Integrand size = 25, antiderivative size = 93 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}+\frac {2 c^2 \sqrt {d \csc (a+b x)} \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{3 b} \]
2/3*c*d*(c*sec(b*x+a))^(3/2)/b/(d*csc(b*x+a))^(1/2)-2/3*c^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))*(d*cs c(b*x+a))^(1/2)*(c*sec(b*x+a))^(1/2)*sin(2*b*x+2*a)^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=-\frac {2 c d \left (-1+\left (-\cot ^2(a+b x)\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\csc ^2(a+b x)\right )\right ) (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}} \]
(-2*c*d*(-1 + (-Cot[a + b*x]^2)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, Csc [a + b*x]^2])*(c*Sec[a + b*x])^(3/2))/(3*b*Sqrt[d*Csc[a + b*x]])
Time = 0.49 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3106, 3042, 3110, 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.
\(\displaystyle \int (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \sec (a+b x))^{5/2} \sqrt {d \csc (a+b x)}dx\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {2}{3} c^2 \int \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} c^2 \int \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3110 |
\(\displaystyle \frac {2}{3} c^2 \sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} c^2 \sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {2}{3} c^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} c^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 c^2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{3 b}+\frac {2 c d (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}\) |
(2*c*d*(c*Sec[a + b*x])^(3/2))/(3*b*Sqrt[d*Csc[a + b*x]]) + (2*c^2*Sqrt[d* Csc[a + b*x]]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2 *a + 2*b*x]])/(3*b)
3.3.50.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(n - 1))), x] + Simp[b^2*((m + n - 2)/(n - 1)) Int[(a*Csc[e + f*x]) ^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(a*Csc[e + f*x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x] )^m*(b*Cos[e + f*x])^n Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[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 = 1.12 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.23
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {c \sec \left (b x +a \right )}\, \sqrt {d \csc \left (b x +a \right )}\, c^{2} \left (2 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )+2 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \tan \left (b x +a \right )\right )}{3 b}\) | \(207\) |
1/3/b*2^(1/2)*(c*sec(b*x+a))^(1/2)*(d*csc(b*x+a))^(1/2)*c^2*(2*(1+csc(b*x+ a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(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*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(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))+2^(1/2)*tan(b*x+a))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28 \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\frac {-i \, \sqrt {-4 i \, c d} c^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + i \, \sqrt {4 i \, c d} c^{2} \cos \left (b x + a\right ) F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, c^{2} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}} \sin \left (b x + a\right )}{3 \, b \cos \left (b x + a\right )} \]
1/3*(-I*sqrt(-4*I*c*d)*c^2*cos(b*x + a)*elliptic_f(arcsin(cos(b*x + a) + I *sin(b*x + a)), -1) + I*sqrt(4*I*c*d)*c^2*cos(b*x + a)*elliptic_f(arcsin(c os(b*x + a) - I*sin(b*x + a)), -1) + 2*c^2*sqrt(c/cos(b*x + a))*sqrt(d/sin (b*x + a))*sin(b*x + a))/(b*cos(b*x + a))
Timed out. \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\int { \sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]
\[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\int { \sqrt {d \csc \left (b x + a\right )} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx=\int {\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}} \,d x \]