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

Optimal. Leaf size=396 \[ -\frac{\sin (c+d x) \left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+6 a^3 b B-9 a b^3 B+12 A b^4\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \cos ^2(c+d x) \left (a^2 (-(A-3 C))-3 a b B+4 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \cos (c+d x) \left (-2 a^2 b (A-C)+a^3 B-3 a b^2 B+4 A b^3\right )}{2 a^3 d \left (a^2-b^2\right )}+\frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C-4 a^3 b B+3 a^4 C+3 a b^3 B-4 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \cos ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x \left (2 a^2 b (A+2 C)+a^3 (-B)-6 a b^2 B+8 A b^3\right )}{2 a^5} \]

[Out]

-((8*A*b^3 - a^3*B - 6*a*b^2*B + 2*a^2*b*(A + 2*C))*x)/(2*a^5) + (2*b^2*(5*a^2*A*b^2 - 4*A*b^4 - 4*a^3*b*B + 3
*a*b^3*B + 3*a^4*C - 2*a^2*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a +
b)^(3/2)*d) - ((12*A*b^4 + 6*a^3*b*B - 9*a*b^3*B - a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Sin[c + d*x])/(3*a^
4*(a^2 - b^2)*d) + ((4*A*b^3 + a^3*B - 3*a*b^2*B - 2*a^2*b*(A - C))*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b
^2)*d) - ((4*A*b^2 - 3*a*b*B - a^2*(A - 3*C))*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d) + ((A*b^2 - a
*(b*B - a*C))*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.75885, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4100, 4104, 3919, 3831, 2659, 208} \[ -\frac{\sin (c+d x) \left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+6 a^3 b B-9 a b^3 B+12 A b^4\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \cos ^2(c+d x) \left (a^2 (-(A-3 C))-3 a b B+4 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \cos (c+d x) \left (-2 a^2 b (A-C)+a^3 B-3 a b^2 B+4 A b^3\right )}{2 a^3 d \left (a^2-b^2\right )}+\frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C-4 a^3 b B+3 a^4 C+3 a b^3 B-4 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \cos ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x \left (2 a^2 b (A+2 C)+a^3 (-B)-6 a b^2 B+8 A b^3\right )}{2 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*A*b^3 - a^3*B - 6*a*b^2*B + 2*a^2*b*(A + 2*C))*x)/(2*a^5) + (2*b^2*(5*a^2*A*b^2 - 4*A*b^4 - 4*a^3*b*B + 3
*a*b^3*B + 3*a^4*C - 2*a^2*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a +
b)^(3/2)*d) - ((12*A*b^4 + 6*a^3*b*B - 9*a*b^3*B - a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Sin[c + d*x])/(3*a^
4*(a^2 - b^2)*d) + ((4*A*b^3 + a^3*B - 3*a*b^2*B - 2*a^2*b*(A - C))*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b
^2)*d) - ((4*A*b^2 - 3*a*b*B - a^2*(A - 3*C))*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d) + ((A*b^2 - a
*(b*B - a*C))*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 4100

