\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [215]

Optimal result

Integrand size = 33, antiderivative size = 244 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {7 (8 A-5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A-2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {7 (8 A-5 B) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \] Output:

7/5*(8*A-5*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d 
*x+c)^(1/2)/a^2/d-5/3*(3*A-2*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1 
/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^2/d+7/15*(8*A-5*B)*sin(d*x+c)/a^2/d/sec(d 
*x+c)^(3/2)-5/3*(3*A-2*B)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)-(3*A-2*B)*sin( 
d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(1+sec(d*x+c))-1/3*(A-B)*sin(d*x+c)/d/sec(d* 
x+c)^(3/2)/(a+a*sec(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 8.72 (sec) , antiderivative size = 946, normalized size of antiderivative = 3.88 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2) 
,x]
 

Output:

(-56*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^ 
((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c 
+ d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/ 
4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(15* 
d*E^(I*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (7*Sqrt[2]*B*Sq 
rt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)) 
]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2* 
I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + 
 d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x]))/(3*d*E^(I*d*x)*(B + A 
*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*Sqrt[C 
os[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2 
)*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x] 
)^2) + (20*B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c 
 + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*Sin[c])/(3* 
d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*Sec 
[c + d*x]^(3/2)*(A + B*Sec[c + d*x])*(((-151*A + 100*B - 73*A*Cos[2*c] + 4 
0*B*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/(10*d) + (4*(-2*A + B)*Cos[2*d*x 
]*Sin[2*c])/(3*d) + (2*A*Cos[3*d*x]*Sin[3*c])/(5*d) + (2*Sec[c/2]*Sec[c/2 
+ (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) - (4*Sec[c/2]*Sec 
[c/2 + (d*x)/2]*(-13*A*Sin[(d*x)/2] + 10*B*Sin[(d*x)/2]))/(3*d) - (2*(-...
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4508, 27, 3042, 4508, 3042, 4274, 3042, 4256, 3042, 4258, 3042, 3119, 3120}

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.

\(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\int \frac {a (11 A-5 B)-7 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (11 A-5 B)-7 a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (11 A-5 B)-7 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {7 a^2 (8 A-5 B)-15 a^2 (3 A-2 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {7 a^2 (8 A-5 B)-15 a^2 (3 A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx-15 a^2 (3 A-2 B) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-15 a^2 (3 A-2 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-15 a^2 (3 A-2 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {7 a^2 (8 A-5 B) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-15 a^2 (3 A-2 B) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )}{a^2}-\frac {6 (3 A-2 B) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

Input:

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]
 

Output:

-1/3*((A - B)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) 
+ ((-6*(3*A - 2*B)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) 
 + (7*a^2*(8*A - 5*B)*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))) - 15*a 
^2*(3*A - 2*B)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c 
 + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/a^2)/(6*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4508

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B 
, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(221)=442\).

Time = 6.00 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.91

method	result	size
default	\(-\frac {\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 (96 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-352 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+80 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-150 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 )-336 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+100 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 )+210 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+266 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-135 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 A -5 B \right )}{30 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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}\)	\(465\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/30/a^2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*co 
s(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/2*c)^8+80*B*cos(1/2*d*x+1/2*c)^8+1 
20*A*cos(1/2*d*x+1/2*c)^6-150*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2 
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^( 
1/2))-336*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2* 
d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+60*B*cos(1/2*d 
*x+1/2*c)^6+100*B*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co 
s(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+210*B*co 
s(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4-2 
40*B*cos(1/2*d*x+1/2*c)^4-135*A*cos(1/2*d*x+1/2*c)^2+105*B*cos(1/2*d*x+1/2 
*c)^2+5*A-5*B)/cos(1/2*d*x+1/2*c)^3/(-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 = 395, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (\sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-8 i \, A + 5 i \, B\right )}\right )} {\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 ) + 21 \, {\left (\sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (8 i \, A - 5 i \, B\right )}\right )} {\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 {2 \, {\left (6 \, A \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (94 \, A - 65 \, B\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorith 
m="fricas")
 

Output:

-1/30*(25*(sqrt(2)*(-3*I*A + 2*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*A + 2 
*I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A + 2*I*B))*weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(3*I*A - 2*I*B)*cos(d*x + c)^ 
2 + 2*sqrt(2)*(3*I*A - 2*I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - 2*I*B))*weie 
rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(sqrt(2)*(-8*I* 
A + 5*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-8*I*A + 5*I*B)*cos(d*x + c) + sqrt 
(2)*(-8*I*A + 5*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(8*I*A - 5*I*B)*cos(d*x + c)^2 
 + 2*sqrt(2)*(8*I*A - 5*I*B)*cos(d*x + c) + sqrt(2)*(8*I*A - 5*I*B))*weier 
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) 
)) - 2*(6*A*cos(d*x + c)^4 - 2*(4*A - 5*B)*cos(d*x + c)^3 - (94*A - 65*B)* 
cos(d*x + c)^2 - 25*(3*A - 2*B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + 
c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorith 
m="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)) 
, x)
 

Giac [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorith 
m="giac")
 

Output:

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2*sec(d*x + c)^(5/2)) 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)),x 
)
 

Output:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2)), 
x)
 

Reduce [F]

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

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

(int(sqrt(sec(c + d*x))/(sec(c + d*x)**5 + 2*sec(c + d*x)**4 + sec(c + d*x 
)**3),x)*a + int(sqrt(sec(c + d*x))/(sec(c + d*x)**4 + 2*sec(c + d*x)**3 + 
 sec(c + d*x)**2),x)*b)/a**2