Integrand size = 25, antiderivative size = 145 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^4 d \sqrt {e \cos (c+d x)}}-\frac {10 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^4 d}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a+a \sin (c+d x))^3}-\frac {12 e^3 (e \cos (c+d x))^{5/2}}{d \left (a^4+a^4 \sin (c+d x)\right )} \]
-4/3*e*(e*cos(d*x+c))^(9/2)/a/d/(a+a*sin(d*x+c))^3-12*e^3*(e*cos(d*x+c))^( 5/2)/d/(a^4+a^4*sin(d*x+c))-10*e^6*(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*c os(d*x+c))^(1/2)-10*e^5*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a^4/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {\sqrt [4]{2} (e \cos (c+d x))^{13/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {13}{4},\frac {17}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{13 a^4 d e (1+\sin (c+d x))^{13/4}} \]
-1/13*(2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[7/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(a^4*d*e*(1 + Sin[c + d*x])^(13/4))
Time = 0.67 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3159, 3042, 3159, 3042, 3115, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {3 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \int (e \cos (c+d x))^{3/2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a \sin (c+d x)+a)^3}\) |
(-4*e*(e*Cos[c + d*x])^(9/2))/(3*a*d*(a + a*Sin[c + d*x])^3) - (3*e^2*((4* e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + a^2*Sin[c + d*x])) + (5*e^2*((2*e^2*Sq rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + ( 2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/a^2))/a^2
3.3.65.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Time = 4.33 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.81
\[-\frac {2 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+48 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{6}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\]
-2/3/(2*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)*(8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-8*sin(1/2*d*x+ 1/2*c)^4*cos(1/2*d*x+1/2*c)-30*(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 )^2-48*sin(1/2*d*x+1/2*c)^5+18*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15* (sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co s(1/2*d*x+1/2*c),2^(1/2))+48*sin(1/2*d*x+1/2*c)^3-20*sin(1/2*d*x+1/2*c))*e ^6/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {15 \, {\left (-i \, \sqrt {2} e^{5} \sin \left (d x + c\right ) - i \, \sqrt {2} e^{5}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, {\left (i \, \sqrt {2} e^{5} \sin \left (d x + c\right ) + i \, \sqrt {2} e^{5}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (e^{5} \cos \left (d x + c\right )^{2} + 11 \, e^{5} \sin \left (d x + c\right ) + 19 \, e^{5}\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, {\left (a^{4} d \sin \left (d x + c\right ) + a^{4} d\right )}} \]
-1/3*(15*(-I*sqrt(2)*e^5*sin(d*x + c) - I*sqrt(2)*e^5)*sqrt(e)*weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*(I*sqrt(2)*e^5*sin(d*x + c) + I*sqrt(2)*e^5)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I *sin(d*x + c)) + 2*(e^5*cos(d*x + c)^2 + 11*e^5*sin(d*x + c) + 19*e^5)*sqr t(e*cos(d*x + c)))/(a^4*d*sin(d*x + c) + a^4*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]