\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [340]

Optimal result

Integrand size = 29, antiderivative size = 237 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {(A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \] Output:

(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c) 
/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d-1/3*(A*b-B*a)*tan(d*x+c)/(a^2-b^2) 
/d/(a+b*sec(d*x+c))^3-1/6*(5*A*a*b-2*B*a^2-3*B*b^2)*tan(d*x+c)/(a^2-b^2)^2 
/d/(a+b*sec(d*x+c))^2-1/6*(11*A*a^2*b+4*A*b^3-2*B*a^3-13*B*a*b^2)*tan(d*x+ 
c)/(a^2-b^2)^3/d/(a+b*sec(d*x+c))
 

Mathematica [A] (verified)

Time = 1.80 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.70 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {(b+a \cos (c+d x)) \sec ^3(c+d x) (A+B \sec (c+d x)) \left (\frac {24 \left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^3}{\sqrt {a^2-b^2}}+18 a^4 A b \sin (c+d x)+39 a^2 A b^3 \sin (c+d x)+18 A b^5 \sin (c+d x)-6 a^5 B \sin (c+d x)-18 a^3 b^2 B \sin (c+d x)-51 a b^4 B \sin (c+d x)+54 a^3 A b^2 \sin (2 (c+d x))+6 a A b^4 \sin (2 (c+d x))-12 a^4 b B \sin (2 (c+d x))-54 a^2 b^3 B \sin (2 (c+d x))+6 b^5 B \sin (2 (c+d x))+18 a^4 A b \sin (3 (c+d x))-5 a^2 A b^3 \sin (3 (c+d x))+2 A b^5 \sin (3 (c+d x))-6 a^5 B \sin (3 (c+d x))-10 a^3 b^2 B \sin (3 (c+d x))+a b^4 B \sin (3 (c+d x))\right )}{24 \left (-a^2+b^2\right )^3 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^4} \] Input:

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]
 

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]^3*(A + B*Sec[c + d*x])*((24*(2*a^3*A + 
3*a*A*b^2 - 4*a^2*b*B - b^3*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^ 
2 - b^2]]*(b + a*Cos[c + d*x])^3)/Sqrt[a^2 - b^2] + 18*a^4*A*b*Sin[c + d*x 
] + 39*a^2*A*b^3*Sin[c + d*x] + 18*A*b^5*Sin[c + d*x] - 6*a^5*B*Sin[c + d* 
x] - 18*a^3*b^2*B*Sin[c + d*x] - 51*a*b^4*B*Sin[c + d*x] + 54*a^3*A*b^2*Si 
n[2*(c + d*x)] + 6*a*A*b^4*Sin[2*(c + d*x)] - 12*a^4*b*B*Sin[2*(c + d*x)] 
- 54*a^2*b^3*B*Sin[2*(c + d*x)] + 6*b^5*B*Sin[2*(c + d*x)] + 18*a^4*A*b*Si 
n[3*(c + d*x)] - 5*a^2*A*b^3*Sin[3*(c + d*x)] + 2*A*b^5*Sin[3*(c + d*x)] - 
 6*a^5*B*Sin[3*(c + d*x)] - 10*a^3*b^2*B*Sin[3*(c + d*x)] + a*b^4*B*Sin[3* 
(c + d*x)]))/(24*(-a^2 + b^2)^3*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x] 
)^4)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.18, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.483, Rules used = {3042, 4491, 25, 3042, 4491, 25, 3042, 4491, 27, 3042, 4318, 3042, 3138, 221}

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.

\(\displaystyle \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

\(\displaystyle -\frac {\int -\frac {\sec (c+d x) (3 (a A-b B)-2 (A b-a B) \sec (c+d x))}{(a+b \sec (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sec (c+d x) (3 (a A-b B)-2 (A b-a B) \sec (c+d x))}{(a+b \sec (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 (a A-b B)-2 (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4491

\(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (2 \left (3 A a^2-5 b B a+2 A b^2\right )-\left (-2 B a^2+5 A b a-3 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 \left (3 A a^2-5 b B a+2 A b^2\right )-\left (-2 B a^2+5 A b a-3 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 \left (3 A a^2-5 b B a+2 A b^2\right )+\left (2 B a^2-5 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4491

\(\displaystyle \frac {\frac {-\frac {\int -\frac {3 \left (2 A a^3-4 b B a^2+3 A b^2 a-b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {6 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {6 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

