Integrand size = 25, antiderivative size = 112 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=-\frac {6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d e^2 \sqrt {\cos (c+d x)}}+\frac {6 \sin (c+d x)}{5 a d e \sqrt {e \cos (c+d x)}}-\frac {2}{5 d e \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))} \]
6/5*sin(d*x+c)/a/d/e/(e*cos(d*x+c))^(1/2)-2/5/d/e/(a+a*sin(d*x+c))/(e*cos( d*x+c))^(1/2)-6/5*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic E(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a/d/e^2/cos(d*x+c)^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {9}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [4]{1+\sin (c+d x)}}{\sqrt [4]{2} a d e \sqrt {e \cos (c+d x)}} \]
(Hypergeometric2F1[-1/4, 9/4, 3/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x] )^(1/4))/(2^(1/4)*a*d*e*Sqrt[e*Cos[c + d*x]])
Time = 0.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3162, 3042, 3116, 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.
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3162 |
| \(\displaystyle \frac {3 \int \frac {1}{(e \cos (c+d x))^{3/2}}dx}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {1}{\left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {3 \left (\frac {2 \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}}-\frac {\int \sqrt {e \cos (c+d x)}dx}{e^2}\right )}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}}-\frac {\int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{e^2}\right )}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {3 \left (\frac {2 \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}}-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{e^2 \sqrt {\cos (c+d x)}}\right )}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}}-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{e^2 \sqrt {\cos (c+d x)}}\right )}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {3 \left (\frac {2 \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d e^2 \sqrt {\cos (c+d x)}}\right )}{5 a}-\frac {2}{5 d e (a \sin (c+d x)+a) \sqrt {e \cos (c+d x)}}\) |
-2/(5*d*e*Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])) + (3*((-2*Sqrt[e*Cos[ c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*e^2*Sqrt[Cos[c + d*x]]) + (2*Sin[c + d*x])/(d*e*Sqrt[e*Cos[c + d*x]])))/(5*a)
3.3.40.3.1 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[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 _)]), 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. \(305\) vs. \(2(124)=248\).
Time = 3.70 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.73
| method | result | size |
| default | \(\frac {\frac {48 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {24 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {48 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {24 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{5}-\frac {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{\left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e d}\) | \(306\) |
2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a/sin(1/2*d*x+1/2*c) /(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d* x+1/2*c)-12*(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)^4-24*sin(1/2*d*x+ 1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(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+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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))- sin(1/2*d*x+1/2*c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=-\frac {3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, \sqrt {2} \cos \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - i \, \sqrt {2} \cos \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) - 2\right )}}{5 \, {\left (a d e^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a d e^{2} \cos \left (d x + c\right )\right )}} \]
-1/5*(3*(I*sqrt(2)*cos(d*x + c)*sin(d*x + c) + I*sqrt(2)*cos(d*x + c))*sqr t(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c))) + 3*(-I*sqrt(2)*cos(d*x + c)*sin(d*x + c) - I*sqrt(2)*cos(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*sqrt(e*cos(d*x + c))*(3*cos(d*x + c)^2 - 3*sin( d*x + c) - 2))/(a*d*e^2*cos(d*x + c)*sin(d*x + c) + a*d*e^2*cos(d*x + c))
\[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=\frac {\int \frac {1}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sin {\left (c + d x \right )} + \left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]
\[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}} \,d x } \]
\[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \,d x \]