Integrand size = 25, antiderivative size = 138 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)}}+\frac {22 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2} \]
22/21*e^3*(e*cos(d*x+c))^(7/2)/a^3/d+22/15*e^5*(e*cos(d*x+c))^(3/2)*sin(d* x+c)/a^3/d+4/3*e*(e*cos(d*x+c))^(11/2)/a/d/(a+a*sin(d*x+c))^2+22/5*e^6*(co s(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^3/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.48 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2\ 2^{3/4} (e \cos (c+d x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {15}{4},\frac {19}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{15 a^3 d e (1+\sin (c+d x))^{15/4}} \]
(-2*2^(3/4)*(e*Cos[c + d*x])^(15/2)*Hypergeometric2F1[1/4, 15/4, 19/4, (1 - Sin[c + d*x])/2])/(15*a^3*d*e*(1 + Sin[c + d*x])^(15/4))
Time = 0.64 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07, 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, 3161, 3042, 3115, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{13/2}}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{13/2}}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {11 e^2 \int \frac {(e \cos (c+d x))^{9/2}}{\sin (c+d x) a+a}dx}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 e^2 \int \frac {(e \cos (c+d x))^{9/2}}{\sin (c+d x) a+a}dx}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \int (e \cos (c+d x))^{5/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \cos (c+d x)}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {11 e^2 \left (\frac {e^2 \left (\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2}\) |
(4*e*(e*Cos[c + d*x])^(11/2))/(3*a*d*(a + a*Sin[c + d*x])^2) + (11*e^2*((2 *e*(e*Cos[c + d*x])^(7/2))/(7*a*d) + (e^2*((6*e^2*Sqrt[e*Cos[c + d*x]]*Ell ipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*e*(e*Cos[c + d*x])^( 3/2)*Sin[c + d*x])/(5*d)))/a))/(3*a^2)
3.3.54.3.1 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[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[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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si mp[g^2/a Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x ] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
Time = 172.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(-\frac {2 e^{7} \left (240 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-480 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-504 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-200 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+126 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-231 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+440 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-125 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(216\) |
-2/105/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^7*(240 *sin(1/2*d*x+1/2*c)^9+504*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-480*sin( 1/2*d*x+1/2*c)^7-504*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-200*sin(1/2*d *x+1/2*c)^5+126*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-231*(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))+440*sin(1/2*d*x+1/2*c)^3-125*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 = 121, normalized size of antiderivative = 0.88 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {231 i \, \sqrt {2} e^{\frac {13}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 231 i \, \sqrt {2} e^{\frac {13}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (15 \, e^{6} \cos \left (d x + c\right )^{3} + 63 \, e^{6} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 140 \, e^{6} \cos \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{105 \, a^{3} d} \]
1/105*(231*I*sqrt(2)*e^(13/2)*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c))) - 231*I*sqrt(2)*e^(13/2)*weierstrass Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2 *(15*e^6*cos(d*x + c)^3 + 63*e^6*cos(d*x + c)*sin(d*x + c) - 140*e^6*cos(d *x + c))*sqrt(e*cos(d*x + c)))/(a^3*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{13/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]