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

Optimal. Leaf size=266 \[ -\frac{\left (11 a^2 A b^2+a^4 (-(2 A-3 C))-6 A b^4\right ) \sin (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-2 a^6 C-6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (6 A+C)-2 a^4 C+3 A b^4\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{3 A b x}{a^4} \]

[Out]

(-3*A*b*x)/a^4 - ((15*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - a^4*b^2*(12*A + C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2
])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((11*a^2*A*b^2 - 6*A*b^4 - a^4*(2*A - 3*C))*Sin[c + d*x
])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*A
*b^4 - 2*a^4*C - a^2*b^2*(6*A + C))*Sin[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 = 0.96297, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4101, 4100, 4104, 3919, 3831, 2659, 208} \[ -\frac{\left (11 a^2 A b^2+a^4 (-(2 A-3 C))-6 A b^4\right ) \sin (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-2 a^6 C-6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (6 A+C)-2 a^4 C+3 A b^4\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{3 A b x}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*b*x)/a^4 - ((15*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - a^4*b^2*(12*A + C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2
])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((11*a^2*A*b^2 - 6*A*b^4 - a^4*(2*A - 3*C))*Sin[c + d*x
])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*A
*b^4 - 2*a^4*C - a^2*b^2*(6*A + C))*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4101

Int[((A_.) + 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^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)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e
+ f*x])^n*Simp[a^2*(A + C)*(m + 1) - (A*b^2 + a^2*C)*(m + n + 1) - a*b*(A + C)*(m + 1)*Csc[e + f*x] + (A*b^2
+ a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && NeQ[a^2 - b^2, 0] &&
LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

