Integrand size = 23, antiderivative size = 121 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {94 a^4 \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {8 a^4 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:
64/5*a^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+136/21*a^4*InverseJacobiA M(1/2*d*x+1/2*c,2^(1/2))/d+94/21*a^4*cos(d*x+c)^(1/2)*sin(d*x+c)/d+8/5*a^4 *cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*a^4*cos(d*x+c)^(5/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.47 (sec) , antiderivative size = 500, normalized size of antiderivative = 4.13 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4 \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {4 \cot (c)}{5 d}+\frac {191 \cos (d x) \sin (c)}{672 d}+\frac {\cos (2 d x) \sin (2 c)}{20 d}+\frac {\cos (3 d x) \sin (3 c)}{224 d}+\frac {191 \cos (c) \sin (d x)}{672 d}+\frac {\cos (2 c) \sin (2 d x)}{20 d}+\frac {\cos (3 c) \sin (3 d x)}{224 d}\right )-\frac {17 (a+a \cos (c+d x))^4 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{42 d \sqrt {1+\cot ^2(c)}}-\frac {2 (a+a \cos (c+d x))^4 \csc (c) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^4/Sqrt[Cos[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*((-4*Cot[c] )/(5*d) + (191*Cos[d*x]*Sin[c])/(672*d) + (Cos[2*d*x]*Sin[2*c])/(20*d) + ( Cos[3*d*x]*Sin[3*c])/(224*d) + (191*Cos[c]*Sin[d*x])/(672*d) + (Cos[2*c]*S in[2*d*x])/(20*d) + (Cos[3*c]*Sin[3*d*x])/(224*d)) - (17*(a + a*Cos[c + d* x])^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]] ]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - Arc Tan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])] *Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (2*(a + a*Cos[c + d*x])^4*Csc[c]*Sec[c/2 + (d*x)/2]^8*((HypergeometricPFQ[{-1/2, - 1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c] )/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]] *Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c] ^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^ 2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqr t[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d)
Time = 0.37 (sec) , antiderivative size = 121, 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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^4}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3236 |
| \(\displaystyle \int \left (a^4 \cos ^{\frac {7}{2}}(c+d x)+4 a^4 \cos ^{\frac {5}{2}}(c+d x)+6 a^4 \cos ^{\frac {3}{2}}(c+d x)+4 a^4 \sqrt {\cos (c+d x)}+\frac {a^4}{\sqrt {\cos (c+d x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^4 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^4/Sqrt[Cos[c + d*x]],x]
Output:
(64*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (136*a^4*EllipticF[(c + d*x)/2, 2])/(21*d) + (94*a^4*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (8*a^4*Cos [c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a^4*Cos[c + d*x]^(5/2)*Sin[c + d* x])/(7*d)
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. \(271\) vs. \(2(108)=216\).
Time = 17.24 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.25
| method | result | size |
| default | \(-\frac {8 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-258 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+448 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-167 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(272\) |
| parts | \(\text {Expression too large to display}\) | \(743\) |
Input:
int((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-8/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(60*cos (1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-258*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-167*sin(1/2*d*x+1/2*c)^ 2*cos(1/2*d*x+1/2*c)+85*(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(cos(1/2*d*x+1/2*c),2^(1/2))-168*(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)))/(-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 complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (15 \, a^{4} \cos \left (d x + c\right )^{2} + 84 \, a^{4} \cos \left (d x + c\right ) + 235 \, a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="fricas")
Output:
-2/105*(170*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( d*x + c)) - 170*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I* sin(d*x + c)) - 336*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 336*I*sqrt(2)*a^4*weierstrassZ eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (1 5*a^4*cos(d*x + c)^2 + 84*a^4*cos(d*x + c) + 235*a^4)*sqrt(cos(d*x + c))*s in(d*x + c))/d
Timed out. \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4/cos(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate((a*cos(d*x + c) + a)^4/sqrt(cos(d*x + c)), x)
\[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x, algorithm="giac")
Output:
integrate((a*cos(d*x + c) + a)^4/sqrt(cos(d*x + c)), x)
Time = 40.86 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,\left (4\,a^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+2\,a^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}-\frac {8\,a^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,a^4\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int((a + a*cos(c + d*x))^4/cos(c + d*x)^(1/2),x)
Output:
(2*(4*a^4*ellipticE(c/2 + (d*x)/2, 2) + 3*a^4*ellipticF(c/2 + (d*x)/2, 2) + 2*a^4*cos(c + d*x)^(1/2)*sin(c + d*x)))/d - (8*a^4*cos(c + d*x)^(7/2)*si n(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x) ^2)^(1/2)) - (2*a^4*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {(a+a \cos (c+d x))^4}{\sqrt {\cos (c+d x)}} \, dx=a^{4} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x +4 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right )+6 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right )+\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x +4 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right )\right ) \] Input:
int((a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2),x)
Output:
a**4*(int(sqrt(cos(c + d*x))/cos(c + d*x),x) + 4*int(sqrt(cos(c + d*x)),x) + 6*int(sqrt(cos(c + d*x))*cos(c + d*x),x) + int(sqrt(cos(c + d*x))*cos(c + d*x)**3,x) + 4*int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x))