac {2 \sqrt [6]{2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},-\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{d (1+\sec (c+d x))^{7/6}} \]
2*2^(1/6)*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))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))^(7/6)
Result contains complex when optimal does not.
Time = 27.27 (sec) , antiderivative size = 1965, normalized size of antiderivative = 24.87 \[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx =\text {Too large to display} \]
(2^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -Tan[(c + d*x)/2]^2]*(a*(1 + Sec [c + d*x]))^(2/3)*Tan[(c + d*x)/2])/(3*d*(1 + Sec[c + d*x])^(2/3)) + (Sec[ c + d*x]^(1/3)*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x ]))^(2/3)*Tan[(c + d*x)/2])/(d*(1 + Sec[c + d*x])^(2/3)) - ((Sec[(c + d*x) /2]^2)^(4/3)*Sec[c + d*x]^(1/3)*(a*(1 + Sec[c + d*x]))^(2/3)*(AppellF1[-2/ 3, -1/3, -1/3, 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)) - AppellF1[-2/3, -1/3 , -1/3, 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))))/(2^(1/3)*d*(-(((Sec[(c + d*x)/2] ^2)^(1/3)*Tan[(c + d*x)/2]*(AppellF1[-2/3, -1/3, -1/3, 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)) - AppellF1[-2/3, -1/3, -1/3, 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))))/2^(1/3)) - (3*(Sec[(c + d*x)/2]^2)^(1/3)*((((1/3 - I/3)*AppellF1[1 /3, -1/3, 2/3, 4/3, (-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 + ((1/3 + I...
Time = 0.38 (sec) , antiderivative size = 79, 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.
\(\displaystyle \int \sec ^{\frac {4}{3}}(c+d x) (a \sec (c+d x)+a)^{2/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{4/3} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{2/3}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {(a \sec (c+d x)+a)^{2/3} \int \sec ^{\frac {4}{3}}(c+d x) (\sec (c+d x)+1)^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a \sec (c+d x)+a)^{2/3} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{4/3} \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{(\sec (c+d x)+1)^{2/3}}\) |
\(\Big \downarrow \) 4312 |
\(\displaystyle \frac {\tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\sqrt [3]{\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1}}{\sqrt {1-\sec (c+d x)}}d(1-\sec (c+d x))}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 \sqrt [6]{2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},-\frac {1}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}}\) |
(2*2^(1/6)*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])^(2/3)*Tan[c + d*x])/(d*(1 + Sec[c + d*x])^(7/ 6))
3.3.85.3.1 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]
\[\int \sec \left (d x +c \right )^{\frac {4}{3}} \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]
Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\text {Timed out} \]
\[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \]
\[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]