\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [338]

Optimal result

Integrand size = 31, antiderivative size = 274 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \] Output:

(A*a^3+4*A*a*b^2-3*B*a^2*b-2*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^2*(A*b-B*a)*tan(d*x+c)/b^2/( 
a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*a*(A*a^2*b-6*A*b^3-4*B*a^3+9*B*a*b^2)*ta 
n(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*(A*a^4*b-10*A*a^2*b^3-6* 
A*b^5+2*B*a^5-5*B*a^3*b^2+18*B*a*b^4)*tan(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*se 
c(d*x+c))
 

Mathematica [A] (verified)

Time = 1.65 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {-\frac {6 \left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {2 (-A b+a B) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))^3}+\frac {\left (3 a^2 A+2 A b^2-5 a b B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))^2}+\frac {\left (-13 a^2 A b-2 A b^3+4 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (b+a \cos (c+d x))}}{6 d} \] Input:

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

Output:

((-6*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[((-a + b)*Tan[(c + 
d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + (2*(-(A*b) + a*B)*Sin[c + d 
*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])^3) + ((3*a^2*A + 2*A*b^2 - 5*a* 
b*B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2) + ((-13*a^ 
2*A*b - 2*A*b^3 + 4*a^3*B + 11*a*b^2*B)*Sin[c + d*x])/((a - b)^3*(a + b)^3 
*(b + a*Cos[c + d*x])))/(6*d)
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.16, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4516, 25, 3042, 4568, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4516

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

\(\displaystyle \frac {\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\int -\frac {\sec (c+d x) \left (2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \sec (c+d x) b\right )}{(a+b \sec (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \sec (c+d x) b\right )}{(a+b \sec (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 \left (B a^3+2 A b a^2-6 b^2 B a+3 A b^3\right ) b^2+\left (2 B a^4+A b a^3-3 b^2 B a^2-6 A b^3 a+6 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4491

\(\displaystyle \frac {\frac {\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {3 b^3 \left (A a^3-3 b B a^2+4 A b^2 a-2 b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {3 b^2 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 b^2 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a^2-b^2}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {6 b^2 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {a \left (-4 a^3 B+a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {6 b^3 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}+\frac {b \left (2 a^5 B+a^4 A b-5 a^3 b^2 B-10 a^2 A b^3+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\)

Input:

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

Output:

-1/3*(a^2*(A*b - a*B)*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x] 
)^3) + ((a*(a^2*A*b - 6*A*b^3 - 4*a^3*B + 9*a*b^2*B)*Tan[c + d*x])/(2*(a^2 
 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((6*b^3*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B 
 - 2*b^3*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - 
 b]*Sqrt[a + b]*(a^2 - b^2)*d) + (b*(a^4*A*b - 10*a^2*A*b^3 - 6*A*b^5 + 2* 
a^5*B - 5*a^3*b^2*B + 18*a*b^4*B)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[ 
c + d*x])))/(2*b*(a^2 - b^2)))/(3*b^2*(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]
 

rule 4516

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B) 
*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x 
] + Simp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x] 
)^(m + 1)*Simp[a*b*(A*b - a*B)*(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Cs 
c[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 
]
 

rule 4568

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] 
)^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) 
 Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m 
+ 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], 
x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 
2, 0]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.37

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 A \,a^{2} b +3 A \,b^{3}-B \,a^{3}-9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 B \,a^{3}-3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(375\)
default	\(\frac {-\frac {2 \left (-\frac {\left (a^{3} A +6 A \,a^{2} b +2 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-3 B \,a^{2} b -6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 A \,a^{2} b +3 A \,b^{3}-B \,a^{3}-9 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (a^{3} A -6 A \,a^{2} b +2 A a \,b^{2}-2 A \,b^{3}+2 B \,a^{3}-3 B \,a^{2} b +6 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (a^{3} A +4 A a \,b^{2}-3 B \,a^{2} b -2 B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(375\)
risch	\(\text {Expression too large to display}\)	\(1180\)

Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(-2*(-1/2*(A*a^3+6*A*a^2*b+2*A*a*b^2+2*A*b^3-2*B*a^3-3*B*a^2*b-6*B*a*b 
^2)/(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*(7*A*a^2*b+3* 
A*b^3-B*a^3-9*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*(A*a^3-6*A*a^2*b+2*A*a*b^2-2*A*b^3+2*B*a^3-3*B*a^2*b+6*B*a*b^2)/(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+(A*a^3+4*A*a*b^2-3*B*a^2*b-2*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. 586 vs. \(2 (260) = 520\).

Time = 0.16 (sec) , antiderivative size = 1230, normalized size of antiderivative = 4.49 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

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

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="m 
axima")
 

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. \(2 (260) = 520\).

Time = 0.25 (sec) , antiderivative size = 693, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="g 
iac")
 

Output:

1/3*(3*(A*a^3 - 3*B*a^2*b + 4*A*a*b^2 - 2*B*b^3)*(pi*floor(1/2*(d*x + c)/p 
i + 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)) + (3*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^5*tan(1/2*d*x + 1/2*c 
)^5 + 12*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 3*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 - 6*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^ 
5 + 12*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 - 6*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 18*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 + 4*B*a^5*tan(1/2*d*x + 1/2*c)^3 - 28*A* 
a^4*b*tan(1/2*d*x + 1/2*c)^3 + 32*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 16*A* 
a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 12*A* 
b^5*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^5*tan(1/2*d*x + 1/2*c) - 6*B*a^5*tan(1/ 
2*d*x + 1/2*c) + 12*A*a^4*b*tan(1/2*d*x + 1/2*c) - 3*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) - 6*B*a^3*b^2*tan(1/2*d*x + 1/ 
2*c) + 12*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) + 6*A*a*b^4*tan(1/2*d*x + 1/2*c) - 18*B*a*b^4*tan(1/2*d*x + 1/2*c) + 6* 
A*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 = 16.43 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-B\,a^3+7\,A\,a^2\,b-9\,B\,a\,b^2+3\,A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A\,a^3+2\,A\,b^3-2\,B\,a^3+2\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^3-2\,A\,b^3+2\,B\,a^3+2\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2-3\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,a^3-3\,B\,a^2\,b+4\,A\,a\,b^2-2\,B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + b/cos(c + d*x))^4),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 626, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x)
 

Output:

(4*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 + 8*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 - 2*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 - 4*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 + 2*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 + 
 4*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 - 3*cos(c + d*x)*sin(c + d*x)*a**4*b + 3*cos(c + d 
*x)*sin(c + d*x)*a**2*b**3 + sin(c + d*x)*a**5 - 5*sin(c + d*x)*a**3*b**2 
+ 4*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 - sin(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))