3.926 \(\int \frac{\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=299 \[ -\frac{\left (a^3 (-(2 A+C))+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\tan (c+d x) \left (-a^2 b^2 (11 A+10 C)+2 a^3 b B+a^4 C+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\tan (c+d x) \left (2 a^2 b B+a^3 C-a b^2 (5 A+6 C)+3 b^3 B\right )}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

[Out]

-(((4*a^2*b*B + b^3*B - a^3*(2*A + C) - a*b^2*(3*A + 4*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]
)/((a - b)^(7/2)*(a + b)^(7/2)*d)) - ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^3) + ((2*a^2*b*B + 3*b^3*B + a^3*C - a*b^2*(5*A + 6*C))*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c
+ d*x])^2) + ((2*a^3*b*B + 13*a*b^3*B + a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*(11*A + 10*C))*Tan[c + d*x])/(6*b*
(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.855812, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4080, 4003, 12, 3831, 2659, 208} \[ -\frac{\left (a^3 (-(2 A+C))+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\tan (c+d x) \left (-a^2 b^2 (11 A+10 C)+2 a^3 b B+a^4 C+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\tan (c+d x) \left (2 a^2 b B+a^3 C-a b^2 (5 A+6 C)+3 b^3 B\right )}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^4,x]

[Out]

-(((4*a^2*b*B + b^3*B - a^3*(2*A + C) - a*b^2*(3*A + 4*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]
)/((a - b)^(7/2)*(a + b)^(7/2)*d)) - ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^3) + ((2*a^2*b*B + 3*b^3*B + a^3*C - a*b^2*(5*A + 6*C))*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c
+ d*x])^2) + ((2*a^3*b*B + 13*a*b^3*B + a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*(11*A + 10*C))*Tan[c + d*x])/(6*b*
(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 4080

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

Rule 4003

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] + Dis
t[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]