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(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)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -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 ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^3(c+d x) \left (4 A b^2-3 a b B-a^2 (A-3 C)+a (A b-a B+b C) \sec (c+d x)-3 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right )+a \left (A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sec (c+d x)-2 b \left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (2 \left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right )+a \left (4 A b^3-3 a^3 B-3 a b^2 B+2 a^2 b (A+3 C)\right ) \sec (c+d x)-3 b \left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-3 \left (a^2-b^2\right ) \left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right )+3 a b \left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) x}{2 a^5}-\frac{\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b^2 \left (4 A b^4+4 a^3 b B-3 a b^3 B-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) x}{2 a^5}-\frac{\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b \left (4 A b^4+4 a^3 b B-3 a b^3 B-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) x}{2 a^5}-\frac{\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 b \left (4 A b^4+4 a^3 b B-3 a b^3 B-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\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^5 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) x}{2 a^5}+\frac{2 b^2 \left (5 a^2 A b^2-4 A b^4-4 a^3 b B+3 a b^3 B+3 a^4 C-2 a^2 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^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.74879, size = 255, normalized size = 0.64 \[ \frac{6 (c+d x) \left (-2 a^2 b (A+2 C)+a^3 B+6 a b^2 B-8 A b^3\right )+3 a \sin (c+d x) \left (a^2 (3 A+4 C)-8 a b B+12 A b^2\right )+\frac{24 b^2 \left (a^2 b^2 (2 C-5 A)+4 a^3 b B-3 a^4 C-3 a b^3 B+4 A b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+3 a^2 (a B-2 A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))-\frac{12 a b^3 \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a \cos (c+d x)+b)}}{12 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(-8*A*b^3 + a^3*B + 6*a*b^2*B - 2*a^2*b*(A + 2*C))*(c + d*x) + (24*b^2*(4*A*b^4 + 4*a^3*b*B - 3*a*b^3*B - 3
*a^4*C + a^2*b^2*(-5*A + 2*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 3*a*(
12*A*b^2 - 8*a*b*B + a^2*(3*A + 4*C))*Sin[c + d*x] - (12*a*b^3*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a -
b)*(a + b)*(b + a*Cos[c + d*x])) + 3*a^2*(-2*A*b + a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)])/(12*a^5*d)

________________________________________________________________________________________

Maple [B]  time = 0.16, size = 1241, normalized size = 3.1 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)

[Out]

