\(\int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [189]

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {40 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {8 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \] Output:

40/3*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+2/3*a^4*sin(d*x+c)/d/sec(d*x+c)^(1/2)+8*a^4*sec(d*x+c)^(1/2)*sin 
(d*x+c)/d+2/3*a^4*sec(d*x+c)^(3/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^4 \sec ^{\frac {3}{2}}(c+d x) \left (80 \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (c+d x)+24 \sin (2 (c+d x))+\sin (3 (c+d x))\right )}{6 d} \] Input:

Integrate[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(3/2),x]
 

Output:

(a^4*Sec[c + d*x]^(3/2)*(80*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 
 5*Sin[c + d*x] + 24*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(6*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 118, 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.

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\)

\(\Big \downarrow \) 4278

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a^4 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {40 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\)

Input:

Int[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(3/2),x]
 

Output:

(40*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/( 
3*d) + (2*a^4*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (8*a^4*Sqrt[Sec[c + 
 d*x]]*Sin[c + d*x])/d + (2*a^4*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*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 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. \(291\) vs. \(2(103)=206\).

Time = 4.16 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.47

method	result	size
default	\(\frac {8 a^{4} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(292\)
parts	\(\text {Expression too large to display}\)	\(852\)

Input:

int((a+a*sec(d*x+c))^4/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

8/3*a^4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1 
/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(2*sin(1/2* 
d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c) 
+10*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+7*cos(1/2*d*x+1/2*c)*si 
n(1/2*d*x+1/2*c)^2-5*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/(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 = 121, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (10 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \frac {{\left (a^{4} \cos \left (d x + c\right )^{2} + 12 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \] Input:

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

-2/3*(10*I*sqrt(2)*a^4*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c 
) + I*sin(d*x + c)) - 10*I*sqrt(2)*a^4*cos(d*x + c)*weierstrassPInverse(-4 
, 0, cos(d*x + c) - I*sin(d*x + c)) - (a^4*cos(d*x + c)^2 + 12*a^4*cos(d*x 
 + c) + a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))
 

Sympy [F]

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

integrate((a+a*sec(d*x+c))**4/sec(d*x+c)**(3/2),x)
 

Output:

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

Maxima [F]

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

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(3/2), x)
 

Giac [F]

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

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(3/2),x)
 

Output:

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

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

Output:

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