\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [506]

Optimal result

Integrand size = 33, antiderivative size = 121 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A+2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 7.64 (sec) , antiderivative size = 801, normalized size of antiderivative = 6.62 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x])*Sec 
[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + 
 Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Co 
t[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x]) 
^2) - (4*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/ 
4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] 
)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqr 
t[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT 
an[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + 
d*x])^2) + (Cos[c/2 + (d*x)/2]^4*(A + B*Sec[c + d*x])*((-4*B*Csc[c])/d - ( 
4*B*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d - (2*Sec[c/2]*Sec[c/2 + (d 
*x)/2]^3*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) - (2*(-A + B)*Sec[c/2 
 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(Sqrt[Cos[c + d*x]]*(B + A*Cos[c + d*x])*( 
a + a*Sec[c + d*x])^2) - (B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*Sec[c + 
 d*x]*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d* 
x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x 
 + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d* 
x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + 
ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcT...
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3433, 3042, 3457, 27, 3042, 3457, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {A \cos (c+d x)+B}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {(A+2 B) a^2+3 B \cos (c+d x) a^2}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {6 B \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {a^2 (A+2 B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 B \int \sqrt {\cos (c+d x)}dx}{a^2}-\frac {6 B \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ( 
((6*a^2*B*EllipticE[(c + d*x)/2, 2])/d + (2*a^2*(A + 2*B)*EllipticF[(c + d 
*x)/2, 2])/d)/a^2 - (6*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(1 + Cos[c + 
d*x])))/(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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3457

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(116)=232\).

Time = 2.91 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.89

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+16 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-3 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+A -B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(350\)

Input:

int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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))-12*B*cos(1/2*d*x+1/2*c)^6+4*B*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))-6*B*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))+2*A*cos(1/2*d*x+1/2*c)^4+16*B*cos(1/2*d*x+1/2*c)^4-3 
*A*cos(1/2*d*x+1/2*c)^2-3*B*cos(1/2*d*x+1/2*c)^2+A-B)/a^2/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 = 0.09 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, B \cos \left (d x + c\right ) - A + 4 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (-i \, \sqrt {2} B \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} B \cos \left (d x + c\right ) - i \, \sqrt {2} B\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} B \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} B \cos \left (d x + c\right ) + i \, \sqrt {2} B\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorith 
m="fricas")
 

Output:

-1/6*(2*(3*B*cos(d*x + c) - A + 4*B)*sqrt(cos(d*x + c))*sin(d*x + c) - (sq 
rt(2)*(-I*A - 2*I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(I*A + 2*I*B)*cos(d*x + c) 
 + sqrt(2)*(-I*A - 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin 
(d*x + c)) - (sqrt(2)*(I*A + 2*I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(-I*A - 2*I 
*B)*cos(d*x + c) + sqrt(2)*(I*A + 2*I*B))*weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*sin(d*x + c)) + 3*(-I*sqrt(2)*B*cos(d*x + c)^2 - 2*I*sqrt(2)*B 
*cos(d*x + c) - I*sqrt(2)*B)*weierstrassZeta(-4, 0, weierstrassPInverse(-4 
, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*B*cos(d*x + c)^2 + 2*I 
*sqrt(2)*B*cos(d*x + c) + I*sqrt(2)*B)*weierstrassZeta(-4, 0, weierstrassP 
Inverse(-4, 0, cos(d*x + c) - I*sin(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. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(3/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(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*cos(d*x + c)^(3/2)) 
, x)
 

Mupad [F(-1)]

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

int((A + B/cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^2),x)
                                                                                    
                                                                                    
 

Output:

int((A + B/cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^2), x)
 

Reduce [F]

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

int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

(int(sqrt(cos(c + d*x))/(cos(c + d*x)**2*sec(c + d*x)**2 + 2*cos(c + d*x)* 
*2*sec(c + d*x) + cos(c + d*x)**2),x)*a + int((sqrt(cos(c + d*x))*sec(c + 
d*x))/(cos(c + d*x)**2*sec(c + d*x)**2 + 2*cos(c + d*x)**2*sec(c + d*x) + 
cos(c + d*x)**2),x)*b)/a**2