\(\int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx\) [312]

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

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {56 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {66 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

-56/5*a^4*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c 
)^(1/2)/d+32/3*a^4*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)) 
*sec(d*x+c)^(1/2)/d+66/5*a^4*sec(d*x+c)^(1/2)*sin(d*x+c)/d+8/3*a^4*sec(d*x 
+c)^(3/2)*sin(d*x+c)/d+2/5*a^4*sec(d*x+c)^(5/2)*sin(d*x+c)/d
 

Mathematica [C] (verified)

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

Time = 5.50 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.73 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {8 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (21 \left (1+e^{2 i (c+d x)}\right )+21 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+20 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\sqrt {\sec (c+d x)} (-3 (-61+5 \cos (2 c)) \cos (d x) \csc (c)+30 \cos (c) \sin (d x)+2 (20+3 \sec (c+d x)) \tan (c+d x))\right )}{240 d} \] Input:

Integrate[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(7/2),x]
 

Output:

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(((-8*I)*Sqrt[2]*Sqrt[E^(I*(c 
 + d*x))/(1 + E^((2*I)*(c + d*x)))]*(21*(1 + E^((2*I)*(c + d*x))) + 21*(-1 
 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 
 3/4, -E^((2*I)*(c + d*x))] + 20*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 
 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d* 
x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + Sqrt[Sec[c + d*x]]*(-3*(-61 
+ 5*Cos[2*c])*Cos[d*x]*Csc[c] + 30*Cos[c]*Sin[d*x] + 2*(20 + 3*Sec[c + d*x 
])*Tan[c + d*x])))/(240*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3717, 3042, 4278, 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.

\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4dx\)

\(\Big \downarrow \) 3717

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^4}{\sqrt {\sec (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4278

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a^4 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {8 a^4 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {66 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {56 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\)

Input:

Int[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(7/2),x]
 

Output:

(-56*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ 
(5*d) + (32*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + 
d*x]])/(3*d) + (66*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (8*a^4*Sec 
[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*a^4*Sec[c + d*x]^(5/2)*Sin[c + d* 
x])/(5*d)
 

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 3717
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4278
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f 
*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I 
GtQ[m, 0] && RationalQ[n]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(140)=280\).

Time = 197.47 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.40

method result size
default \(-\frac {32 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{4} \left (\frac {41 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{60 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{20 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{320 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}-\frac {33 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{40 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{24 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(386\)
parts \(\text {Expression too large to display}\) \(1031\)

Input:

int((a+a*cos(d*x+c))^4*sec(d*x+c)^(7/2),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*(41/60*( 
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))-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)))-1/320*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)^3-33/40*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)-1/24*cos(1/2*d*x+1/2*c)*(-2*s 
in(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)/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 = 202, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \, {\left (40 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 40 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 42 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 42 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (99 \, a^{4} \cos \left (d x + c\right )^{2} + 20 \, a^{4} \cos \left (d x + c\right ) + 3 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

-2/15*(40*I*sqrt(2)*a^4*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) - 40*I*sqrt(2)*a^4*cos(d*x + c)^2*weierstrassPInver 
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 42*I*sqrt(2)*a^4*cos(d*x + c)^2 
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d* 
x + c))) - 42*I*sqrt(2)*a^4*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (99*a^4*cos(d*x + c)^ 
2 + 20*a^4*cos(d*x + c) + 3*a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d 
*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*cos(d*x+c))**4*sec(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^4 \,d x \] Input:

int((1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^4,x)
 

Output:

int((1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^4, x)
 

Reduce [F]

\[ \int (a+a \cos (c+d x))^4 \sec ^{\frac {7}{2}}(c+d x) \, dx=a^{4} \left (4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x +4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right )+6 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) \] Input:

int((a+a*cos(d*x+c))^4*sec(d*x+c)^(7/2),x)
                                                                                    
                                                                                    
 

Output:

a**4*(4*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x) + int(sqrt( 
sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**3,x) + 4*int(sqrt(sec(c + d*x) 
)*cos(c + d*x)**3*sec(c + d*x)**3,x) + 6*int(sqrt(sec(c + d*x))*cos(c + d* 
x)**2*sec(c + d*x)**3,x) + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x))