Integrand size = 25, antiderivative size = 191 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{33 a^4 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4+a^4 \sin (c+d x)\right )} \]
2/33*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x +1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^4/d/(e*cos(d*x+c))^(1/2)-2/15*(e*cos(d *x+c))^(1/2)/d/e/(a+a*sin(d*x+c))^4-14/165*(e*cos(d*x+c))^(1/2)/a/d/e/(a+a *sin(d*x+c))^3-2/33*(e*cos(d*x+c))^(1/2)/d/e/(a^2+a^2*sin(d*x+c))^2-2/33*( e*cos(d*x+c))^(1/2)/d/e/(a^4+a^4*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=-\frac {\sqrt {e \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {19}{4},\frac {5}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{4\ 2^{3/4} a^4 d e \sqrt [4]{1+\sin (c+d x)}} \]
-1/4*(Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[1/4, 19/4, 5/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^4*d*e*(1 + Sin[c + d*x])^(1/4))
Time = 0.86 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3160, 3042, 3160, 3042, 3160, 3042, 3162, 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.
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^4 \sqrt {e \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^4 \sqrt {e \cos (c+d x)}}dx\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)^3}dx}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)^3}dx}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)^2}dx}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)^2}dx}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)}dx}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)}dx}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3162 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 a}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 a \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\right )}{11 a}-\frac {2 \sqrt {e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3}\right )}{15 a}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4}\) |
(-2*Sqrt[e*Cos[c + d*x]])/(15*d*e*(a + a*Sin[c + d*x])^4) + (7*((-2*Sqrt[e *Cos[c + d*x]])/(11*d*e*(a + a*Sin[c + d*x])^3) + (5*((-2*Sqrt[e*Cos[c + d *x]])/(7*d*e*(a + a*Sin[c + d*x])^2) + (3*((2*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2])/(3*a*d*Sqrt[e*Cos[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]])/ (3*d*e*(a + a*Sin[c + d*x]))))/(7*a)))/(11*a)))/(15*a)
3.3.71.3.1 Defintions of rubi rules used
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 _)])^(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. \(761\) vs. \(2(195)=390\).
Time = 12.18 (sec) , antiderivative size = 762, normalized size of antiderivative = 3.99
-2/165/(128*sin(1/2*d*x+1/2*c)^14-448*sin(1/2*d*x+1/2*c)^12+672*sin(1/2*d* x+1/2*c)^10-560*sin(1/2*d*x+1/2*c)^8+280*sin(1/2*d*x+1/2*c)^6-84*sin(1/2*d *x+1/2*c)^4+14*sin(1/2*d*x+1/2*c)^2-1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2* d*x+1/2*c)^2*e+e)^(1/2)*(640*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^ 14+640*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-2240*(sin(1/2*d*x+1/2*c)^2 )^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^( 1/2)*sin(1/2*d*x+1/2*c)^12-1920*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^12+3 360*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^10+2496*sin(1/2*d*x+1/2*c )^10*cos(1/2*d*x+1/2*c)-2800*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^ 8-1792*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+1400*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*sin(1/2*d*x+1/2*c)^6+616*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-420*( sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1 /2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-40*sin(1/2*d*x+1/2*c)^4*cos( 1/2*d*x+1/2*c)+70*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c ),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+240*sin(1 /2*d*x+1/2*c)^5+160*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(sin(1/2*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\frac {5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 8 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 4 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) - 8 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 8 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 4 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) + 8 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (20 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{2} - 37\right )} \sin \left (d x + c\right ) - 48\right )}}{165 \, {\left (a^{4} d e \cos \left (d x + c\right )^{4} - 8 \, a^{4} d e \cos \left (d x + c\right )^{2} + 8 \, a^{4} d e - 4 \, {\left (a^{4} d e \cos \left (d x + c\right )^{2} - 2 \, a^{4} d e\right )} \sin \left (d x + c\right )\right )}} \]
1/165*(5*(-I*sqrt(2)*cos(d*x + c)^4 + 8*I*sqrt(2)*cos(d*x + c)^2 + 4*(I*sq rt(2)*cos(d*x + c)^2 - 2*I*sqrt(2))*sin(d*x + c) - 8*I*sqrt(2))*sqrt(e)*we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*cos (d*x + c)^4 - 8*I*sqrt(2)*cos(d*x + c)^2 + 4*(-I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2))*sin(d*x + c) + 8*I*sqrt(2))*sqrt(e)*weierstrassPInverse(-4, 0 , cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(e*cos(d*x + c))*(20*cos(d*x + c) ^2 + (5*cos(d*x + c)^2 - 37)*sin(d*x + c) - 48))/(a^4*d*e*cos(d*x + c)^4 - 8*a^4*d*e*cos(d*x + c)^2 + 8*a^4*d*e - 4*(a^4*d*e*cos(d*x + c)^2 - 2*a^4* d*e)*sin(d*x + c))
Timed out. \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\int { \frac {1}{\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\int { \frac {1}{\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx=\int \frac {1}{\sqrt {e\,\cos \left (c+d\,x\right )}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]