\(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^4} \, dx\) [700]

Optimal result

Integrand size = 33, antiderivative size = 261 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {b \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]

[Out]

-b*(b^2*(A+2*C)+a^2*(4*A+3*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*(A*b^2+C*a^2)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^3+1/6*(3*A*b^4-4*a^4*C+a^2*b^2*(2*A+9*C))*tan(d
*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^2+1/6*a*(a^2*b^2*(2*A-5*C)+2*a^4*C+b^4*(13*A+18*C))*tan(d*x+c)/b^2/(a
^2-b^2)^3/d/(a+b*sec(d*x+c))

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4176, 4165, 4088, 12, 3916, 2738, 214} \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {a \left (2 a^4 C+a^2 b^2 (2 A-5 C)+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac {\left (-4 a^4 C+a^2 b^2 (2 A+9 C)+3 A b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2} \]

[In]

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

[Out]

-((b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a
+ b)^(7/2)*d)) + (a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((3*A*b^4 - 4
*a^4*C + a^2*b^2*(2*A + 9*C))*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + (a*(a^2*b^2*(2*A
- 5*C) + 2*a^4*C + b^4*(13*A + 18*C))*Tan[c + d*x])/(6*b^2*(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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

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 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4088

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] + D
ist[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 4165

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 4176

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

Rubi steps \begin{align*} \text {integral}& = \frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) \left (-3 b \left (A b^2+a^2 C\right )+a \left (2 A b^2-\left (a^2-3 b^2\right ) C\right ) \sec (c+d x)+3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = \frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (-2 a b^2 \left (a^2 C-b^2 (5 A+6 C)\right )-b \left (a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2} \\ & = \frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\int -\frac {3 b^4 \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3} \\ & = \frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (b \left (b^2 (A+2 C)+a^2 (4 A+3 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 {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = \frac {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \text {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 )}{\left (a^2-b^2\right )^3 d} \\ & = -\frac {b \left (4 a^2 A+A b^2+3 a^2 C+2 b^2 C\right ) \text {arctanh}\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 {a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {\frac {24 b \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {2 \left (6 b \left (-A b^4+9 a^2 b^2 (A+C)+a^4 (2 A+C)\right ) \cos (c+d x)+a \left (a^2 b^2 (14 A+C)+a^4 (6 A+8 C)+b^4 (25 A+36 C)+\left (-A b^4+a^4 (6 A+4 C)+a^2 b^2 (10 A+11 C)\right ) \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{(b+a \cos (c+d x))^3}}{24 \left (-a^2+b^2\right )^3 d} \]

[In]

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

[Out]

-1/24*((24*b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2
- b^2] + (2*(6*b*(-(A*b^4) + 9*a^2*b^2*(A + C) + a^4*(2*A + C))*Cos[c + d*x] + a*(a^2*b^2*(14*A + C) + a^4*(6*
A + 8*C) + b^4*(25*A + 36*C) + (-(A*b^4) + a^4*(6*A + 4*C) + a^2*b^2*(10*A + 11*C))*Cos[2*(c + d*x)]))*Sin[c +
d*x])/(b + a*Cos[c + d*x])^3)/((-a^2 + b^2)^3*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}+2 a^{3} C +3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (3 a^{2} A +7 A \,b^{2}+C \,a^{2}+9 C \,b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+2 a^{3} C -3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {b \left (4 a^{2} A +A \,b^{2}+3 C \,a^{2}+2 C \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(374\)
default	\(\frac {\frac {-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}+2 a^{3} C +3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (3 a^{2} A +7 A \,b^{2}+C \,a^{2}+9 C \,b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+2 a^{3} C -3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {b \left (4 a^{2} A +A \,b^{2}+3 C \,a^{2}+2 C \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(374\)
risch	\(\text {Expression too large to display}\)	\(1265\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (247) = 494\).

Time = 0.34 (sec) , antiderivative size = 1113, normalized size of antiderivative = 4.26 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]

[In]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(d*x+c)^2*(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 de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (247) = 494\).

Time = 0.37 (sec) , antiderivative size = 693, normalized size of antiderivative = 2.66 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (4 \, A a^{2} b + 3 \, C a^{2} b + A b^{3} + 2 \, C b^{3}\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 {6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A 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} \]

[In]

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

[Out]

-1/3*(3*(4*A*a^2*b + 3*C*a^2*b + A*b^3 + 2*C*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*A*a^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2
*c)^5 - 3*C*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*b^2*tan(1/2*d*x + 1/2
*c)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1
/2*c)^5 + 18*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^5*tan(1/2*d*x + 1/2*c)^3
- 4*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 16*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 32*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3
+ 28*A*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 36*C*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^5*tan(1/2*d*x + 1/2*c) + 6*C*a
^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b^2*tan(1
/2*d*x + 1/2*c) + 6*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 27*C*a^2*b^3*tan(1/2*
d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) + 18*C*a*b^4*tan(1/2*d*x + 1/2*c) - 3*A*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

Mupad [B] (verification not implemented)

Time = 20.99 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+2\,A\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3+C\,a^3+7\,A\,a\,b^2+9\,C\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-2\,A\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}+\frac {b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b+6{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^4}{\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (4\,A\,a^2+A\,b^2+3\,C\,a^2+2\,C\,b^2\right )\,1{}\mathrm {i}}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

[In]

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

[Out]

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