\(\int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [173]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 a^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {64 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \] Output:

-64/5*a^4*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+136/21*a^4*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))/d+2/7*a^4*sin(d*x+c)/d/cos(d*x+c)^(7/2)+8/5*a^4* 
sin(d*x+c)/d/cos(d*x+c)^(5/2)+94/21*a^4*sin(d*x+c)/d/cos(d*x+c)^(3/2)+64/5 
*a^4*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.69 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^4 \csc (c+d x) \left (5 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},\cos ^2(c+d x)\right )+7 \cos (c+d x) \left (4 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\cos ^2(c+d x)\right )+4 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\cos ^2(c+d x)\right )-\cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cos ^2(c+d x)\right )\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{35 d \cos ^{\frac {7}{2}}(c+d x)} \] Input:

Integrate[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(9/2),x]
 

Output:

(2*a^4*Csc[c + d*x]*(5*Hypergeometric2F1[-7/4, 1/2, -3/4, Cos[c + d*x]^2] 
+ 7*Cos[c + d*x]*(4*Hypergeometric2F1[-5/4, 1/2, -1/4, Cos[c + d*x]^2] + 5 
*Cos[c + d*x]*(2*Hypergeometric2F1[-3/4, 1/2, 1/4, Cos[c + d*x]^2] + 4*Cos 
[c + d*x]*Hypergeometric2F1[-1/4, 1/2, 3/4, Cos[c + d*x]^2] - Cos[c + d*x] 
^2*Hypergeometric2F1[1/4, 1/2, 5/4, Cos[c + d*x]^2])))*Sqrt[Sin[c + d*x]^2 
])/(35*d*Cos[c + d*x]^(7/2))
 

Rubi [A] (verified)

Time = 0.40 (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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\)

\(\Big \downarrow \) 3236

\(\displaystyle \int \left (\frac {4 a^4}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a^4}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^4}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {a^4}{\cos ^{\frac {9}{2}}(c+d x)}+\frac {a^4}{\sqrt {\cos (c+d x)}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {136 a^4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {64 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {94 a^4 \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {64 a^4 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\)

Input:

Int[(a + a*Cos[c + d*x])^4/Cos[c + d*x]^(9/2),x]
 

Output:

(-64*a^4*EllipticE[(c + d*x)/2, 2])/(5*d) + (136*a^4*EllipticF[(c + d*x)/2 
, 2])/(21*d) + (2*a^4*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (8*a^4*Sin[ 
c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (94*a^4*Sin[c + d*x])/(21*d*Cos[c + d 
*x]^(3/2)) + (64*a^4*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3236
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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(438\) vs. \(2(130)=260\).

Time = 14.23 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.99

method result size
default \(-\frac {32 \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 (\frac {253 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )}{420 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{896 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{4}}-\frac {47 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{672 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{80 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}-\frac {4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \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}}}-\frac {2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\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}\) \(439\)
parts \(\text {Expression too large to display}\) \(1026\)

Input:

int((a+a*cos(d*x+c))^4/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-32*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(253/420 
*(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))-1/896*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)^4-47/672*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- 
1/80*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)^3-4/5*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)-2/5*(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))-Ell 
ipticE(cos(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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.46 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (672 \, a^{4} \cos \left (d x + c\right )^{3} + 235 \, a^{4} \cos \left (d x + c\right )^{2} + 84 \, a^{4} \cos \left (d x + c\right ) + 15 \, a^{4}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(9/2),x, algorithm="fricas")
 

Output:

-2/105*(170*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c)) - 170*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassPIn 
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 336*I*sqrt(2)*a^4*cos(d*x + 
c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si 
n(d*x + c))) - 336*I*sqrt(2)*a^4*cos(d*x + c)^4*weierstrassZeta(-4, 0, wei 
erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (672*a^4*cos(d*x 
 + c)^3 + 235*a^4*cos(d*x + c)^2 + 84*a^4*cos(d*x + c) + 15*a^4)*sqrt(cos( 
d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**4/cos(d*x+c)**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(9/2),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(9/2), x)
 

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^4/cos(d*x+c)^(9/2),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)^4/cos(d*x + c)^(9/2), x)
 

Mupad [B] (verification not implemented)

Time = 41.62 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.35 \[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int((a + a*cos(c + d*x))^4/cos(c + d*x)^(9/2),x)
 

Output:

(2*a^4*ellipticF(c/2 + (d*x)/2, 2))/d + (8*a^4*sin(c + d*x)*hypergeom([-1/ 
4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2 
)) + (4*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*c 
os(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (8*a^4*sin(c + d*x)*hypergeom( 
[-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^ 
2)^(1/2)) + (2*a^4*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^ 
2))/(7*d*cos(c + d*x)^(7/2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+a \cos (c+d x))^4}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=a^{4} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x +\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}}d x +4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right )+6 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right )+4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right )\right ) \] Input:

int((a+a*cos(d*x+c))^4/cos(d*x+c)^(9/2),x)
 

Output:

a**4*(int(sqrt(cos(c + d*x))/cos(c + d*x),x) + int(sqrt(cos(c + d*x))/cos( 
c + d*x)**5,x) + 4*int(sqrt(cos(c + d*x))/cos(c + d*x)**4,x) + 6*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))