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

Optimal. Leaf size=138 \[ \frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {A x}{a^2}-\frac {2 \left (a^3 (-B)+2 a^2 A b+a^2 b C-A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4060, 3919, 3831, 2659, 208} \[ -\frac {2 \left (2 a^2 A b+a^2 b C+a^3 (-B)-A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {A x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/a^2 - (2*(2*a^2*A*b - A*b^3 - a^3*B + a^2*b*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2
*(a - b)^(3/2)*(a + b)^(3/2)*d) + ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*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 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 4060

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

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {-A \left (a^2-b^2\right )+a (A b-a B+b C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {A x}{a^2}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (A b^3+a^3 B-a^2 b (2 A+C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {A x}{a^2}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (A b^3+a^3 B-a^2 b (2 A+C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^2 b \left (a^2-b^2\right )}\\ &=\frac {A x}{a^2}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (2 \left (A b^3+a^3 B-a^2 b (2 A+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 )}{a^2 b \left (a^2-b^2\right ) d}\\ &=\frac {A x}{a^2}-\frac {2 \left (2 a^2 A b-A b^3-a^3 B+a^2 b C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 2.29, size = 299, normalized size = 2.17 \[ \frac {2 (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {2 i (\cos (c)-i \sin (c)) \left (a^3 B-a^2 b (2 A+C)+A b^3\right ) (a \cos (c+d x)+b) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{d \left (a^2-b^2\right )^{3/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {\left (a (a C-b B)+A b^2\right ) (a \sin (d x)-b \sin (c))}{d (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )}+A x (a \cos (c+d x)+b)\right )}{a^2 (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*x*(b + a*Cos[c + d*x]) - ((2*I)*(A*b^3 + a^
3*B - a^2*b*(2*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[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^(3/2)*d*Sqrt[(Cos[c] - I
*Sin[c])^2]) + ((A*b^2 + a*(-(b*B) + a*C))*(-(b*Sin[c]) + a*Sin[d*x]))/((a - b)*(a + b)*d*(Cos[c/2] - Sin[c/2]
)*(Cos[c/2] + Sin[c/2]))))/(a^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)

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fricas [B]  time = 0.73, size = 593, normalized size = 4.30 \[ \left [\frac {2 \, {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x + {\left (B a^{3} b - {\left (2 \, A + C\right )} a^{2} b^{2} + A b^{4} + {\left (B a^{4} - {\left (2 \, A + C\right )} a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (C a^{5} - B a^{4} b + {\left (A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x + {\left (B a^{3} b - {\left (2 \, A + C\right )} a^{2} b^{2} + A b^{4} + {\left (B a^{4} - {\left (2 \, A + C\right )} a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{5} - B a^{4} b + {\left (A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(A*a^5 - 2*A*a^3*b^2 + A*a*b^4)*d*x*cos(d*x + c) + 2*(A*a^4*b - 2*A*a^2*b^3 + A*b^5)*d*x + (B*a^3*b -
(2*A + C)*a^2*b^2 + A*b^4 + (B*a^4 - (2*A + C)*a^3*b + A*a*b^3)*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^5 - B*a^4*b + (A - C)*a^3*b^2 + B*a^2*b^3 - A*a*b^4)*s
in(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), ((A*a^5 - 2*A*a^3
*b^2 + A*a*b^4)*d*x*cos(d*x + c) + (A*a^4*b - 2*A*a^2*b^3 + A*b^5)*d*x + (B*a^3*b - (2*A + C)*a^2*b^2 + A*b^4
+ (B*a^4 - (2*A + C)*a^3*b + A*a*b^3)*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^5 - B*a^4*b + (A - C)*a^3*b^2 + B*a^2*b^3 - A*a*b^4)*sin(d*x + c))/((a
^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)]

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giac [A]  time = 0.25, size = 222, normalized size = 1.61 \[ \frac {\frac {2 \, {\left (B a^{3} - 2 \, A a^{2} b - C a^{2} b + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {{\left (d x + c\right )} A}{a^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(B*a^3 - 2*A*a^2*b - C*a^2*b + A*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^4 - a^2*b^2)*sqrt(-a^2 + b^2)) + (d*x + c)*A/a
^2 - 2*(C*a^2*tan(1/2*d*x + 1/2*c) - B*a*b*tan(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(
a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d