Input:

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]
 

Output:

-1/3*((A*b - a*B)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ( 
-1/2*((5*a*A*b - 2*a^2*B - 3*b^2*B)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Se 
c[c + d*x])^2) + ((6*(2*a^3*A + 3*a*A*b^2 - 4*a^2*b*B - b^3*B)*ArcTanh[(Sq 
rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - 
b^2)*d) - ((11*a^2*A*b + 4*A*b^3 - 2*a^3*B - 13*a*b^2*B)*Tan[c + d*x])/((a 
^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*(a^2 - b^2)))/(3*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4491

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e 
 + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 
/((m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp 
[(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free 
Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m 
, -1]
 

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 A a \,b^{2}-4 B \,a^{2} b -B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(376\)
default	\(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 A a \,b^{2}-4 B \,a^{2} b -B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(376\)
risch	\(\text {Expression too large to display}\)	\(1378\)

Input:

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE 
)
 

Output:

1/d*(-2*(-1/2*(6*A*a^2*b+3*A*a*b^2+2*A*b^3-2*B*a^3-2*B*a^2*b-6*B*a*b^2-B*b 
^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5+2/3*(9*A*a^2*b+A* 
b^3-3*B*a^3-7*B*a*b^2)/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^ 
3-1/2*(6*A*a^2*b-3*A*a*b^2+2*A*b^3-2*B*a^3+2*B*a^2*b-6*B*a*b^2+B*b^3)/(a+b 
)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-ta 
n(1/2*d*x+1/2*c)^2*b-a-b)^3+(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3)/(a^6-3*a^4 
*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/( 
(a+b)*(a-b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (222) = 444\).

Time = 0.16 (sec) , antiderivative size = 1238, normalized size of antiderivative = 5.22 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="fri 
cas")
 

Output:

[1/12*(3*(2*A*a^3*b^3 - 4*B*a^2*b^4 + 3*A*a*b^5 - B*b^6 + (2*A*a^6 - 4*B*a 
^5*b + 3*A*a^4*b^2 - B*a^3*b^3)*cos(d*x + c)^3 + 3*(2*A*a^5*b - 4*B*a^4*b^ 
2 + 3*A*a^3*b^3 - B*a^2*b^4)*cos(d*x + c)^2 + 3*(2*A*a^4*b^2 - 4*B*a^3*b^3 
 + 3*A*a^2*b^4 - B*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x 
 + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + 
 a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + 
 b^2)) + 2*(2*B*a^5*b^2 - 11*A*a^4*b^3 + 11*B*a^3*b^4 + 7*A*a^2*b^5 - 13*B 
*a*b^6 + 4*A*b^7 + (6*B*a^7 - 18*A*a^6*b + 4*B*a^5*b^2 + 23*A*a^4*b^3 - 11 
*B*a^3*b^4 - 7*A*a^2*b^5 + B*a*b^6 + 2*A*b^7)*cos(d*x + c)^2 + 3*(2*B*a^6* 
b - 9*A*a^5*b^2 + 7*B*a^4*b^3 + 8*A*a^3*b^4 - 10*B*a^2*b^5 + A*a*b^6 + B*b 
^7)*cos(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 
 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b 
^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^ 
3*b^8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2* 
b^9 + b^11)*d), 1/6*(3*(2*A*a^3*b^3 - 4*B*a^2*b^4 + 3*A*a*b^5 - B*b^6 + (2 
*A*a^6 - 4*B*a^5*b + 3*A*a^4*b^2 - B*a^3*b^3)*cos(d*x + c)^3 + 3*(2*A*a^5* 
b - 4*B*a^4*b^2 + 3*A*a^3*b^3 - B*a^2*b^4)*cos(d*x + c)^2 + 3*(2*A*a^4*b^2 
 - 4*B*a^3*b^3 + 3*A*a^2*b^4 - B*a*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*arc 
tan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + ( 
2*B*a^5*b^2 - 11*A*a^4*b^3 + 11*B*a^3*b^4 + 7*A*a^2*b^5 - 13*B*a*b^6 + ...
 

Sympy [F]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**4,x)
 

Output:

Integral((A + B*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**4, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="max 
ima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (222) = 444\).