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 (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (3 A b^2-a^2 (2 A-C)+2 a b (A+C) \sec (c+d x)-2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (-11 a^2 A b^2+6 A b^4+a^4 (2 A-3 C)+a b \left (A b^2-a^2 (4 A+3 C)\right ) \sec (c+d x)-\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{6 A b \left (a^2-b^2\right )^2+a \left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{3 A b x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 A b x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 a^4 b \left (a^2-b^2\right )^2}\\ &=-\frac{3 A b x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+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^4 b \left (a^2-b^2\right )^2 d}\\ &=-\frac{3 A b x}{a^4}+\frac{\left (12 a^4 A b^2-15 a^2 A b^4+6 A b^6+2 a^6 C+a^4 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^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.58112, size = 902, normalized size = 3.39 \[ \frac{(b+a \cos (c+d x)) \sec (c+d x) \left (C \sec ^2(c+d x)+A\right ) \left (\frac{\sec (c) \left (A \sin (d x) a^7+A \sin (2 c+d x) a^7+A \sin (2 c+3 d x) a^7+A \sin (4 c+3 d x) a^7-6 A b d x \cos (c+2 d x) a^6-6 A b d x \cos (3 c+2 d x) a^6+8 b C \sin (c) a^6+4 A b \sin (c+2 d x) a^6-8 b C \sin (c+2 d x) a^6+4 A b \sin (3 c+2 d x) a^6-24 A b^2 d x \cos (2 c+d x) a^5+2 A b^2 \sin (d x) a^5-22 b^2 C \sin (d x) a^5+2 A b^2 \sin (2 c+d x) a^5+10 b^2 C \sin (2 c+d x) a^5-2 A b^2 \sin (2 c+3 d x) a^5-2 A b^2 \sin (4 c+3 d x) a^5+12 A b^3 d x \cos (c+2 d x) a^4+12 A b^3 d x \cos (3 c+2 d x) a^4+16 A b^3 \sin (c) a^4+14 b^3 C \sin (c) a^4-24 A b^3 \sin (c+2 d x) a^4+2 b^3 C \sin (c+2 d x) a^4-8 A b^3 \sin (3 c+2 d x) a^4+48 A b^4 d x \cos (2 c+d x) a^3-53 A b^4 \sin (d x) a^3+4 b^4 C \sin (d x) a^3+11 A b^4 \sin (2 c+d x) a^3-4 b^4 C \sin (2 c+d x) a^3+A b^4 \sin (2 c+3 d x) a^3+A b^4 \sin (4 c+3 d x) a^3-6 A b^5 d x \cos (c+2 d x) a^2-6 A b^5 d x \cos (3 c+2 d x) a^2+22 A b^5 \sin (c) a^2-4 b^5 C \sin (c) a^2+14 A b^5 \sin (c+2 d x) a^2+4 A b^5 \sin (3 c+2 d x) a^2-24 A b^2 \left (a^2-b^2\right )^2 d x \cos (d x) a-24 A b^6 d x \cos (2 c+d x) a+32 A b^6 \sin (d x) a-8 A b^6 \sin (2 c+d x) a-12 A b \left (a^2-b^2\right )^2 \left (a^2+2 b^2\right ) d x \cos (c)-20 A b^7 \sin (c)\right )}{\left (a^2-b^2\right )^2}-\frac{8 i \left (2 C a^6+b^2 (12 A+C) a^4-15 A b^4 a^2+6 A b^6\right ) \tan ^{-1}\left (\frac{(i \cos (c)+\sin (c)) \left (a \sin (c)+(a \cos (c)-b) \tan \left (\frac{d x}{2}\right )\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^2 (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{5/2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{4 a^4 d (\cos (2 (c+d x)) A+A+2 C) (a+b \sec (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(((-8*I)*(-15*a^2*A*b^4 + 6*A*b^6 + 2*a^6*C + a^4*b^
2*(12*A + C))*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Co
s[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^2*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^(5/2)*Sqrt[(Cos[c] - I*Sin[c])^
2]) + (Sec[c]*(-12*A*b*(a^2 - b^2)^2*(a^2 + 2*b^2)*d*x*Cos[c] - 24*a*A*b^2*(a^2 - b^2)^2*d*x*Cos[d*x] - 24*a^5
*A*b^2*d*x*Cos[2*c + d*x] + 48*a^3*A*b^4*d*x*Cos[2*c + d*x] - 24*a*A*b^6*d*x*Cos[2*c + d*x] - 6*a^6*A*b*d*x*Co
s[c + 2*d*x] + 12*a^4*A*b^3*d*x*Cos[c + 2*d*x] - 6*a^2*A*b^5*d*x*Cos[c + 2*d*x] - 6*a^6*A*b*d*x*Cos[3*c + 2*d*
x] + 12*a^4*A*b^3*d*x*Cos[3*c + 2*d*x] - 6*a^2*A*b^5*d*x*Cos[3*c + 2*d*x] + 16*a^4*A*b^3*Sin[c] + 22*a^2*A*b^5
*Sin[c] - 20*A*b^7*Sin[c] + 8*a^6*b*C*Sin[c] + 14*a^4*b^3*C*Sin[c] - 4*a^2*b^5*C*Sin[c] + a^7*A*Sin[d*x] + 2*a
^5*A*b^2*Sin[d*x] - 53*a^3*A*b^4*Sin[d*x] + 32*a*A*b^6*Sin[d*x] - 22*a^5*b^2*C*Sin[d*x] + 4*a^3*b^4*C*Sin[d*x]
+ a^7*A*Sin[2*c + d*x] + 2*a^5*A*b^2*Sin[2*c + d*x] + 11*a^3*A*b^4*Sin[2*c + d*x] - 8*a*A*b^6*Sin[2*c + d*x]
+ 10*a^5*b^2*C*Sin[2*c + d*x] - 4*a^3*b^4*C*Sin[2*c + d*x] + 4*a^6*A*b*Sin[c + 2*d*x] - 24*a^4*A*b^3*Sin[c + 2
*d*x] + 14*a^2*A*b^5*Sin[c + 2*d*x] - 8*a^6*b*C*Sin[c + 2*d*x] + 2*a^4*b^3*C*Sin[c + 2*d*x] + 4*a^6*A*b*Sin[3*
c + 2*d*x] - 8*a^4*A*b^3*Sin[3*c + 2*d*x] + 4*a^2*A*b^5*Sin[3*c + 2*d*x] + a^7*A*Sin[2*c + 3*d*x] - 2*a^5*A*b^
2*Sin[2*c + 3*d*x] + a^3*A*b^4*Sin[2*c + 3*d*x] + a^7*A*Sin[4*c + 3*d*x] - 2*a^5*A*b^2*Sin[4*c + 3*d*x] + a^3*
A*b^4*Sin[4*c + 3*d*x]))/(a^2 - b^2)^2))/(4*a^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.138, size = 1132, normalized size = 4.3 \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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

