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

Optimal. Leaf size=252 \[ \frac {a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a \left (a^2 (-C)+5 A b^2+6 b^2 C\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {\left (a^4 C-a^2 b^2 (11 A+10 C)-2 b^4 (2 A+3 C)\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]

[Out]

a*(a^2*(2*A+C)+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+C*a^2)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^3-1/6*a*(5*A*b^2-C*a^2+6*C*b^2)*tan(d*x+c)/b/(a^2-b^
2)^2/d/(a+b*sec(d*x+c))^2+1/6*(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 = 0.55, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4081, 4003, 12, 3831, 2659, 208} \[ \frac {a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\left (-a^2 b^2 (11 A+10 C)+a^4 C-2 b^4 (2 A+3 C)\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac {a \left (a^2 (-C)+5 A b^2+6 b^2 C\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2*(2*A + C) + 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^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) - (a*(5*A*b^2 - a^2*C
+ 6*b^2*C)*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((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 4081

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=-\frac {\left (A b^2+a^2 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 a b (A+C)+\left (2 A b^2-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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 \left (b^3 (2 A+3 C)+\frac {1}{2} a^2 (6 A b+4 b C)\right )-a \left (5 A b^2-\left (a^2-6 b^2\right ) 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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 (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 a b \left (a^2 (2 A+C)+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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 (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 (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 (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 (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\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^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 (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 (a \left (a^2 (2 A+C)+b^2 (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 )}{b \left (a^2-b^2\right )^3 d}\\ &=\frac {a \left (2 a^2 A+3 A b^2+a^2 C+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-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 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 (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 = 6.94, size = 438, normalized size = 1.74 \[ \frac {\sec ^2(c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {6 i a (\cos (c)-i \sin (c)) \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) (a \cos (c+d x)+b)^3 \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{7/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {2 b \sec (c) \left (a^2 C+A b^2\right ) (b \sin (c)-a \sin (d x))}{a^5-a^3 b^2}+\frac {\sec (c) (a \cos (c+d x)+b) \left (\sin (c) \left (-5 a^4 b C-11 a^2 A b^3+6 A b^5\right )+a \sin (d x) \left (3 a^4 C+a^2 b^2 (9 A+2 C)-4 A b^4\right )\right )}{a^3 \left (a^2-b^2\right )^2}+\frac {\sec (c) (a \cos (c+d x)+b)^2 \left (3 \sin (c) \left (a^6 C+a^4 b^2 (9 A+4 C)-6 a^2 A b^4+2 A b^6\right )-a b \sin (d x) \left (a^4 (18 A+13 C)+a^2 b^2 (2 C-5 A)+2 A b^4\right )\right )}{\left (a^3-a b^2\right )^3}\right )}{3 d (a+b \sec (c+d x))^4 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(((-6*I)*a*(a^2*(2*A + C) + b^2*(3*A + 4*C))*ArcTa
n[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]
)]*(b + a*Cos[c + d*x])^3*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^(7/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (2*b*(A*b^2 +
a^2*C)*Sec[c]*(b*Sin[c] - a*Sin[d*x]))/(a^5 - a^3*b^2) + ((b + a*Cos[c + d*x])*Sec[c]*((-11*a^2*A*b^3 + 6*A*b
^5 - 5*a^4*b*C)*Sin[c] + a*(-4*A*b^4 + 3*a^4*C + a^2*b^2*(9*A + 2*C))*Sin[d*x]))/(a^3*(a^2 - b^2)^2) + ((b + a
*Cos[c + d*x])^2*Sec[c]*(3*(-6*a^2*A*b^4 + 2*A*b^6 + a^6*C + a^4*b^2*(9*A + 4*C))*Sin[c] - a*b*(2*A*b^4 + a^2*
b^2*(-5*A + 2*C) + a^4*(18*A + 13*C))*Sin[d*x]))/(a^3 - a*b^2)^3))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*
Sec[c + d*x])^4)

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fricas [B]  time = 0.61, size = 1114, normalized size = 4.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+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 + (3*A + 4*C)*a*b^5 + ((2*A + C)*a^6 + (3*A + 4*C)*a^4*b^2)*cos(d*x + c)^3 + 3*((
2*A + C)*a^5*b + (3*A + 4*C)*a^3*b^3)*cos(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 + (3*A + 4*C)*a^2*b^4)*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 - 11*(A + C)*a^
4*b^3 + (7*A + 4*C)*a^2*b^5 + 2*(2*A + 3*C)*b^7 - ((18*A + 13*C)*a^6*b - (23*A + 11*C)*a^4*b^3 + (7*A - 2*C)*a
^2*b^5 - 2*A*b^7)*cos(d*x + c)^2 + 3*(C*a^7 - (9*A + 10*C)*a^5*b^2 + (8*A + 7*C)*a^3*b^4 + (A + 2*C)*a*b^6)*co
s(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 + (3*A + 4*C)*a*b^5 + ((2*A + C)*a^6 + (3*A + 4*C)*a^4*b^2)*cos(d*x + c)^3 + 3*((2*A + C)*a^5*b + (3*A + 4*C
)*a^3*b^3)*cos(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 + (3*A + 4*C)*a^2*b^4)*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 - 11*(A + C)*a^4*b^3 + (7*A + 4*
C)*a^2*b^5 + 2*(2*A + 3*C)*b^7 - ((18*A + 13*C)*a^6*b - (23*A + 11*C)*a^4*b^3 + (7*A - 2*C)*a^2*b^5 - 2*A*b^7)
*cos(d*x + c)^2 + 3*(C*a^7 - (9*A + 10*C)*a^5*b^2 + (8*A + 7*C)*a^3*b^4 + (A + 2*C)*a*b^6)*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 = 0.44, size = 693, normalized size = 2.75 \[ -\frac {\frac {3 \, {\left (2 \, A a^{3} + C a^{3} + 3 \, A a b^{2} + 4 \, C a b^{2}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {3 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+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 + 3*A*a*b^2 + 4*C*a*b^2)*(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)) - (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 + 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 - 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 + 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 - 6*C*a*b^4*tan(1/2*d*x +
1/2*c)^5 + 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^5*tan(1/2*d*x + 1/2*c)^5 - 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 + 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 + 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 - 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) + 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) + 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) + 12*C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 3*A*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) + 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 = 0.74, size = 373, normalized size = 1.48 \[ \frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}+C \,a^{3}+6 C \,a^{2} b +2 C a \,b^{2}+2 b^{3} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {2 \left (9 a^{2} A +A \,b^{2}+7 a^{2} C +3 b^{2} C \right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-C \,a^{3}+6 C \,a^{2} b -2 C a \,b^{2}+2 b^{3} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{3}}+\frac {a \left (2 a^{2} A +3 A \,b^{2}+a^{2} C +4 b^{2} C \right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+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+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+A*b^2+7*C*a^2+3*C*b^2)*b/(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-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+a*(2*A*a^2+3*A*b^2+C*a^2+4*C*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+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 = 8.34, size = 483, normalized size = 1.92 \[ -\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3+3\,C\,b^3+9\,A\,a^2\,b+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {a\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b+6{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^4}{\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^2+3\,A\,b^2+C\,a^2+4\,C\,b^2\right )\,1{}\mathrm {i}}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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