Rule 12

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

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{\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{\int \frac{\sec (c+d x) \left (3 b (b B-a (A+C))+\left (2 A b^2-2 a b B-a^2 C+3 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\int \frac{\sec (c+d x) \left (-2 b \left (5 a b B-a^2 (3 A+2 C)-b^2 (2 A+3 C)\right )+\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\int \frac{3 b \left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 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 )}{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+a^3 C+4 a 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-b)^{7/2} (a+b)^{7/2} d}-\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 7.55916, size = 1069, normalized size = 3.58 \[ \frac{\left (-2 A a^3-C a^3+4 b B a^2-3 A b^2 a-4 b^2 C a+b^3 B\right ) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{2 i \tan ^{-1}\left (\sec \left (\frac{d x}{2}\right ) \left (\frac{\cos (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{i \sin (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) \left (i a \sin \left (c+\frac{d x}{2}\right )-i b \sin \left (\frac{d x}{2}\right )\right )\right ) \cos (c)}{\sqrt{a^2-b^2} d \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{2 \tan ^{-1}\left (\sec \left (\frac{d x}{2}\right ) \left (\frac{\cos (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{i \sin (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) \left (i a \sin \left (c+\frac{d x}{2}\right )-i b \sin \left (\frac{d x}{2}\right )\right )\right ) \sin (c)}{\sqrt{a^2-b^2} d \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) (b+a \cos (c+d x))^4}{\left (b^2-a^2\right )^3 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^4}+\frac{\sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (3 C \sin (c) a^6+6 B \sin (d x) a^6-12 b B \sin (c) a^5-18 A b \sin (d x) a^5-13 b C \sin (d x) a^5+27 A b^2 \sin (c) a^4+12 b^2 C \sin (c) a^4+10 b^2 B \sin (d x) a^4-3 b^3 B \sin (c) a^3+5 A b^3 \sin (d x) a^3-2 b^3 C \sin (d x) a^3-18 A b^4 \sin (c) a^2-b^4 B \sin (d x) a^2-2 A b^5 \sin (d x) a+6 A b^6 \sin (c)\right ) (b+a \cos (c+d x))^3}{3 a^3 \left (a^2-b^2\right )^3 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^4}+\frac{\sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (3 C \sin (d x) a^5-5 b C \sin (c) a^4-6 b B \sin (d x) a^4+8 b^2 B \sin (c) a^3+9 A b^2 \sin (d x) a^3+2 b^2 C \sin (d x) a^3-11 A b^3 \sin (c) a^2+b^3 B \sin (d x) a^2-3 b^4 B \sin (c) a-4 A b^4 \sin (d x) a+6 A b^5 \sin (c)\right ) (b+a \cos (c+d x))^2}{3 a^3 \left (a^2-b^2\right )^2 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^4}+\frac{2 \sec (c) \sec ^2(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (A \sin (c) b^4-a B \sin (c) b^3-a A \sin (d x) b^3+a^2 C \sin (c) b^2+a^2 B \sin (d x) b^2-a^3 C \sin (d x) b\right ) (b+a \cos (c+d x))}{3 a^3 \left (a^2-b^2\right ) d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(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 - a^3*C - 4*a*b^2*C)*(b + a*Cos[c + d*x])^4*Sec[c + d*x]^2*(A + B*S
ec[c + d*x] + C*Sec[c + d*x]^2)*(((-2*I)*ArcTan[Sec[(d*x)/2]*(Cos[c]/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*
c]]) - (I*Sin[c])/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*c]]))*((-I)*b*Sin[(d*x)/2] + I*a*Sin[c + (d*x)/2])]
*Cos[c])/(Sqrt[a^2 - b^2]*d*Sqrt[Cos[2*c] - I*Sin[2*c]]) - (2*ArcTan[Sec[(d*x)/2]*(Cos[c]/(Sqrt[a^2 - b^2]*Sqr
t[Cos[2*c] - I*Sin[2*c]]) - (I*Sin[c])/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*c]]))*((-I)*b*Sin[(d*x)/2] + I
*a*Sin[c + (d*x)/2])]*Sin[c])/(Sqrt[a^2 - b^2]*d*Sqrt[Cos[2*c] - I*Sin[2*c]])))/((-a^2 + b^2)^3*(A + 2*C + 2*B
*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^4) + (2*(b + a*Cos[c + d*x])*Sec[c]*Sec[c + d*x]^2*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*b^4*Sin[c] - a*b^3*B*Sin[c] + a^2*b^2*C*Sin[c] - a*A*b^3*Sin[d*x] + a
^2*b^2*B*Sin[d*x] - a^3*b*C*Sin[d*x]))/(3*a^3*(a^2 - b^2)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*
(a + b*Sec[c + d*x])^4) + ((b + a*Cos[c + d*x])^2*Sec[c]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2
)*(-11*a^2*A*b^3*Sin[c] + 6*A*b^5*Sin[c] + 8*a^3*b^2*B*Sin[c] - 3*a*b^4*B*Sin[c] - 5*a^4*b*C*Sin[c] + 9*a^3*A*
b^2*Sin[d*x] - 4*a*A*b^4*Sin[d*x] - 6*a^4*b*B*Sin[d*x] + a^2*b^3*B*Sin[d*x] + 3*a^5*C*Sin[d*x] + 2*a^3*b^2*C*S
in[d*x]))/(3*a^3*(a^2 - b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^4) + (
(b + a*Cos[c + d*x])^3*Sec[c]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(27*a^4*A*b^2*Sin[c] - 18
*a^2*A*b^4*Sin[c] + 6*A*b^6*Sin[c] - 12*a^5*b*B*Sin[c] - 3*a^3*b^3*B*Sin[c] + 3*a^6*C*Sin[c] + 12*a^4*b^2*C*Si
n[c] - 18*a^5*A*b*Sin[d*x] + 5*a^3*A*b^3*Sin[d*x] - 2*a*A*b^5*Sin[d*x] + 6*a^6*B*Sin[d*x] + 10*a^4*b^2*B*Sin[d
*x] - a^2*b^4*B*Sin[d*x] - 13*a^5*b*C*Sin[d*x] - 2*a^3*b^3*C*Sin[d*x]))/(3*a^3*(a^2 - b^2)^3*d*(A + 2*C + 2*B*
Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^4)

