Integrand size = 25, antiderivative size = 134 \[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}+\frac {4 \sqrt {c \sin (a+b x)}}{7 b c d^3 (d \cos (a+b x))^{3/2}}+\frac {4 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{7 b d^4 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \]
2/7*(c*sin(b*x+a))^(1/2)/b/c/d/(d*cos(b*x+a))^(7/2)+4/7*(c*sin(b*x+a))^(1/ 2)/b/c/d^3/(d*cos(b*x+a))^(3/2)-4/7*(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)/b/d^4/(d *cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\frac {2 \cos ^3(a+b x) \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{4},\frac {5}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)}}{b c (d \cos (a+b x))^{9/2}} \]
(2*Cos[a + b*x]^3*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[1/4, 11/4, 5/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]])/(b*c*(d*Cos[a + b*x])^(9/2))
Time = 0.53 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3051, 3042, 3051, 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.
\(\displaystyle \int \frac {1}{\sqrt {c \sin (a+b x)} (d \cos (a+b x))^{9/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {c \sin (a+b x)} (d \cos (a+b x))^{9/2}}dx\) |
\(\Big \downarrow \) 3051 |
\(\displaystyle \frac {6 \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}}dx}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6 \int \frac {1}{(d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}}dx}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3051 |
\(\displaystyle \frac {6 \left (\frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\right )}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6 \left (\frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx}{3 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\right )}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {6 \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\right )}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6 \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{3 d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\right )}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {6 \left (\frac {2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {2 \sqrt {c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}\right )}{7 d^2}+\frac {2 \sqrt {c \sin (a+b x)}}{7 b c d (d \cos (a+b x))^{7/2}}\) |
(2*Sqrt[c*Sin[a + b*x]])/(7*b*c*d*(d*Cos[a + b*x])^(7/2)) + (6*((2*Sqrt[c* Sin[a + b*x]])/(3*b*c*d*(d*Cos[a + b*x])^(3/2)) + (2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])))/(7*d^2)
3.3.95.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(b*Sin[e + f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1) /(a*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1)) Int[(b*Sin[e + f*x ])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m , -1] && IntegersQ[2*m, 2*n]
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[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 = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.69
method | result | size |
default | \(\frac {\sqrt {2}\, \left (4 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right )+4 \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \tan \left (b x +a \right )+\sqrt {2}\, \tan \left (b x +a \right ) \left (\sec ^{2}\left (b x +a \right )\right )\right )}{7 b \sqrt {c \sin \left (b x +a \right )}\, \sqrt {d \cos \left (b x +a \right )}\, d^{4}}\) | \(226\) |
1/7/b*2^(1/2)/(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)/d^4*(4*(-cot(b*x+a )+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+ a))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*cos(b*x+ a)+4*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot (b*x+a)-csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2 ^(1/2))+2*2^(1/2)*tan(b*x+a)+2^(1/2)*tan(b*x+a)*sec(b*x+a)^2)
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {i \, c d} \cos \left (b x + a\right )^{4} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {-i \, c d} \cos \left (b x + a\right )^{4} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {d \cos \left (b x + a\right )} {\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {c \sin \left (b x + a\right )}\right )}}{7 \, b c d^{5} \cos \left (b x + a\right )^{4}} \]
-2/7*(2*sqrt(I*c*d)*cos(b*x + a)^4*elliptic_f(arcsin(cos(b*x + a) + I*sin( b*x + a)), -1) + 2*sqrt(-I*c*d)*cos(b*x + a)^4*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - sqrt(d*cos(b*x + a))*(2*cos(b*x + a)^2 + 1)*s qrt(c*sin(b*x + a)))/(b*c*d^5*cos(b*x + a)^4)
Timed out. \[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\int { \frac {1}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \sqrt {c \sin \left (b x + a\right )}} \,d x } \]
\[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\int { \frac {1}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \sqrt {c \sin \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \cos (a+b x))^{9/2} \sqrt {c \sin (a+b x)}} \, dx=\int \frac {1}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}\,\sqrt {c\,\sin \left (a+b\,x\right )}} \,d x \]