Integrand size = 21, antiderivative size = 86 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=-\frac {3 \sqrt {2} a \operatorname {AppellF1}\left (\frac {13}{6},\frac {1}{2},2,\frac {19}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{13 d \sqrt {1-\sec (c+d x)}} \]
-3/13*a*AppellF1(13/6,2,1/2,19/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(1+sec(d *x+c))*(a+a*sec(d*x+c))^(2/3)*2^(1/2)*tan(d*x+c)/d/(1-sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2700\) vs. \(2(86)=172\).
Time = 16.01 (sec) , antiderivative size = 2700, normalized size of antiderivative = 31.40 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\text {Result too large to show} \]
(((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*(Sin [c + d*x] - Tan[(c + d*x)/2]))/(d*(1 + Sec[c + d*x])^(5/3)) - (2^(2/3)*(Co s[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*((2*Sec[ (c + d*x)/2]^2*(1 + Sec[c + d*x])^(2/3))/3 + (5*Cos[c + d*x]*Sec[(c + d*x) /2]^2*(1 + Sec[c + d*x])^(2/3))/6)*Tan[(c + d*x)/2]*(AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x) /2]^2)^(2/3)*Tan[(c + d*x)/2]^2 + (243*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(-9*AppellF1[1/2, 2/3 , 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 2/ 3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)))/( 9*d*(1 + Sec[c + d*x])^(5/3)*(-1/9*(Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2 *Sec[c + d*x])^(2/3)*(AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[ (c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2]^2 + (243*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2 ]*Cos[(c + d*x)/2]^2)/(-9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, - Tan[(c + d*x)/2]^2] + 2*(3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -T an[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)))/2^(1/3) - (2^(2/3)*(Cos[(c + d*x )/2]^2*Sec[c + d*x])^(2/3)*Tan[(c + d*x)/2]*(AppellF1[3/2, 2/3, 1, 5/2,...
Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4315, 3042, 4314, 149, 1012}
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.
\(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^{5/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/3}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {a (a \sec (c+d x)+a)^{2/3} \int \cos (c+d x) (\sec (c+d x)+1)^{5/3}dx}{(\sec (c+d x)+1)^{2/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (a \sec (c+d x)+a)^{2/3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{5/3}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{(\sec (c+d x)+1)^{2/3}}\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {a \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos ^2(c+d x) (\sec (c+d x)+1)^{7/6}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
\(\Big \downarrow \) 149 |
\(\displaystyle -\frac {6 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos ^2(c+d x) (\sec (c+d x)+1)^2}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {3 \sqrt {2} a \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {13}{6},2,\frac {1}{2},\frac {19}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{13 d \sqrt {1-\sec (c+d x)}}\) |
(-3*Sqrt[2]*a*AppellF1[13/6, 2, 1/2, 19/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x])/2]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(13*d *Sqrt[1 - Sec[c + d*x]])
3.2.52.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1 ) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[2*n] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x ]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 )/Sqrt[a - b*x]), x], x, 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]
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]
\[\int \cos \left (d x +c \right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}d x\]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\text {Timed out} \]
\[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \cos \left (d x + c\right ) \,d x } \]
\[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/3} \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3} \,d x \]