Integrand size = 25, antiderivative size = 150 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{33 a^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )} \]
10/33*sin(d*x+c)/a^2/d/e/(e*cos(d*x+c))^(3/2)-2/11/d/e/(e*cos(d*x+c))^(3/2 )/(a+a*sin(d*x+c))^2-2/11/d/e/(e*cos(d*x+c))^(3/2)/(a^2+a^2*sin(d*x+c))+10 /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^2/d/e^2/(e*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {15}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{6\ 2^{3/4} a^2 d e (e \cos (c+d x))^{3/2}} \]
(Hypergeometric2F1[-3/4, 15/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x ])^(3/4))/(6*2^(3/4)*a^2*d*e*(e*Cos[c + d*x])^(3/2))
Time = 0.66 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3160, 3042, 3162, 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.
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7 \int \frac {1}{(e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)}dx}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \int \frac {1}{(e \cos (c+d x))^{5/2} (\sin (c+d x) a+a)}dx}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3162 |
| \(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{(e \cos (c+d x))^{5/2}}dx}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{\left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {7 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\right )}{11 a}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}}\) |
-2/(11*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2) + (7*(-2/(7*d*e* (e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])) + (5*((2*Sqrt[Cos[c + d*x]]*E llipticF[(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (2*Sin[c + d*x] )/(3*d*e*(e*Cos[c + d*x])^(3/2))))/(7*a)))/(11*a)
3.3.51.3.1 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 _)])^(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. \(556\) vs. \(2(158)=316\).
Time = 8.59 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.71
| method | result | size |
| default | \(-\frac {2 \left (160 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-400 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-320 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+400 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+264 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-200 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-104 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+50 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+28 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{33 \left (32 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \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}\, e^{2} d}\) | \(557\) |
-2/33/(32*sin(1/2*d*x+1/2*c)^10-80*sin(1/2*d*x+1/2*c)^8+80*sin(1/2*d*x+1/2 *c)^6-40*sin(1/2*d*x+1/2*c)^4+10*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)/e^2*(160*(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+160*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-400* (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)^8-320*cos(1/2*d*x+1/2*c)*sin( 1/2*d*x+1/2*c)^8+400*(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+264*si n(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-200*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(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)^4-104*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+50*(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+28*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2 *c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt icF(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\frac {5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (10 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{2} - 7\right )} \sin \left (d x + c\right ) - 4\right )}}{33 \, {\left (a^{2} d e^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2}\right )}} \]
1/33*(5*(-I*sqrt(2)*cos(d*x + c)^4 + 2*I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) + 2*I*sqrt(2)*cos(d*x + c)^2)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*cos(d*x + c)^4 - 2*I*sqrt(2)*cos(d *x + c)^2*sin(d*x + c) - 2*I*sqrt(2)*cos(d*x + c)^2)*sqrt(e)*weierstrassPI nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(e*cos(d*x + c))*(10* cos(d*x + c)^2 + (5*cos(d*x + c)^2 - 7)*sin(d*x + c) - 4))/(a^2*d*e^3*cos( d*x + c)^4 - 2*a^2*d*e^3*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*e^3*cos(d*x + c)^2)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]