Integrand size = 19, antiderivative size = 82 \[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}-\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{3 b d^2 \sqrt {d \tan (a+b x)}} \]
1/3*(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^2/(d*tan(b*x+a))^(1/2)-2/ 3*sec(b*x+a)/b/d/(d*tan(b*x+a))^(3/2)
Result contains complex when optimal does not.
Time = 0.74 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.38 \[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {2 \cos (2 (a+b x)) \csc (a+b x) \sqrt {\sec ^2(a+b x)} \left (\sqrt {\sec ^2(a+b x)}-\sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \tan ^{\frac {3}{2}}(a+b x)\right )}{3 b d^2 \sqrt {d \tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \]
(2*Cos[2*(a + b*x)]*Csc[a + b*x]*Sqrt[Sec[a + b*x]^2]*(Sqrt[Sec[a + b*x]^2 ] - (-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Tan [a + b*x]^(3/2)))/(3*b*d^2*Sqrt[d*Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
Time = 0.46 (sec) , antiderivative size = 82, 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, 3089, 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.
\(\displaystyle \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3089 |
\(\displaystyle -\frac {\int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx}{3 d^2}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}}dx}{3 d^2}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3094 |
\(\displaystyle -\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {d \tan (a+b x)}}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {d \tan (a+b x)}}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b d^2 \sqrt {d \tan (a+b x)}}-\frac {2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}\) |
(-2*Sec[a + b*x])/(3*b*d*(d*Tan[a + b*x])^(3/2)) - (EllipticF[a - Pi/4 + b *x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b*d^2*Sqrt[d*Tan[a + b*x]])
3.3.69.3.1 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[(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Simp[(m + n + 1)/(b^2*(n + 1)) Int[(a*Sec[e + f*x])^m*(b*Ta n[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && I ntegersQ[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. \(298\) vs. \(2(97)=194\).
Time = 1.10 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.65
method | result | size |
default | \(\frac {\left (\csc ^{2}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{2} \left (2 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {2-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )-\left (\csc ^{4}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{4}+1\right ) \sqrt {2}}{6 b \sqrt {\left (\csc ^{3}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{3}-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \sqrt {\csc \left (b x +a \right ) \left (1-\cos \left (b x +a \right )\right ) \left (\left (\csc ^{2}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{2}-1\right )}\, {\left (\left (\csc ^{2}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{2}-1\right )}^{2} {\left (-\frac {d \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{\left (\csc ^{2}\left (b x +a \right )\right ) \left (1-\cos \left (b x +a \right )\right )^{2}-1}\right )}^{\frac {5}{2}}}\) | \(299\) |
1/6/b*csc(b*x+a)^2*(1-cos(b*x+a))^2*(2*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(2- 2*csc(b*x+a)+2*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))*(csc(b*x+a)-cot(b*x+a))-csc(b* x+a)^4*(1-cos(b*x+a))^4+1)/(csc(b*x+a)^3*(1-cos(b*x+a))^3-csc(b*x+a)+cot(b *x+a))^(1/2)/(csc(b*x+a)*(1-cos(b*x+a))*(csc(b*x+a)^2*(1-cos(b*x+a))^2-1)) ^(1/2)/(csc(b*x+a)^2*(1-cos(b*x+a))^2-1)^2/(-d/(csc(b*x+a)^2*(1-cos(b*x+a) )^2-1)*(csc(b*x+a)-cot(b*x+a)))^(5/2)*2^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45 \[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{3 \, {\left (b d^{3} \cos \left (b x + a\right )^{2} - b d^{3}\right )}} \]
1/3*((cos(b*x + a)^2 - 1)*sqrt(I*d)*elliptic_f(arcsin(cos(b*x + a) + I*sin (b*x + a)), -1) + (cos(b*x + a)^2 - 1)*sqrt(-I*d)*elliptic_f(arcsin(cos(b* x + a) - I*sin(b*x + a)), -1) + 2*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b* x + a))/(b*d^3*cos(b*x + a)^2 - b*d^3)
\[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {\sec {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int { \frac {\sec \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int { \frac {\sec \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {1}{\cos \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]