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Maple [A]  time = 0.103, size = 452, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 6\,A{a}^{2}b+3\,Aa{b}^{2}+2\,A{b}^{3}-2\,B{a}^{3}-2\,B{a}^{2}b-6\,Ba{b}^{2}-B{b}^{3}+{a}^{3}C+6\,{a}^{2}bC+2\,Ca{b}^{2}+2\,C{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ \left ( a-b \right ) \left ({a}^{3}+3\,{a}^{2}b+3\,a{b}^{2}+{b}^{3} \right ) }}+2/3\,{\frac{ \left ( 9\,A{a}^{2}b+A{b}^{3}-3\,B{a}^{3}-7\,Ba{b}^{2}+7\,{a}^{2}bC+3\,C{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}-1/2\,{\frac{ \left ( 6\,A{a}^{2}b-3\,Aa{b}^{2}+2\,A{b}^{3}-2\,B{a}^{3}+2\,B{a}^{2}b-6\,Ba{b}^{2}+B{b}^{3}-{a}^{3}C+6\,{a}^{2}bC-2\,Ca{b}^{2}+2\,C{b}^{3} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }} \right ) }+{\frac{2\,A{a}^{3}+3\,Aa{b}^{2}-4\,B{a}^{2}b-B{b}^{3}+{a}^{3}C+4\,Ca{b}^{2}}{{a}^{6}-3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}-{b}^{6}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x)

[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+C*a^3+6*C*a^2*b+2*C*a*b^2+2*C*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+7*C*a^2*b+3*C*b^
3)/(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-C*a^3+6*C*a^2*b-2*C*a*b^2+2*C*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+C*a^3+4*C*a*b^2)/(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)))

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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(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.30993, size = 3109, normalized size = 10.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*((2*A + C)*a^3*b^3 - 4*B*a^2*b^4 + (3*A + 4*C)*a*b^5 - B*b^6 + ((2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*
C)*a^4*b^2 - B*a^3*b^3)*cos(d*x + c)^3 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - B*a^2*b^4)*c
os(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A + 4*C)*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*(C*a^6*b + 2*B*a^5*b^2 - 11*(A + C)*a^4
*b^3 + 11*B*a^3*b^4 + (7*A + 4*C)*a^2*b^5 - 13*B*a*b^6 + 2*(2*A + 3*C)*b^7 + (6*B*a^7 - (18*A + 13*C)*a^6*b +
4*B*a^5*b^2 + (23*A + 11*C)*a^4*b^3 - 11*B*a^3*b^4 - (7*A - 2*C)*a^2*b^5 + B*a*b^6 + 2*A*b^7)*cos(d*x + c)^2 +
3*(C*a^7 + 2*B*a^6*b - (9*A + 10*C)*a^5*b^2 + 7*B*a^4*b^3 + (8*A + 7*C)*a^3*b^4 - 10*B*a^2*b^5 + (A + 2*C)*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 + C)*a^3*b^3 - 4*B*a^2*b^4 + (3*A + 4*C)*a*b^5 - B*b^6 + ((2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*C)*a^4*b^2
- B*a^3*b^3)*cos(d*x + c)^3 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - B*a^2*b^4)*cos(d*x + c)
^2 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A + 4*C)*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))) + (C*a^6*b + 2*B*a^5*b^2 - 11*(A + C)*a^4*
b^3 + 11*B*a^3*b^4 + (7*A + 4*C)*a^2*b^5 - 13*B*a*b^6 + 2*(2*A + 3*C)*b^7 + (6*B*a^7 - (18*A + 13*C)*a^6*b + 4
*B*a^5*b^2 + (23*A + 11*C)*a^4*b^3 - 11*B*a^3*b^4 - (7*A - 2*C)*a^2*b^5 + B*a*b^6 + 2*A*b^7)*cos(d*x + c)^2 +
3*(C*a^7 + 2*B*a^6*b - (9*A + 10*C)*a^5*b^2 + 7*B*a^4*b^3 + (8*A + 7*C)*a^3*b^4 - 10*B*a^2*b^5 + (A + 2*C)*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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**4,x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/(a + b*sec(c + d*x))**4, x)

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Giac [B]  time = 1.51201, size = 1307, normalized size = 4.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*A*a^3 + C*a^3 - 4*B*a^2*b + 3*A*a*b^2 + 4*C*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 - 3*C*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 - 12*C*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 + 27*C*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 - 12*C*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*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*A*b^5*t
an(1/2*d*x + 1/2*c)^5 + 3*B*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*C*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 + 28*C*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 - 16*C*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 - 12*C*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^5*tan(1/2*d*x + 1/2*c
) + 3*C*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) - 12*C*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*tan(1/2*d*x + 1/2*c) - 27*C*a^3*b^2
*tan(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) - 12*C*a^2*b^3*ta
n(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*C*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) - 6*C*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