Integrand size = 23, antiderivative size = 356 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=-\frac {3 \left (15 a^2-64 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{440 b d}-\frac {9 a (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{88 b d}+\frac {3 (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{11 b d}-\frac {a \left (30 a^2-373 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{220 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (15 a^4-79 a^2 b^2+64 b^4\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{110 \sqrt {2} b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
-3/440*(15*a^2-64*b^2)*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d-9/88*a*(a+b*s ec(d*x+c))^(5/3)*tan(d*x+c)/b/d+3/11*(a+b*sec(d*x+c))^(8/3)*tan(d*x+c)/b/d -1/440*a*(30*a^2-373*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)+1/220*(15*a^4-79*a^2*b^2+64 *b^4)*AppellF1(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*sec(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. \(21890\) vs. \(2(356)=712\).
Time = 46.60 (sec) , antiderivative size = 21890, normalized size of antiderivative = 61.49 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\text {Result too large to show} \]
Time = 1.23 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4327, 27, 3042, 4490, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx\) |
| \(\Big \downarrow \) 4327 |
| \(\displaystyle \frac {3 \int \frac {1}{3} \sec (c+d x) (8 b-3 a \sec (c+d x)) (a+b \sec (c+d x))^{5/3}dx}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (8 b-3 a \sec (c+d x)) (a+b \sec (c+d x))^{5/3}dx}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (8 b-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {3}{8} \int \frac {1}{3} \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (49 a b-\left (15 a^2-64 b^2\right ) \sec (c+d x)\right )dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{8} \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (49 a b-\left (15 a^2-64 b^2\right ) \sec (c+d x)\right )dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (49 a b+\left (64 b^2-15 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {3}{5} \int \frac {\sec (c+d x) \left (b \left (215 a^2+128 b^2\right )-a \left (30 a^2-373 b^2\right ) \sec (c+d x)\right )}{3 \sqrt [3]{a+b \sec (c+d x)}}dx-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \int \frac {\sec (c+d x) \left (b \left (215 a^2+128 b^2\right )-a \left (30 a^2-373 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}}dx-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (215 a^2+128 b^2\right )-a \left (30 a^2-373 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 4495 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \left (\frac {2 \left (15 a^4-79 a^2 b^2+64 b^4\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}}dx}{b}-\frac {a \left (30 a^2-373 b^2\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3}dx}{b}\right )-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \left (\frac {2 \left (15 a^4-79 a^2 b^2+64 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \left (30 a^2-373 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}\right )-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 4321 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \left (\frac {a \left (30 a^2-373 b^2\right ) \tan (c+d x) \int \frac {(a+b \sec (c+d x))^{2/3}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}-\frac {2 \left (15 a^4-79 a^2 b^2+64 b^4\right ) \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \left (\frac {a \left (30 a^2-373 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {2 \left (15 a^4-79 a^2 b^2+64 b^4\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}}}d\sec (c+d x)}{b d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}\right )-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{5} \left (\frac {2 \sqrt {2} \left (15 a^4-79 a^2 b^2+64 b^4\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {\sqrt {2} a \left (30 a^2-373 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}\right )-\frac {3 \left (15 a^2-64 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d}\right )-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 d}}{11 b}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{11 b d}\) |
(3*(a + b*Sec[c + d*x])^(8/3)*Tan[c + d*x])/(11*b*d) + ((-9*a*(a + b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*d) + ((-3*(15*a^2 - 64*b^2)*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d) + (-((Sqrt[2]*a*(30*a^2 - 373*b^2)*Appell F1[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))) + (2*Sqrt[2]*(15*a^4 - 79*a^2*b^2 + 64*b^4)*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*S qrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3)))/5)/8)/(11*b)
3.7.95.3.1 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]
\[\int \sec \left (d x +c \right )^{3} \left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}d x\]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\text {Timed out} \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/3} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]