Integrand size = 23, antiderivative size = 138 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {7 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {10 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}-\frac {7 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
-7*(cos(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))/a^2/d+10/3*(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))/a^2/d-7/3*cos(d*x+c)^(3/2)*sin(d*x+c )/a^2/d/(1+cos(d*x+c))-1/3*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^ 2+10/3*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.61 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\sqrt {\cos (c+d x)} \csc ^3(c+d x) \left (15+76 \cos (c+d x)-24 \cos (2 (c+d x))-12 \cos (3 (c+d x))+\cos (4 (c+d x))-40 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right ) \sin ^2(c+d x)^{3/2}-112 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\cos ^2(c+d x)\right ) \sin ^2(c+d x)^{3/2}\right )}{12 a^2 d} \]
(Sqrt[Cos[c + d*x]]*Csc[c + d*x]^3*(15 + 76*Cos[c + d*x] - 24*Cos[2*(c + d *x)] - 12*Cos[3*(c + d*x)] + Cos[4*(c + d*x)] - 40*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2]*(Sin[c + d*x]^2)^(3/2) - 112*Cos[c + d*x]*Hyperg eometric2F1[3/4, 5/2, 7/4, Cos[c + d*x]^2]*(Sin[c + d*x]^2)^(3/2)))/(12*a^ 2*d)
Time = 0.78 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3244, 27, 3042, 3456, 27, 3042, 3227, 3042, 3115, 3042, 3119, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (5 a-9 a \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (5 a-9 a \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (5 a-9 a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int 3 \sqrt {\cos (c+d x)} \left (7 a^2-10 a^2 \cos (c+d x)\right )dx}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \sqrt {\cos (c+d x)} \left (7 a^2-10 a^2 \cos (c+d x)\right )dx}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 a^2-10 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \sqrt {\cos (c+d x)}dx-10 a^2 \int \cos ^{\frac {3}{2}}(c+d x)dx\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-10 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-10 a^2 \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (7 a^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-10 a^2 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {14 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-10 a^2 \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {14 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-10 a^2 \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {14 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*(Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - ((14*C os[c + d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*((14*a^2*Ellip ticE[(c + d*x)/2, 2])/d - 10*a^2*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2 *Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))))/a^2)/(6*a^2)
3.2.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[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[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Time = 6.20 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.96
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+42 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}{6 a^{2} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(270\) |
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*cos(1/2*d *x+1/2*c)^8+12*cos(1/2*d*x+1/2*c)^6+20*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co s(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2* d*x+1/2*c)^3+42*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^( 1/2)*cos(1/2*d*x+1/2*c)^3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-48*cos(1/2 *d*x+1/2*c)^4+21*cos(1/2*d*x+1/2*c)^2-1)/a^2/(-2*sin(1/2*d*x+1/2*c)^4+sin( 1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^3/sin(1/2*d*x+1/2*c)/(2*cos(1/2 *d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 10\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 10 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/6*(2*(2*cos(d*x + c)^2 + 13*cos(d*x + c) + 10)*sqrt(cos(d*x + c))*sin(d* x + c) - 10*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt( 2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 10*(-I*sqr t(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c )^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]