-4/d/a^3*C*arctan(tan(1/2*d*x+1/2*c))*b-8/d*b^6/a^5/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+
1/2*c)/((a+b)*(a-b))^(1/2))*A+6/d*b^2/a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b
)*(a-b))^(1/2))*C-4/d*b^4/a^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(
1/2))*C+2/d*b^3/a^2/(a^2-b^2)*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)*C-8/d/a^2
/(a+b)/(a-b)/((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+10/d/a^3/(a+b)/(a
-b)/((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^4+6/d*b^5/a^4/(a+b)/(a-b)/((
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*B/a^2*arctan(tan(1/2*d*x+1/2*c))
-2/d*A/a^3*b*arctan(tan(1/2*d*x+1/2*c))+6/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B*b^2-2/d*b^4/a^3/(a^2-b^2)*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)*B+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)^5*A*b+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A*b^2-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)
^3*tan(1/2*d*x+1/2*c)^5*B*b+12/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A*b^2+2/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A+2/d*b^5/a^4/(a^2-b^2)*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)*A-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B+2/d/a^2/(1+tan(1/2*d*x+1/2*c
)^2)^3*tan(1/2*d*x+1/2*c)^5*C+4/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/a^2/(1+tan(1/2*d
*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A+2/d/a^2/(1+tan(1
/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*C+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B-8/d/a^5*arctan
(tan(1/2*d*x+1/2*c))*A*b^3-8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*B*b+6/d/a^4/(1+tan(1/2*d*x+
1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b^2-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B*b-2/d/a^3/(1+tan(
1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.930769, size = 3000, normalized size = 7.58 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^8 - 2*(A + 2*C)*a^7*b + 4*B*a^6*b^2 - 4*(A - 2*C)*a^5*b^3 - 11*B*a^4*b^4 + 2*(7*A - 2*C)*a^3*b^5
+ 6*B*a^2*b^6 - 8*A*a*b^7)*d*x*cos(d*x + c) + 3*(B*a^7*b - 2*(A + 2*C)*a^6*b^2 + 4*B*a^5*b^3 - 4*(A - 2*C)*a^4
*b^4 - 11*B*a^3*b^5 + 2*(7*A - 2*C)*a^2*b^6 + 6*B*a*b^7 - 8*A*b^8)*d*x + 3*(3*C*a^4*b^3 - 4*B*a^3*b^4 + (5*A -
2*C)*a^2*b^5 + 3*B*a*b^6 - 4*A*b^7 + (3*C*a^5*b^2 - 4*B*a^4*b^3 + (5*A - 2*C)*a^3*b^4 + 3*B*a^2*b^5 - 4*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*(2*A + 3
*C)*a^7*b - 12*B*a^6*b^2 + 2*(5*A - 9*C)*a^5*b^3 + 30*B*a^4*b^4 - 2*(19*A - 6*C)*a^3*b^5 - 18*B*a^2*b^6 + 24*A
*a*b^7 + 2*(A*a^8 - 2*A*a^6*b^2 + A*a^4*b^4)*cos(d*x + c)^3 + (3*B*a^8 - 4*A*a^7*b - 6*B*a^6*b^2 + 8*A*a^5*b^3
+ 3*B*a^4*b^4 - 4*A*a^3*b^5)*cos(d*x + c)^2 + (2*(2*A + 3*C)*a^8 - 9*B*a^7*b + 4*(A - 3*C)*a^6*b^2 + 18*B*a^5
*b^3 - 2*(10*A - 3*C)*a^4*b^4 - 9*B*a^3*b^5 + 12*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a
^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b^3 + a^5*b^5)*d), 1/6*(3*(B*a^8 - 2*(A + 2*C)*a^7*b + 4*B*a^6*b^2 - 4
*(A - 2*C)*a^5*b^3 - 11*B*a^4*b^4 + 2*(7*A - 2*C)*a^3*b^5 + 6*B*a^2*b^6 - 8*A*a*b^7)*d*x*cos(d*x + c) + 3*(B*a
^7*b - 2*(A + 2*C)*a^6*b^2 + 4*B*a^5*b^3 - 4*(A - 2*C)*a^4*b^4 - 11*B*a^3*b^5 + 2*(7*A - 2*C)*a^2*b^6 + 6*B*a*
b^7 - 8*A*b^8)*d*x + 6*(3*C*a^4*b^3 - 4*B*a^3*b^4 + (5*A - 2*C)*a^2*b^5 + 3*B*a*b^6 - 4*A*b^7 + (3*C*a^5*b^2 -
4*B*a^4*b^3 + (5*A - 2*C)*a^3*b^4 + 3*B*a^2*b^5 - 4*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))) + (2*(2*A + 3*C)*a^7*b - 12*B*a^6*b^2 + 2*(5*A - 9*C)
*a^5*b^3 + 30*B*a^4*b^4 - 2*(19*A - 6*C)*a^3*b^5 - 18*B*a^2*b^6 + 24*A*a*b^7 + 2*(A*a^8 - 2*A*a^6*b^2 + A*a^4*
b^4)*cos(d*x + c)^3 + (3*B*a^8 - 4*A*a^7*b - 6*B*a^6*b^2 + 8*A*a^5*b^3 + 3*B*a^4*b^4 - 4*A*a^3*b^5)*cos(d*x +
c)^2 + (2*(2*A + 3*C)*a^8 - 9*B*a^7*b + 4*(A - 3*C)*a^6*b^2 + 18*B*a^5*b^3 - 2*(10*A - 3*C)*a^4*b^4 - 9*B*a^3*
b^5 + 12*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*
b^3 + a^5*b^5)*d)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.29786, size = 761, normalized size = 1.92 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(12*(3*C*a^4*b^2 - 4*B*a^3*b^3 + 5*A*a^2*b^4 - 2*C*a^2*b^4 + 3*B*a*b^5 - 4*A*b^6)*(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
- a^5*b^2)*sqrt(-a^2 + b^2)) + 12*(C*a^2*b^3*tan(1/2*d*x + 1/2*c) - B*a*b^4*tan(1/2*d*x + 1/2*c) + A*b^5*tan(
1/2*d*x + 1/2*c))/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) + 3*(B*a^3 -
2*A*a^2*b - 4*C*a^2*b + 6*B*a*b^2 - 8*A*b^3)*(d*x + c)/a^5 + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(
1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*b*tan(1/2*d*x +
1/2*c)^5 + 18*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3
- 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 36*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*t
an(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 6*A*a*b*tan(1/2*d*x + 1/2*c) - 12*B*a*b*tan(1/2*d*x + 1/2
*c) + 18*A*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^4))/d