Integrand size = 23, antiderivative size = 102 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}} \]
-2/3*b/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d/e/(e*sin(d*x+c))^(3/2)- 2/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Ellipt icF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*sin(d*x+c ))^(1/2)
Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (b+a \cos (c+d x)+a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 d e (e \sin (c+d x))^{3/2}} \]
(-2*(b + a*Cos[c + d*x] + a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d* x]^(3/2)))/(3*d*e*(e*Sin[c + d*x])^(3/2))
Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3148, 3042, 3116, 3042, 3121, 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.
\(\displaystyle \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-b \sin \left (c+d x-\frac {\pi }{2}\right )}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle a \left (\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}\) |
(-2*b)/(3*d*e*(e*Sin[c + d*x])^(3/2)) + a*((-2*Cos[c + d*x])/(3*d*e*(e*Sin [c + d*x])^(3/2)) + (2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]] )/(3*d*e^2*Sqrt[e*Sin[c + d*x]]))
3.1.39.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Time = 2.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {-\frac {2 b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(124\) |
parts | \(-\frac {a \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b}{3 d e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\) | \(126\) |
(-2/3*b/e/(e*sin(d*x+c))^(3/2)-1/3*a/e^2*((1-sin(d*x+c))^(1/2)*(2*sin(d*x+ c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2 *sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/ d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\frac {{\left (\sqrt {2} a \cos \left (d x + c\right )^{2} - \sqrt {2} a\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} a \cos \left (d x + c\right )^{2} - \sqrt {2} a\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {e \sin \left (d x + c\right )}}{3 \, {\left (d e^{3} \cos \left (d x + c\right )^{2} - d e^{3}\right )}} \]
1/3*((sqrt(2)*a*cos(d*x + c)^2 - sqrt(2)*a)*sqrt(-I*e)*weierstrassPInverse (4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*a*cos(d*x + c)^2 - sqrt(2 )*a)*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(a*cos(d*x + c) + b)*sqrt(e*sin(d*x + c)))/(d*e^3*cos(d*x + c)^2 - d*e^3 )
\[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {a + b \cos {\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {b \cos \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int { \frac {b \cos \left (d x + c\right ) + a}{\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\int \frac {a+b\,\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]