Integrand size = 23, antiderivative size = 147 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {28 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {52 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {52 a^3 \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {28 a^3 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]
-28/5*a^3*(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))/d+52/21*a^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))/d+2/7*a^3*sin(d*x+c)/d/cos(d *x+c)^(7/2)+6/5*a^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)+52/21*a^3*sin(d*x+c)/d/c os(d*x+c)^(3/2)+28/5*a^3*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.64 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^3 \csc (c+d x) \left (5 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},\cos ^2(c+d x)\right )+7 \cos (c+d x) \left (3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\cos ^2(c+d x)\right )+\cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\cos ^2(c+d x)\right )\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{35 d \cos ^{\frac {7}{2}}(c+d x)} \]
(2*a^3*Csc[c + d*x]*(5*Hypergeometric2F1[-7/4, 1/2, -3/4, Cos[c + d*x]^2] + 7*Cos[c + d*x]*(3*Hypergeometric2F1[-5/4, 1/2, -1/4, Cos[c + d*x]^2] + 5 *Cos[c + d*x]*(Hypergeometric2F1[-3/4, 1/2, 1/4, Cos[c + d*x]^2] + Cos[c + d*x]*Hypergeometric2F1[-1/4, 1/2, 3/4, Cos[c + d*x]^2])))*Sqrt[Sin[c + d* x]^2])/(35*d*Cos[c + d*x]^(7/2))
Time = 0.37 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3236, 2009}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (\frac {a^3}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {3 a^3}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {3 a^3}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {a^3}{\cos ^{\frac {9}{2}}(c+d x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {52 a^3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {28 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {52 a^3 \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {28 a^3 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\) |
(-28*a^3*EllipticE[(c + d*x)/2, 2])/(5*d) + (52*a^3*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a^3*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (6*a^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (52*a^3*Sin[c + d*x])/(21*d*Cos[c + d* x]^(3/2)) + (28*a^3*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
3.2.66.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(438\) vs. \(2(179)=358\).
Time = 10.41 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.99
| method | result | size |
| default | \(-\frac {16 \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 )}\, a^{3} \left (-\frac {13 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{168 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {53 \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 )}{105 \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 )}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{448 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{4}}-\frac {7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{10 \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 )}}-\frac {7 \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 (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{20 \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 )}}-\frac {3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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 )}}{160 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}\right )}{\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}\) | \(439\) |
| parts | \(\text {Expression too large to display}\) | \(1006\) |
-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-13/168 *cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^2+53/105*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/ 2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/ 2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/448*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^4- 7/10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1) *sin(1/2*d*x+1/2*c)^2)^(1/2)-7/20*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2 *d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2 )*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2)))-3/160*cos(1/2*d*x+1/2*c)*(-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)^2-1/2)^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 = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.46 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (65 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 65 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 147 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 147 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (294 \, a^{3} \cos \left (d x + c\right )^{3} + 130 \, a^{3} \cos \left (d x + c\right )^{2} + 63 \, a^{3} \cos \left (d x + c\right ) + 15 \, a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \]
-2/105*(65*I*sqrt(2)*a^3*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 65*I*sqrt(2)*a^3*cos(d*x + c)^4*weierstrassPInve rse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 147*I*sqrt(2)*a^3*cos(d*x + c) ^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( d*x + c))) - 147*I*sqrt(2)*a^3*cos(d*x + c)^4*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (294*a^3*cos(d*x + c)^3 + 130*a^3*cos(d*x + c)^2 + 63*a^3*cos(d*x + c) + 15*a^3)*sqrt(cos(d* x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Time = 15.57 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \cos (c+d x))^3}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\frac {2\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+\frac {6\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}+2\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+2\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]
((2*a^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + (6* a^3*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) )/5 + 2*a^3*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 2*a^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))