Integrand size = 25, antiderivative size = 116 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a+a \sin (c+d x))^2}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e \left (a^2+a^2 \sin (c+d x)\right )} \]
-2/5*(e*cos(d*x+c))^(3/2)/d/e/(a+a*sin(d*x+c))^2-2/5*(e*cos(d*x+c))^(3/2)/ d/e/(a^2+a^2*sin(d*x+c))-2/5*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2* c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a^2/d/cos(d* x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=-\frac {(e \cos (c+d x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{3 \sqrt [4]{2} a^2 d e (1+\sin (c+d x))^{3/4}} \]
-1/3*((e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 9/4, 7/4, (1 - Sin[c + d*x])/2])/(2^(1/4)*a^2*d*e*(1 + Sin[c + d*x])^(3/4))
Time = 0.50 (sec) , antiderivative size = 116, 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, 3160, 3042, 3162, 3042, 3121, 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.
| \(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3162 |
| \(\displaystyle \frac {-\frac {\int \sqrt {e \cos (c+d x)}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a d \sqrt {\cos (c+d x)}}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}\) |
(-2*(e*Cos[c + d*x])^(3/2))/(5*d*e*(a + a*Sin[c + d*x])^2) + ((-2*Sqrt[e*C os[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]) - (2*(e*C os[c + d*x])^(3/2))/(d*e*(a + a*Sin[c + d*x])))/(5*a)
3.3.48.3.1 Defintions of rubi rules used
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]
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 _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1)) Int[ (g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & & IntegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[b*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*S in[e + f*x]))), x] + Simp[p/(a*(p - 1)) Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && !GeQ[p, 1] && Intege rQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(303\) vs. \(2(128)=256\).
Time = 4.39 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.62
| method | result | size |
| default | \(\frac {2 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(304\) |
2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^2/sin(1/2*d*x+1/2* c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x +1/2*c)-4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^ (1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2 *c)^4*cos(1/2*d*x+1/2*c)+4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos( 1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+ 6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*sin( 1/2*d*x+1/2*c))*e/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.29 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - 2 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) - i \, \sqrt {2} \cos \left (d x + c\right ) - 2 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + 2 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) + i \, \sqrt {2} \cos \left (d x + c\right ) + 2 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 1\right )}}{5 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]
-1/5*((I*sqrt(2)*cos(d*x + c)^2 + (-I*sqrt(2)*cos(d*x + c) - 2*I*sqrt(2))* sin(d*x + c) - I*sqrt(2)*cos(d*x + c) - 2*I*sqrt(2))*sqrt(e)*weierstrassZe ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + (-I *sqrt(2)*cos(d*x + c)^2 + (I*sqrt(2)*cos(d*x + c) + 2*I*sqrt(2))*sin(d*x + c) + I*sqrt(2)*cos(d*x + c) + 2*I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*sqrt(e*cos (d*x + c))*(cos(d*x + c)^2 + (cos(d*x + c) - 1)*sin(d*x + c) + 2*cos(d*x + c) + 1))/(a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*co s(d*x + c) + 2*a^2*d)*sin(d*x + c))
Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sqrt {e \cos \left (d x + c\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int \frac {\sqrt {e\,\cos \left (c+d\,x\right )}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]