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maple [B]  time = 0.65, size = 448, normalized size = 3.25 \[ -\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A \,b^{2}}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B b}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {4 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A \,b^{3}}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C b}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a/(a^2-b^2)*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)*A*b^2+2/d/(a^2-b^2)*ta
n(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)*B*b-2/d*a/(a^2-b^2)*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)*C-4/d*b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+
1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+2/d/a^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/(
(a-b)*(a+b))^(1/2))*A*b^3+2/d/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(
1/2))*B*a-2/d/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C*b+2/d/a^
2*arctan(tan(1/2*d*x+1/2*c))*A

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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((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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 = 13.32, size = 4544, normalized size = 32.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*atan(((A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b
^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 +
4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (A*((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*
b^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6
- a^3*b^3 - a^4*b^2) - (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8
*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1i)/a^2))/a^2 + (A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^
2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 4*
A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) -
(A*((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 -
C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (A*tan(c/2 + (d*x)/2)*(2*a^9*
b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1
i)/a^2))/a^2)/((64*(A^3*b^5 + A*B^2*a^5 - A^2*B*a^5 - A^3*a*b^4 + 2*A^3*a^4*b - 3*A^3*a^2*b^3 + 2*A^3*a^3*b^2
- 3*A^2*B*a^4*b - A^2*C*a*b^4 + A^2*C*a^4*b + A^2*B*a^2*b^3 + A^2*B*a^3*b^2 + A*C^2*a^3*b^2 - A^2*C*a^2*b^3 +
3*A^2*C*a^3*b^2 - 2*A*B*C*a^4*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*
A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 -
4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2)
+ (A*((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3
- C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (A*tan(c/2 + (d*x)/2)*(2*a^
9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))
*1i)/a^2)*1i)/a^2 - (A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*
A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*
a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (A*((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3
+ A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^
5*b + a^6 - a^3*b^3 - a^4*b^2) + (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^
3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1i)/a^2)*1i)/a^2)))/(a^2*d) + (atan(((((32*tan(c/
2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^
2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b +
a^5 - a^2*b^3 - a^3*b^2) + (((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2
+ C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32
*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*(2*a^9*b - 2*a^4*b^6 + 2
*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 -
3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 -
3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*1i)/(a^8 - a^2*b^6 + 3*a^4*b^4 -
3*a^6*b^2) + (((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*
b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4
+ 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b
^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6
- a^3*b^3 - a^4*b^2) + (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b
)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a
^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^
8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*1i)/(a
^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))/((64*(A^3*b^5 + A*B^2*a^5 - A^2*B*a^5 - A^3*a*b^4 + 2*A^3*a^4*b - 3*A^3
*a^2*b^3 + 2*A^3*a^3*b^2 - 3*A^2*B*a^4*b - A^2*C*a*b^4 + A^2*C*a^4*b + A^2*B*a^2*b^3 + A^2*B*a^3*b^2 + A*C^2*a
^3*b^2 - A^2*C*a^2*b^3 + 3*A^2*C*a^3*b^2 - 2*A*B*C*a^4*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (((32*tan(c/2 +
(d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a
^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^
5 - a^2*b^3 - a^3*b^2) + (((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2 +
C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*ta
n(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*(2*a^9*b - 2*a^4*b^6 + 2*a^
5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*
a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a
^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6
*b^2) - (((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 +
4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4*A*
C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B
*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*
b^3 - a^4*b^2) + (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*(2*a
^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a
^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^8 - a^
2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b))/(a^8 - a^2*
b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*A*a^2*b - C*a^2*b)*2i)/(d*(a^8 -
a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)) - (2*tan(c/2 + (d*x)/2)*(A*b^2 + C*a^2 - B*a*b))/(d*(a + b)*(a*b - a^2)*(a
+ b - tan(c/2 + (d*x)/2)^2*(a - b)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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