Integrand size = 19, antiderivative size = 80 \[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{3 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b} \]
1/3*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)+2/ 3*d*sec(b*x+a)*(d*tan(b*x+a))^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86 \[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 d \cos (a+b x) \left (\sec ^2(a+b x)-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{3 b} \]
(2*d*Cos[a + b*x]*(Sec[a + b*x]^2 - Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[ a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[d*Tan[a + b*x]])/(3*b)
Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3091, 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.
\(\displaystyle \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (a+b x) (d \tan (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {1}{3} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {1}{3} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3094 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{3 b}-\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b \sqrt {d \tan (a+b x)}}\) |
-1/3*(d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]] )/(b*Sqrt[d*Tan[a + b*x]]) + (2*d*Sec[a + b*x]*Sqrt[d*Tan[a + b*x]])/(3*b)
3.3.43.3.1 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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(220\) vs. \(2(95)=190\).
Time = 1.39 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.76
method | result | size |
default | \(-\frac {\left (-\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (b x +a \right )\right )-\sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )+\sin \left (b x +a \right ) \sqrt {2}\right ) \tan \left (b x +a \right ) d \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}}{3 b \left (\cos ^{2}\left (b x +a \right )-1\right )}\) | \(221\) |
-1/3/b*(-(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)* (cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/ 2*2^(1/2))*cos(b*x+a)^2-(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))*cos(b*x+a)+sin(b*x+a)*2^(1/2))*tan(b*x+a)*d*(d*t an(b*x+a))^(1/2)/(cos(b*x+a)^2-1)*2^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.21 \[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {\sqrt {i \, d} d \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-i \, d} d \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right )} \]
1/3*(sqrt(I*d)*d*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + sqrt(-I*d)*d*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) - I*s in(b*x + a)), -1) + 2*d*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*cos(b*x + a) )
\[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec {\left (a + b x \right )}\, dx \]
\[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right ) \,d x } \]
\[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right ) \,d x } \]
Timed out. \[ \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{\cos \left (a+b\,x\right )} \,d x \]