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

Optimal result

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

[Out]

(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))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4088, 12, 3916, 2738, 214} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac {(A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {\left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 214

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 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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 3916

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

Rule 4088

Int[csc[(e_.) + (f_.)*(x_)]*(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 + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[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] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\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 {\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 \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 {\int \frac {\sec (c+d x) \left (2 \left (3 a^2 A+2 A b^2-5 a b B\right )-\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2} \\ & = -\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))}-\frac {\int -\frac {3 \left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3} \\ & = -\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))}+\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\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))}+\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3} \\ & = -\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))}+\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^3 d} \\ & = \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))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.43 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.70 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {(b+a \cos (c+d x)) \sec ^3(c+d x) (A+B \sec (c+d x)) \left (\frac {24 \left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^3}{\sqrt {a^2-b^2}}+18 a^4 A b \sin (c+d x)+39 a^2 A b^3 \sin (c+d x)+18 A b^5 \sin (c+d x)-6 a^5 B \sin (c+d x)-18 a^3 b^2 B \sin (c+d x)-51 a b^4 B \sin (c+d x)+54 a^3 A b^2 \sin (2 (c+d x))+6 a A b^4 \sin (2 (c+d x))-12 a^4 b B \sin (2 (c+d x))-54 a^2 b^3 B \sin (2 (c+d x))+6 b^5 B \sin (2 (c+d x))+18 a^4 A b \sin (3 (c+d x))-5 a^2 A b^3 \sin (3 (c+d x))+2 A b^5 \sin (3 (c+d x))-6 a^5 B \sin (3 (c+d x))-10 a^3 b^2 B \sin (3 (c+d x))+a b^4 B \sin (3 (c+d x))\right )}{24 \left (-a^2+b^2\right )^3 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^4} \]

[In]

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

[Out]

((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)*ArcTa
nh[((-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] - 5
1*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)]))/(24*(-a^2 + b^2)^3*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^4)

Maple [A] (verified)

Time = 1.27 (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	\(\text {Expression too large to display}\)	\(1386\)

[In]

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

[Out]

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-tan(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 = 0.37 (sec) , antiderivative size = 1238, normalized size of antiderivative = 5.22 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[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)*arctan(-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 + 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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for 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 = 0.38 (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=-\frac {\frac {3 \, {\left (2 \, A a^{3} - 4 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

-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*t
an(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 = 19.25 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.85 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-3\,B\,a^3+9\,A\,a^2\,b-7\,B\,a\,b^2+A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,B\,a^3-2\,A\,b^3+B\,b^3-3\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-4\,B\,a^2\,b+3\,A\,a\,b^2-B\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

[In]

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

[Out]

((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*(tan(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))