Integrand size = 23, antiderivative size = 173 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\frac {128 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {904 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{231 d}+\frac {904 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {128 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {150 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {8 a^4 \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {2 a^4 \cos ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{11 d} \]
128/15*a^4*(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+904/231*a^4*(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+128/45*a^4*cos(d*x+c)^( 3/2)*sin(d*x+c)/d+150/77*a^4*cos(d*x+c)^(5/2)*sin(d*x+c)/d+8/9*a^4*cos(d*x +c)^(7/2)*sin(d*x+c)/d+2/11*a^4*cos(d*x+c)^(9/2)*sin(d*x+c)/d+904/231*a^4* sin(d*x+c)*cos(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.09 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.57 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\frac {a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (-108480 \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\cos (c+d x) (-236544 \cot (c)+122610 \sin (c+d x)+45584 \sin (2 (c+d x))+14445 \sin (3 (c+d x))+3080 \sin (4 (c+d x))+315 \sin (5 (c+d x)))+\frac {59136 \sec (c) \left (-2 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))+(3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}}\right )}{443520 d \sqrt {\cos (c+d x)}} \]
(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(-108480*Cos[c + d*x]*Sqrt[Co s[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5 /4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Cos[c + d*x]*(-236544*Cot[c] + 122610*Sin[c + d*x] + 45584*Sin[2*(c + d*x)] + 1 4445*Sin[3*(c + d*x)] + 3080*Sin[4*(c + d*x)] + 315*Sin[5*(c + d*x)]) + (5 9136*Sec[c]*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Ta n[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + C os[c + d*x + ArcTan[Tan[c]]])*Csc[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/( Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])))/(443520*d*Sqrt[Cos[c + d*x]])
Time = 0.44 (sec) , antiderivative size = 173, 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.
| \(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (a^4 \cos ^{\frac {11}{2}}(c+d x)+4 a^4 \cos ^{\frac {9}{2}}(c+d x)+6 a^4 \cos ^{\frac {7}{2}}(c+d x)+4 a^4 \cos ^{\frac {5}{2}}(c+d x)+a^4 \cos ^{\frac {3}{2}}(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {904 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{231 d}+\frac {128 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x)}{11 d}+\frac {8 a^4 \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{9 d}+\frac {150 a^4 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{77 d}+\frac {128 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {904 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}\) |
(128*a^4*EllipticE[(c + d*x)/2, 2])/(15*d) + (904*a^4*EllipticF[(c + d*x)/ 2, 2])/(231*d) + (904*a^4*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (128* a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (150*a^4*Cos[c + d*x]^(5/2)* Sin[c + d*x])/(77*d) + (8*a^4*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + (2* a^4*Cos[c + d*x]^(9/2)*Sin[c + d*x])/(11*d)
3.2.67.3.1 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]
Time = 15.95 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.58
| method | result | size |
| default | \(-\frac {8 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{4} \left (5040 \left (\cos ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5320 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1740 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+326 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+678 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4465 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1695 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3696 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2001 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3465 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(273\) |
| parts | \(\text {Expression too large to display}\) | \(1029\) |
-8/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(5040* cos(1/2*d*x+1/2*c)^13-5320*cos(1/2*d*x+1/2*c)^11+1740*cos(1/2*d*x+1/2*c)^9 +326*cos(1/2*d*x+1/2*c)^7+678*cos(1/2*d*x+1/2*c)^5-4465*cos(1/2*d*x+1/2*c) ^3+1695*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3696*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c os(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2001*co s(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)/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 = 188, normalized size of antiderivative = 1.09 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=-\frac {2 \, {\left (3390 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 3390 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 7392 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 7392 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (315 \, a^{4} \cos \left (d x + c\right )^{4} + 1540 \, a^{4} \cos \left (d x + c\right )^{3} + 3375 \, a^{4} \cos \left (d x + c\right )^{2} + 4928 \, a^{4} \cos \left (d x + c\right ) + 6780 \, a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3465 \, d} \]
-2/3465*(3390*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c)) - 3390*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 7392*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 7392*I*sqrt(2)*a^4*weierst rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (315*a^4*cos(d*x + c)^4 + 1540*a^4*cos(d*x + c)^3 + 3375*a^4*cos(d*x + c)^2 + 4928*a^4*cos(d*x + c) + 6780*a^4)*sqrt(cos(d*x + c))*sin(d*x + c))/ d
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\text {Timed out} \]
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
Time = 15.03 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.28 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx=\frac {2\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {8\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,a^4\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,a^4\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^4\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*a^4*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*a^4*cos(c + d*x)^(1/2)*sin( c + d*x))/(3*d) - (8*a^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7 /4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (4*a^4*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d* (sin(c + d*x)^2)^(1/2)) - (8*a^4*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeo m([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (2*a ^4*cos(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13/4], 17/4, cos(c + d *x)^2))/(13*d*(sin(c + d*x)^2)^(1/2))