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

Optimal. Leaf size=205 \[ -\frac{\left (-5 a^2 A b^3+6 a^4 A b-a^3 b^2 B-2 a^5 B+2 A b^5\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b \left (5 a^2 A b-3 a^3 B-2 A b^3\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{A x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.536315, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3923, 4060, 3919, 3831, 2659, 208} \[ -\frac{\left (-5 a^2 A b^3+6 a^4 A b-a^3 b^2 B-2 a^5 B+2 A b^5\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b \left (5 a^2 A b-3 a^3 B-2 A b^3\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{A x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4060

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

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

Rule 2659

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 208

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

Rubi steps

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

Mathematica [A]  time = 1.41674, size = 267, normalized size = 1.3 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) (A+B \sec (c+d x)) \left (-\frac{a b \left (-6 a^2 A b+4 a^3 B-a b^2 B+3 A b^3\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}-\frac{2 \left (5 a^2 A b^3-6 a^4 A b+a^3 b^2 B+2 a^5 B-2 A b^5\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^2 (a B-A b) \sin (c+d x)}{(a-b) (a+b)}+2 A (c+d x) (a \cos (c+d x)+b)^2\right )}{2 a^3 d (a+b \sec (c+d x))^3 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.099, size = 1063, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*A/a^3*arctan(tan(1/2*d*x+1/2*c))-6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^2/(a-b)/(a^2+
2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-1/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^3/(a-b)/(a^2+2
*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^4/(a-b)/(a^2+
2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+4/d*a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b/(a-b)/(a^2+2*a
*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+1/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^2/(a-b)/(a^2+2*a*b+
b^2)*tan(1/2*d*x+1/2*c)^3*B+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^2/(a+b)/(a-b)^2*tan(1/
2*d*x+1/2*c)*A-1/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^3/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c
)*A-2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^4/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A-4/d*a
/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B+1/d/(tan(1/2*d*x+1
/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-6/d*a*b/(a^4-2*a^2*b^2+b^4)/((a
+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+5/d/a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b
))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^3-2/d/a^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))
^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^5+2/d*a^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(
1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+1/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arcta
nh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.750819, size = 2479, normalized size = 12.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6)*d*x*cos(d*x + c)^2 + 8*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^
3*b^5 - A*a*b^7)*d*x*cos(d*x + c) + 4*(A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*d*x - (2*B*a^5*b^2 - 6*A
*a^4*b^3 + B*a^3*b^4 + 5*A*a^2*b^5 - 2*A*b^7 + (2*B*a^7 - 6*A*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5)*c
os(d*x + c)^2 + 2*(2*B*a^6*b - 6*A*a^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*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*(3*B*a^6*b^2 - 5*A*a^5*b^3 - 3*B*a^4*b^4
+ 7*A*a^3*b^5 - 2*A*a*b^7 + (4*B*a^7*b - 6*A*a^6*b^2 - 5*B*a^5*b^3 + 9*A*a^4*b^4 + B*a^3*b^5 - 3*A*a^2*b^6)*co
s(d*x + c))*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 +
3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d), 1/2*(2*(A*a^8 - 3*A*a^6
*b^2 + 3*A*a^4*b^4 - A*a^2*b^6)*d*x*cos(d*x + c)^2 + 4*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*d*x*cos
(d*x + c) + 2*(A*a^6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*d*x + (2*B*a^5*b^2 - 6*A*a^4*b^3 + B*a^3*b^4 + 5
*A*a^2*b^5 - 2*A*b^7 + (2*B*a^7 - 6*A*a^6*b + B*a^5*b^2 + 5*A*a^4*b^3 - 2*A*a^2*b^5)*cos(d*x + c)^2 + 2*(2*B*a
^6*b - 6*A*a^5*b^2 + B*a^4*b^3 + 5*A*a^3*b^4 - 2*A*a*b^6)*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))) - (3*B*a^6*b^2 - 5*A*a^5*b^3 - 3*B*a^4*b^4 + 7*A*a^3*b^5
- 2*A*a*b^7 + (4*B*a^7*b - 6*A*a^6*b^2 - 5*B*a^5*b^3 + 9*A*a^4*b^4 + B*a^3*b^5 - 3*A*a^2*b^6)*cos(d*x + c))*si
n(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a
^4*b^7)*d*cos(d*x + c) + (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x))/(a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.43989, size = 617, normalized size = 3.01 \begin{align*} \frac{\frac{{\left (2 \, B a^{5} - 6 \, A a^{4} b + B a^{3} b^{2} + 5 \, A a^{2} b^{3} - 2 \, A b^{5}\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^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{{\left (d x + c\right )} A}{a^{3}} + \frac{4 \, B a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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 ) - 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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 )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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