Integrand size = 19, antiderivative size = 82 \[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{3 b d^2} \]
-2/3*csc(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)+1/3*csc(b*x+a)*(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))*sin(2*b*x +2*a)^(1/2)*(d*tan(b*x+a))^(1/2)/b/d^2
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {2 \cos (2 (a+b x)) \sec (a+b x) \sqrt {\sec ^2(a+b x)} \left (\sqrt {\sec ^2(a+b x)}-\sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \tan ^{\frac {3}{2}}(a+b x)\right )}{3 b (d \tan (a+b x))^{3/2} \left (-1+\tan ^2(a+b x)\right )} \]
(2*Cos[2*(a + b*x)]*Sec[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*Tan[a + b*x])^(3/2)*(-1 + Tan[a + b*x]^2))
Time = 0.47 (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, 3077, 3042, 3081, 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.
\(\displaystyle \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) (d \tan (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3077 |
\(\displaystyle -\frac {\int \csc (a+b x) \sqrt {d \tan (a+b x)}dx}{3 d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx}{3 d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle -\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 d^2 \sqrt {\sin (a+b x)}}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{3 d^2 \sqrt {\sin (a+b x)}}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{3 b d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}\) |
(-2*Csc[a + b*x])/(3*b*d*Sqrt[d*Tan[a + b*x]]) - (Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b*d^2)
3.2.1.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n + 1))), x] - Simp[(n + 1)/(b^2*(m + n + 1)) Int[(a*Sin[e + f*x])^m*( b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && IntegersQ[2*m, 2*n] && !(EqQ[n, -3/2] && EqQ[m, 1] )
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
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 = 0.79 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.38
method | result | size |
default | \(-\frac {\left (\sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\sec \left (b x +a \right ) \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {-\csc \left (b x +a \right )+1+\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\csc \left (b x +a \right ) \sqrt {2}\right ) \sqrt {2}}{3 b \sqrt {d \tan \left (b x +a \right )}\, d}\) | \(195\) |
-1/3/b/(d*tan(b*x+a))^(1/2)/d*((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))+sec(b*x+a)*(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))+csc(b*x+a)*2^(1/2))*2^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45 \[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {i \, d} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-i \, d} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{3 \, {\left (b d^{2} \cos \left (b x + a\right )^{2} - b d^{2}\right )}} \]
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^2*cos(b*x + a)^2 - b*d^2)
\[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\csc {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]