Integrand size = 23, antiderivative size = 305 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=-\frac {9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac {3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}-\frac {\left (6 a^2-25 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{20 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{10 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
-9/40*a*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d+3/8*(a+b*sec(d*x+c))^(5/3)*t an(d*x+c)/b/d-1/40*(6*a^2-25*b^2)*AppellF1(1/2,-2/3,1/2,3/2,b*(1-sec(d*x+c ))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b^2/d/((a+b *sec(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+sec(d*x+c))^(1/2)+3/20*a*(a^2-b^2)*Ap pellF1(1/2,1/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b*se c(d*x+c))/(a+b))^(1/3)*tan(d*x+c)/b^2/d/(a+b*sec(d*x+c))^(1/3)*2^(1/2)/(1+ sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(7783\) vs. \(2(305)=610\).
Time = 45.20 (sec) , antiderivative size = 7783, normalized size of antiderivative = 25.52 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\text {Result too large to show} \]
Time = 0.93 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4327, 27, 3042, 4490, 27, 3042, 4495, 3042, 4321, 156, 155}
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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx\) |
| \(\Big \downarrow \) 4327 |
| \(\displaystyle \frac {3 \int \frac {1}{3} \sec (c+d x) (5 b-3 a \sec (c+d x)) (a+b \sec (c+d x))^{2/3}dx}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (5 b-3 a \sec (c+d x)) (a+b \sec (c+d x))^{2/3}dx}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (5 b-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {3}{5} \int \frac {\sec (c+d x) \left (19 a b-\left (6 a^2-25 b^2\right ) \sec (c+d x)\right )}{3 \sqrt [3]{a+b \sec (c+d x)}}dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\sec (c+d x) \left (19 a b-\left (6 a^2-25 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}}dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (19 a b+\left (25 b^2-6 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 4495 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {6 a \left (a^2-b^2\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}}dx}{b}-\frac {\left (6 a^2-25 b^2\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3}dx}{b}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {6 a \left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (6 a^2-25 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 4321 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {\left (6 a^2-25 b^2\right ) \tan (c+d x) \int \frac {(a+b \sec (c+d x))^{2/3}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}-\frac {6 a \left (a^2-b^2\right ) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {\left (6 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {6 a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {6 \sqrt {2} a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {\sqrt {2} \left (6 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}}{8 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}\) |
(3*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*b*d) + ((-9*a*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d) + (-((Sqrt[2]*(6*a^2 - 25*b^2)*AppellF1[ 1/2, 1/2, -2/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)] *(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3))) + (6*Sqrt[2]*a*(a^2 - b^2)*AppellF1[1/2 , 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(1/3)*Tan[c + d*x])/(b*d*Sqrt[1 + Sec[c + d*x] ]*(a + b*Sec[c + d*x])^(1/3)))/5)/(8*b)
3.7.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x ]]) Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f*x]] , x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b Int[ Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] + Simp[B/b Int[Csc[e + f*x]*( a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && N eQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
\[\int \sec \left (d x +c \right )^{3} \left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]