Integrand size = 21, antiderivative size = 136 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {4 d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b} \]
4/77*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4 *Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*b*x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)-4 /77*d*sec(b*x+a)*(d*tan(b*x+a))^(1/2)/b-2/77*d*sec(b*x+a)^3*(d*tan(b*x+a)) ^(1/2)/b+2/11*d*sec(b*x+a)^5*(d*tan(b*x+a))^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.77 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.66 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d \sec ^5(a+b x) \left (-23+6 \cos (2 (a+b x))+\cos (4 (a+b x))+16 \cos ^6(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{154 b} \]
-1/154*(d*Sec[a + b*x]^5*(-23 + 6*Cos[2*(a + b*x)] + Cos[4*(a + b*x)] + 16 *Cos[a + b*x]^6*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec [a + b*x]^2])*Sqrt[d*Tan[a + b*x]])/b
Time = 0.81 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3091, 3042, 3093, 3042, 3093, 3042, 3094, 3042, 3053, 3042, 3120}
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.
| \(\displaystyle \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (a+b x)^5 (d \tan (a+b x))^{3/2}dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \int \frac {\sec ^5(a+b x)}{\sqrt {d \tan (a+b x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \int \frac {\sec (a+b x)^5}{\sqrt {d \tan (a+b x)}}dx\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \int \frac {\sec ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \int \frac {\sec (a+b x)^3}{\sqrt {d \tan (a+b x)}}dx+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3093 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2}{3} \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3094 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3053 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 \sqrt {d \tan (a+b x)}}+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {6}{7} \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 \sqrt {d \tan (a+b x)}}+\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}\right )+\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \left (\frac {2 \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{7 b d}+\frac {6}{7} \left (\frac {2 \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b d}+\frac {2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b \sqrt {d \tan (a+b x)}}\right )\right )\) |
(2*d*Sec[a + b*x]^5*Sqrt[d*Tan[a + b*x]])/(11*b) - (d^2*((2*Sec[a + b*x]^3 *Sqrt[d*Tan[a + b*x]])/(7*b*d) + (6*((2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b*Sqrt[d*Tan[a + b*x]]) + (2*Sec[a + b* x]*Sqrt[d*Tan[a + b*x]])/(3*b*d)))/7))/11
3.3.41.3.1 Defintions of rubi rules used
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[a^2*((m - 2)/(m + n - 1)) Int[(a*Sec[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && ( GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[ 2*m, 2*n]
Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) Int[ 1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Time = 1.67 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.95
| method | result | size |
| default | \(\frac {d \sqrt {d \tan \left (b x +a \right )}\, \left (4 \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+4 \sin \left (b x +a \right ) \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right ) \tan \left (b x +a \right ) \sqrt {2}+\left (\tan ^{2}\left (b x +a \right )\right ) \sec \left (b x +a \right ) \sqrt {2}-7 \left (\tan ^{2}\left (b x +a \right )\right ) \left (\sec ^{3}\left (b x +a \right )\right ) \sqrt {2}\right ) \sqrt {2}}{77 b \left (\cos ^{2}\left (b x +a \right )-1\right )}\) | \(265\) |
1/77/b*d*(d*tan(b*x+a))^(1/2)/(cos(b*x+a)^2-1)*(4*sin(b*x+a)*cos(b*x+a)*(1 +csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1 +cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))+ 4*sin(b*x+a)*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2 )*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2 ),1/2*2^(1/2))+2*sin(b*x+a)*tan(b*x+a)*2^(1/2)+tan(b*x+a)^2*sec(b*x+a)*2^( 1/2)-7*tan(b*x+a)^2*sec(b*x+a)^3*2^(1/2))*2^(1/2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.93 \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 \, {\left (2 \, \sqrt {i \, d} d \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {-i \, d} d \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - {\left (2 \, d \cos \left (b x + a\right )^{4} + d \cos \left (b x + a\right )^{2} - 7 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{77 \, b \cos \left (b x + a\right )^{5}} \]
2/77*(2*sqrt(I*d)*d*cos(b*x + a)^5*elliptic_f(arcsin(cos(b*x + a) + I*sin( b*x + a)), -1) + 2*sqrt(-I*d)*d*cos(b*x + a)^5*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - (2*d*cos(b*x + a)^4 + d*cos(b*x + a)^2 - 7*d) *sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*cos(b*x + a)^5)
\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{5}{\left (a + b x \right )}\, dx \]
\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5} \,d x } \]
\[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5} \,d x } \]
Timed out. \[ \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{{\cos \left (a+b\,x\right )}^5} \,d x \]