Integrand size = 21, antiderivative size = 90 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\frac {3 \sqrt {2} \operatorname {AppellF1}\left (-\frac {7}{6},\frac {1}{2},2,-\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{7 a d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) (a+a \sec (c+d x))^{2/3}} \]
3/7*AppellF1(-7/6,2,1/2,-1/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan( d*x+c)/a/d/(1+sec(d*x+c))/(a+a*sec(d*x+c))^(2/3)/(1-sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3011\) vs. \(2(90)=180\).
Time = 16.20 (sec) , antiderivative size = 3011, normalized size of antiderivative = 33.46 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]
(((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(5/3)*((-48*Si n[c + d*x])/7 + (51*Tan[(c + d*x)/2])/7 - (3*Sec[(c + d*x)/2]^2*Tan[(c + d *x)/2])/14))/(d*(a*(1 + Sec[c + d*x]))^(5/3)) - (5*2^(1/3)*(1 + Sec[c + d* x])^(5/3)*((-30*(1 + Sec[c + d*x])^(1/3))/7 + (55*Cos[c + d*x]*(1 + Sec[c + d*x])^(1/3))/7)*Tan[(c + d*x)/2]*((-11*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d*x)/2]^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) + 9*Cos[(c + d*x)/2]^2*(-11 - AppellF1[1/2, 1/3, 1, 3 /2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]/((-1 + Tan[(c + d*x)/2]^2)*(A ppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(- 3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + Ap pellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)))))/(63*d*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*((-5*Sec[(c + d*x)/2]^2*((-11*AppellF1[3/2, 1/3, 1, 5 /2, Tan[(c + d*x)/2]^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) + 9*Cos[(c + d*x)/2]^2*(-11 - AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]/((-1 + Tan[(c + d*x )/2]^2)*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^ 2] + (2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/ 2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2 ])*Tan[(c + d*x)/2]^2)/9)))))/(63*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d...
Time = 0.42 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a \sec (c+d x)+a)^{5/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/3}}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {(\sec (c+d x)+1)^{2/3} \int \frac {\cos (c+d x)}{(\sec (c+d x)+1)^{5/3}}dx}{a (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(\sec (c+d x)+1)^{2/3} \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{5/3}}dx}{a (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {\tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos ^2(c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{13/6}}d\sec (c+d x)}{a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 149 |
\(\displaystyle -\frac {6 \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos ^{10}(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {3 \sqrt {2} \sin (c+d x) \cos ^6(c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {7}{6},2,\frac {1}{2},-\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
(3*Sqrt[2]*AppellF1[-7/6, 2, 1/2, -1/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x ])/2]*Cos[c + d*x]^6*(1 + Sec[c + d*x])^(1/6)*Sin[c + d*x])/(7*a*d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2/3))
3.2.64.3.1 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]
\[\int \frac {\cos \left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]
Timed out. \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{3}}}\, dx \]
\[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
\[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
Timed out. \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]