Time = 0.24 (sec) , antiderivative size = 693, normalized size of antiderivative = 2.92 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="gia 
c")
 

Output:

-1/3*(3*(2*A*a^3 - 4*B*a^2*b + 3*A*a*b^2 - B*b^3)*(pi*floor(1/2*(d*x + c)/ 
pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x 
+ 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-a^ 
2 + b^2)) + (6*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 18*A*a^4*b*tan(1/2*d*x + 1/2 
*c)^5 - 6*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 27*A*a^3*b^2*tan(1/2*d*x + 1/2* 
c)^5 + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b^3*tan(1/2*d*x + 1/2 
*c)^5 - 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b^4*tan(1/2*d*x + 1/2* 
c)^5 + 12*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 
+ 3*B*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^5*tan(1/2*d*x + 1/2*c)^3 + 36*A* 
a^4*b*tan(1/2*d*x + 1/2*c)^3 - 16*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 32*A* 
a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 28*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*A*b 
^5*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^5*tan(1/2*d*x + 1/2*c) - 18*A*a^4*b*tan( 
1/2*d*x + 1/2*c) + 6*B*a^4*b*tan(1/2*d*x + 1/2*c) - 27*A*a^3*b^2*tan(1/2*d 
*x + 1/2*c) + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b^3*tan(1/2*d*x 
+ 1/2*c) + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/2*d*x + 1/2 
*c) + 12*B*a*b^4*tan(1/2*d*x + 1/2*c) - 6*A*b^5*tan(1/2*d*x + 1/2*c) - 3*B 
*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2 
*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3))/d
 

Mupad [B] (verification not implemented)

Time = 16.06 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.85 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-3\,B\,a^3+9\,A\,a^2\,b-7\,B\,a\,b^2+A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,B\,a^3-2\,A\,b^3+B\,b^3-3\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-4\,B\,a^2\,b+3\,A\,a\,b^2-B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)*(a + b/cos(c + d*x))^4),x)
 

Output:

((4*tan(c/2 + (d*x)/2)^3*(A*b^3 - 3*B*a^3 + 9*A*a^2*b - 7*B*a*b^2))/(3*(a 
+ b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5*(2*B*a^3 - 2*A*b^3 + B 
*b^3 - 3*A*a*b^2 - 6*A*a^2*b + 6*B*a*b^2 + 2*B*a^2*b))/((a + b)^3*(a - b)) 
 - (tan(c/2 + (d*x)/2)*(2*A*b^3 - 2*B*a^3 + B*b^3 - 3*A*a*b^2 + 6*A*a^2*b 
- 6*B*a*b^2 + 2*B*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(d*(t 
an(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(c/2 + (d*x)/ 
2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - 
 tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3))) + (atanh((tan(c/2 
+ (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)*( 
a - b)^(7/2)))*(2*A*a^3 - B*b^3 + 3*A*a*b^2 - 4*B*a^2*b))/(d*(a + b)^(7/2) 
*(a - b)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 633, normalized size of antiderivative = 2.67 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x)
 

Output:

(8*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2))*cos(c + d*x)*a**3*b + 4*sqrt( - a**2 + b**2)*atan((tan( 
(c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a* 
b**3 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)* 
b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**4 - 2*sqrt( - a**2 + b**2)*ata 
n((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + 
d*x)**2*a**2*b**2 + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan( 
(c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**4 + 6*sqrt( - a**2 + b**2)*atan(( 
tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**2*b**2 + 
 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2))*b**4 - 4*cos(c + d*x)*sin(c + d*x)*a**4*b + 5*cos(c + d 
*x)*sin(c + d*x)*a**2*b**3 - cos(c + d*x)*sin(c + d*x)*b**5 - 3*sin(c + d* 
x)*a**3*b**2 + 3*sin(c + d*x)*a*b**4)/(2*d*(2*cos(c + d*x)*a**7*b - 6*cos( 
c + d*x)*a**5*b**3 + 6*cos(c + d*x)*a**3*b**5 - 2*cos(c + d*x)*a*b**7 - si 
n(c + d*x)**2*a**8 + 3*sin(c + d*x)**2*a**6*b**2 - 3*sin(c + d*x)**2*a**4* 
b**4 + sin(c + d*x)**2*a**2*b**6 + a**8 - 2*a**6*b**2 + 2*a**2*b**6 - b**8 
))