2/d*A/a^3*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d*A/a^4*b*arctan(tan(1/2*d*x+1/2*c))+8/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/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^3+1/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/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^4-4/d/a^3/(tan(1/2*
d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^5/(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/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b*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/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C*b^2-8/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)*A+1/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)*A+4/d/a^3/(tan(1/2*d*x+1/2*c
)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*b^5/(a+b)/(a^2-2*a*b+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^2-2*a*b+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^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+12/d*b^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))*A-15/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))*A*b^4+6/d/a^4/(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^6+2/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*a
rctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2+1/d/(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^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)*(A+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 = 0.786619, size = 2654, normalized size = 9.98 \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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(12*(A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*d*x*cos(d*x + c)^2 + 24*(A*a^7*b^2 - 3*A*a^5*b^4 +
3*A*a^3*b^6 - A*a*b^8)*d*x*cos(d*x + c) + 12*(A*a^6*b^3 - 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*d*x - (2*C*a^6*b
^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8 + (2*C*a^8 + (12*A + C)*a^6*b^2 - 15*A*a^4*b^4 + 6*A*a^2*b^6)
*cos(d*x + c)^2 + 2*(2*C*a^7*b + (12*A + C)*a^5*b^3 - 15*A*a^3*b^5 + 6*A*a*b^7)*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^2 - (13*A - 3*C)*a^5*b^4
+ 17*A*a^3*b^6 - 6*A*a*b^8 + 2*(A*a^9 - 3*A*a^7*b^2 + 3*A*a^5*b^4 - A*a^3*b^6)*cos(d*x + c)^2 + (4*(A - C)*a^
8*b - 5*(4*A - C)*a^6*b^3 + (25*A - C)*a^4*b^5 - 9*A*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^12 - 3*a^10*b^2
+ 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) + (a^10*
b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d), -1/2*(6*(A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*d*x*cos(d
*x + c)^2 + 12*(A*a^7*b^2 - 3*A*a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)*d*x*cos(d*x + c) + 6*(A*a^6*b^3 - 3*A*a^4*b^5
+ 3*A*a^2*b^7 - A*b^9)*d*x - (2*C*a^6*b^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8 + (2*C*a^8 + (12*A +
C)*a^6*b^2 - 15*A*a^4*b^4 + 6*A*a^2*b^6)*cos(d*x + c)^2 + 2*(2*C*a^7*b + (12*A + C)*a^5*b^3 - 15*A*a^3*b^5 + 6
*A*a*b^7)*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*A - 3*C)*a^7*b^2 - (13*A - 3*C)*a^5*b^4 + 17*A*a^3*b^6 - 6*A*a*b^8 + 2*(A*a^9 - 3*A*a^7*b^2 + 3*A*a
^5*b^4 - A*a^3*b^6)*cos(d*x + c)^2 + (4*(A - C)*a^8*b - 5*(4*A - C)*a^6*b^3 + (25*A - C)*a^4*b^5 - 9*A*a^2*b^7
)*cos(d*x + c))*sin(d*x + c))/((a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^11*b - 3*a^9*
b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) + (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*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)*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29381, size = 663, normalized size = 2.49 \begin{align*} \frac{\frac{{\left (2 \, C a^{6} + 12 \, A a^{4} b^{2} + C a^{4} b^{2} - 15 \, A a^{2} b^{4} + 6 \, A b^{6}\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^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{3 \,{\left (d x + c\right )} A b}{a^{4}} + \frac{4 \, C a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, A a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, A a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, A a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, A b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, A a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, A a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, A a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, A b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} 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}} + \frac{2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*C*a^6 + 12*A*a^4*b^2 + C*a^4*b^2 - 15*A*a^2*b^4 + 6*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^8 - 2*a^6*b^2 + a^4*b^4)
*sqrt(-a^2 + b^2)) - 3*(d*x + c)*A*b/a^4 + (4*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c
)^3 + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)
^3 - 5*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^6*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^5*b*tan(1/2*d*x + 1/2*c) - 3*C*
a^4*b^2*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c) + C*a^3*b^3*tan(1/2*d*x + 1/2*c) - 7*A*a^2*b^4
*tan(1/2*d*x + 1/2*c) + 5*A*a*b^5*tan(1/2*d*x + 1/2*c) + 4*A*b^6*tan(1/2*d*x + 1/2*c))/((a^7 - 2*a^5*b^2 + a^3
*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) + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*
x + 1/2*c)^2 + 1)*a^3))/d