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

[Out]

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

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Rubi [A]  time = 0.86, antiderivative size = 299, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 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 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]

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*}

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Mathematica [C]  time = 7.55, 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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fricas [B]  time = 0.62, size = 1414, normalized size = 4.73 \[ \text {result too large to display} \]

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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giac [B]  time = 0.43, size = 968, normalized size = 3.24 \[ \text {result too large to display} \]

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

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maple [A]  time = 0.79, size = 452, normalized size = 1.51 \[ \frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}-2 a^{3} B -2 a^{2} b B -6 B a \,b^{2}-b^{3} B +C \,a^{3}+6 C \,a^{2} b +2 C a \,b^{2}+2 b^{3} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 a^{3} B -7 B a \,b^{2}+7 C \,a^{2} b +3 b^{3} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-2 a^{3} B +2 a^{2} b B -6 B a \,b^{2}+b^{3} B -C \,a^{3}+6 C \,a^{2} b -2 C a \,b^{2}+2 b^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{3}}+\frac {\left (2 A \,a^{3}+3 A a \,b^{2}-4 a^{2} b B -b^{3} B +C \,a^{3}+4 C a \,b^{2}\right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 8.40, size = 516, normalized size = 1.73 \[ \frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-B\,b^3+C\,a^3+3\,A\,a\,b^2-4\,B\,a^2\,b+4\,C\,a\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3-3\,B\,a^3+3\,C\,b^3+9\,A\,a^2\,b-7\,B\,a\,b^2+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3-2\,B\,a^3-B\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-2\,B\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)*(a - b)^(7/2)))*(2*A*
a^3 - B*b^3 + C*a^3 + 3*A*a*b^2 - 4*B*a^2*b + 4*C*a*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((tan(c/2 + (d*x)/
2)*(2*A*b^3 - 2*B*a^3 + B*b^3 - C*a^3 + 2*C*b^3 - 3*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 + 2*B*a^2*b - 2*C*a*b^2 +
6*C*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) - (4*tan(c/2 + (d*x)/2)^3*(A*b^3 - 3*B*a^3 + 3*C*b^3 + 9
*A*a^2*b - 7*B*a*b^2 + 7*C*a^2*b))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5*(2*A*b^3 - 2*B*a^
3 - B*b^3 + C*a^3 + 2*C*b^3 + 3*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 - 2*B*a^2*b + 2*C*a*b^2 + 6*C*a^2*b))/((a + b)
^3*(a - b)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*
a^2*b - 3*a^3 - 3*b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3))
)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]

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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