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

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

[Out]

(B*x)/a^3 - ((6*a^4*b*B - 5*a^2*b^3*B + 2*b^5*B - 2*a^5*C - a^3*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/
Sqrt[a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (b*(b*B - a*C)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec
[c + d*x])^2) + (b*(5*a^2*b*B - 2*b^3*B - 3*a^3*C)*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 = 0.580986, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {4072, 3923, 4060, 3919, 3831, 2659, 208} \[ -\frac{\left (-5 a^2 b^3 B-a^3 b^2 C+6 a^4 b B-2 a^5 C+2 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{b \left (5 a^2 b B-3 a^3 C-2 b^3 B\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b (b B-a C) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{B x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4072

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

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{\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=\int \frac{B+C \sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx\\ &=\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{-2 \left (a^2-b^2\right ) B+2 a (b B-a C) \sec (c+d x)-b (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b (b B-a C) \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 b B-2 b^3 B-3 a^3 C\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 \left (a^2-b^2\right )^2 B-a \left (4 a^2 b B-b^3 B-2 a^3 C-a b^2 C\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{B x}{a^3}+\frac{b (b B-a C) \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 b B-2 b^3 B-3 a^3 C\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 b B-5 a^2 b^3 B+2 b^5 B-2 a^5 C-a^3 b^2 C\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{B x}{a^3}+\frac{b (b B-a C) \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 b B-2 b^3 B-3 a^3 C\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 b B-5 a^2 b^3 B+2 b^5 B-2 a^5 C-a^3 b^2 C\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{B x}{a^3}+\frac{b (b B-a C) \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 b B-2 b^3 B-3 a^3 C\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 b B-5 a^2 b^3 B+2 b^5 B-2 a^5 C-a^3 b^2 C\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{B x}{a^3}-\frac{\left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-2 a^5 C-a^3 b^2 C\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 (b B-a C) \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 b B-2 b^3 B-3 a^3 C\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 = 1.38194, size = 203, normalized size = 0.99 \[ \frac{\frac{a b \left (6 a^2 b B-4 a^3 C+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}-\frac{2 \left (5 a^2 b^3 B+a^3 b^2 C-6 a^4 b B+2 a^5 C-2 b^5 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^2 (a C-b B) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)^2}+2 B (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.125, 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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

2/d*B/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*B-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*B+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*B+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*C+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*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-b)^2*tan(1/
2*d*x+1/2*c)*B-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
)*B-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)*B-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)*C+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)*C-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))*B+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))*B*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))*B*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))*C+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^2*C

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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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.751762, 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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(4*(B*a^8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*d*x*cos(d*x + c)^2 + 8*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a^
3*b^5 - B*a*b^7)*d*x*cos(d*x + c) + 4*(B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b^6 - B*b^8)*d*x - (2*C*a^5*b^2 - 6*B
*a^4*b^3 + C*a^3*b^4 + 5*B*a^2*b^5 - 2*B*b^7 + (2*C*a^7 - 6*B*a^6*b + C*a^5*b^2 + 5*B*a^4*b^3 - 2*B*a^2*b^5)*c
os(d*x + c)^2 + 2*(2*C*a^6*b - 6*B*a^5*b^2 + C*a^4*b^3 + 5*B*a^3*b^4 - 2*B*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*C*a^6*b^2 - 5*B*a^5*b^3 - 3*C*a^4*b^4
+ 7*B*a^3*b^5 - 2*B*a*b^7 + (4*C*a^7*b - 6*B*a^6*b^2 - 5*C*a^5*b^3 + 9*B*a^4*b^4 + C*a^3*b^5 - 3*B*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*(B*a^8 - 3*B*a^6
*b^2 + 3*B*a^4*b^4 - B*a^2*b^6)*d*x*cos(d*x + c)^2 + 4*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a^3*b^5 - B*a*b^7)*d*x*cos
(d*x + c) + 2*(B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b^6 - B*b^8)*d*x + (2*C*a^5*b^2 - 6*B*a^4*b^3 + C*a^3*b^4 + 5
*B*a^2*b^5 - 2*B*b^7 + (2*C*a^7 - 6*B*a^6*b + C*a^5*b^2 + 5*B*a^4*b^3 - 2*B*a^2*b^5)*cos(d*x + c)^2 + 2*(2*C*a
^6*b - 6*B*a^5*b^2 + C*a^4*b^3 + 5*B*a^3*b^4 - 2*B*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*C*a^6*b^2 - 5*B*a^5*b^3 - 3*C*a^4*b^4 + 7*B*a^3*b^5
- 2*B*a*b^7 + (4*C*a^7*b - 6*B*a^6*b^2 - 5*C*a^5*b^3 + 9*B*a^4*b^4 + C*a^3*b^5 - 3*B*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.61489, size = 617, normalized size = 3.01 \begin{align*} \frac{\frac{{\left (2 \, C a^{5} - 6 \, B a^{4} b + C a^{3} b^{2} + 5 \, B a^{2} b^{3} - 2 \, B 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 )} B}{a^{3}} + \frac{4 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, B a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B 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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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