\(\int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [191]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {64 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^4 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {94 a^4 \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \] Output:

64/5*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+136/21*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+2/7*a^4*sin(d*x+c)/d/sec(d*x+c)^(5/2)+8/5*a^4*sin(d*x 
+c)/d/sec(d*x+c)^(3/2)+94/21*a^4*sin(d*x+c)/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 3.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^4 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \left (-5376 i+\frac {10752 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-2720 i \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right ) \sec (c+d x)+1910 \sin (c+d x)+336 \sin (2 (c+d x))+30 \sin (3 (c+d x))\right )}{420 d \sqrt {\sec (c+d x)}} \] Input:

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

Output:

(a^4*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(-5376*I + ((10752*I) 
*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I 
)*(c + d*x))] - (2720*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1 
/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]*Sec[c + d*x] + 1910*Sin[c + d*x] + 336 
*Sin[2*(c + d*x)] + 30*Sin[3*(c + d*x)]))/(420*d*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 161, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4278

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Time = 7.00 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {8 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{4} \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-258 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+448 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-167 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+85 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(272\)
parts	\(\text {Expression too large to display}\)	\(858\)

Input:

int((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),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*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+ 
1/2*c)+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-167*cos(1/2*d*x+1/2*c)* 
sin(1/2*d*x+1/2*c)^2+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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(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)/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 = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 336 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (15 \, a^{4} \cos \left (d x + c\right )^{3} + 84 \, a^{4} \cos \left (d x + c\right )^{2} + 235 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d} \] Input:

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(7/2),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)^3 + 84*a^4*cos(d*x + c)^2 + 235*a^4*cos(d*x + c))*sin(d 
*x + c)/sqrt(cos(d*x + c)))/d
 

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=a^{4} \left (\int \frac {1}{\sec ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {6}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \] Input:

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

Output:

a**4*(Integral(sec(c + d*x)**(-7/2), x) + Integral(4/sec(c + d*x)**(5/2), 
x) + Integral(6/sec(c + d*x)**(3/2), x) + Integral(4/sqrt(sec(c + d*x)), x 
) + Integral(sqrt(sec(c + d*x)), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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