Integrand size = 25, antiderivative size = 95 \[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=-\frac {c}{3 b d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}+\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b d^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \]
-1/3*c/b/d/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(3/2)-1/2*(sin(a+1/4*Pi+b*x )^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b/d^2/(d *csc(b*x+a))^(1/2)/(c*sec(b*x+a))^(1/2)/sin(2*b*x+2*a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=-\frac {\left (1+\cos (2 (a+b x))-3 \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\csc ^2(a+b x)\right )\right ) \tan (a+b x)}{6 b d^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \]
-1/6*((1 + Cos[2*(a + b*x)] - 3*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[ -1/2, 1/4, 1/2, Csc[a + b*x]^2])*Tan[a + b*x])/(b*d^2*Sqrt[d*Csc[a + b*x]] *Sqrt[c*Sec[a + b*x]])
Time = 0.50 (sec) , antiderivative size = 95, 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, 3107, 3042, 3110, 3042, 3052, 3042, 3119}
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.
\(\displaystyle \int \frac {1}{\sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {c \sec (a+b x)} (d \csc (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3107 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx}{2 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}dx}{2 d^2}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3110 |
\(\displaystyle \frac {\int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{2 d^2 \sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}dx}{2 d^2 \sqrt {c \cos (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {\int \sqrt {\sin (2 a+2 b x)}dx}{2 d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\sin (2 a+2 b x)}dx}{2 d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c}{3 b d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}\) |
-1/3*c/(b*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(3/2)) + EllipticE[a - Pi/4 + b*x, 2]/(2*b*d^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Si n[2*a + 2*b*x]])
3.3.61.3.1 Defintions of rubi rules used
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[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[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1) /(a*f*(m + n))), x] + Simp[(m + 1)/(a^2*(m + n)) Int[(a*Csc[e + f*x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n, 0] && 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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. \(409\) vs. \(2(106)=212\).
Time = 1.24 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.32
method | result | size |
default | \(-\frac {\sqrt {2}\, \left (6 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )-3 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )-2 \cos \left (b x +a \right )^{4} \sqrt {2}+6 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1+\csc \left (b x +a \right )-\cot \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {2}\, \cos \left (b x +a \right )^{2}-3 \sqrt {2}\, \cos \left (b x +a \right )\right ) \sec \left (b x +a \right ) \csc \left (b x +a \right )}{12 b \sqrt {d \csc \left (b x +a \right )}\, \sqrt {c \sec \left (b x +a \right )}\, d^{2}}\) | \(410\) |
-1/12/b*2^(1/2)*(6*(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)*EllipticE((1+csc(b*x+a)-cot(b*x+a)) ^(1/2),1/2*2^(1/2))*cos(b*x+a)-3*(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*cos(b*x+a)^4*2^(1/2)+6*(1+c sc(b*x+a)-cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-cs c(b*x+a))^(1/2)*EllipticE((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))-3*( 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))+ 5*2^(1/2)*cos(b*x+a)^2-3*2^(1/2)*cos(b*x+a))/(d*csc(b*x+a))^(1/2)/(c*sec(b *x+a))^(1/2)/d^2*sec(b*x+a)*csc(b*x+a)
\[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=\int { \frac {1}{\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}} \sqrt {c \sec \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=\int { \frac {1}{\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}} \sqrt {c \sec \left (b x + a\right )}} \,d x } \]
\[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=\int { \frac {1}{\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}} \sqrt {c \sec \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \csc (a+b x))^{5/2} \sqrt {c \sec (a+b x)}} \, dx=\int \frac {1}{\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]