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

Optimal. Leaf size=131 \[ -\frac {2 \left (a^3 B-2 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {B \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4009, 3998, 3770, 3831, 2659, 208} \[ -\frac {2 \left (a^3 B-2 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {a (A b-a B) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {B \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(b^2*d) - (2*(A*b^3 + a^3*B - 2*a*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/((a - b)^(3/2)*b^2*(a + b)^(3/2)*d) + (a*(A*b - a*B)*Tan[c + d*x])/(b*(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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4009

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

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 191, normalized size = 1.46 \[ \frac {\cos (c+d x) (A+B \sec (c+d x)) \left (\frac {2 \left (a B \left (a^2-2 b^2\right )+A b^3\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a b (a B-A b) \sin (c+d x)}{(b-a) (a+b) (a \cos (c+d x)+b)}-B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b^2 d (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 4.06, size = 694, normalized size = 5.30 \[ \left [\frac {{\left (B a^{3} b - 2 \, B a b^{3} + A b^{4} + {\left (B a^{4} - 2 \, B a^{2} b^{2} + 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 ) + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5} + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5} + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}, -\frac {2 \, {\left (B a^{3} b - 2 \, B a b^{3} + A b^{4} + {\left (B a^{4} - 2 \, B a^{2} b^{2} + 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 (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5} + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5} + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((B*a^3*b - 2*B*a*b^3 + A*b^4 + (B*a^4 - 2*B*a^2*b^2 + 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)) + (B*a^4*b - 2*B*a^2*b^3 + B*b^5 + (B*a^5 - 2*B*a^3*b^2 +
B*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (B*a^4*b - 2*B*a^2*b^3 + B*b^5 + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4
)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(B*a^4*b - A*a^3*b^2 - B*a^2*b^3 + A*a*b^4)*sin(d*x + c))/((a^5*b^2
- 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d), -1/2*(2*(B*a^3*b - 2*B*a*b^3 + A*b^4 +
(B*a^4 - 2*B*a^2*b^2 + 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))) - (B*a^4*b - 2*B*a^2*b^3 + B*b^5 + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4)*cos(d*x + c))*lo
g(sin(d*x + c) + 1) + (B*a^4*b - 2*B*a^2*b^3 + B*b^5 + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4)*cos(d*x + c))*log(-sin(
d*x + c) + 1) + 2*(B*a^4*b - A*a^3*b^2 - B*a^2*b^3 + A*a*b^4)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*c
os(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d)]

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giac [A]  time = 0.31, size = 231, normalized size = 1.76 \[ -\frac {\frac {2 \, {\left (B a^{3} - 2 \, B a b^{2} + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} - \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(B*a^3 - 2*B*a*b^2 + 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^2*b^2 - b^4)*sqrt(-a^2 + b^2)) - B*log(abs(tan(1/2*d*x
+ 1/2*c) + 1))/b^2 + B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 - 2*(B*a^2*tan(1/2*d*x + 1/2*c) - A*a*b*tan(1/2*
d*x + 1/2*c))/((a^2*b - b^3)*(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 = 0.75, size = 350, normalized size = 2.67 \[ -\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d b \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,b^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(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+2/d/b*a^2/(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)*B-2/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^3/b^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(t
an(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B+4/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-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*B+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*B

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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)^2*(A+B*sec(d*x+c))/(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 = 9.65, size = 3751, normalized size = 28.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (B*atan(((B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8
+ 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 +
4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 - (32*tan(c/2 + (d*x)/2)*(A^
2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*
B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 - (B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9
- A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) +
(32*B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 +
b^5 - a^2*b^3 - a^3*b^2))))/b^2 + (32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*
a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 -
a^3*b^2))*1i)/b^2)/((B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4
+ A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*B*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*
a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 - (32*tan(c/2 + (d
*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*
b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (64*(B^3*a^5 - A*B^2*b^5 + A^2*B
*b^5 + 2*B^3*a*b^4 - B^3*a^4*b + 2*B^3*a^2*b^3 - 3*B^3*a^3*b^2 - 3*A*B^2*a*b^4 + A*B^2*a^2*b^3 + A*B^2*a^3*b^2
))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (B*((B*((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^
3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*B*tan(c/2 + (d*x)/2)*(2*a*b^
9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2
+ (32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^
3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2))*2i)/(b^2*d) -
(atan(((((32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4
*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (((32*(A*a^2*
b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a
^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*(2*a*b^9 -
2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^
6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b
^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*
b^2) + (((32*tan(c/2 + (d*x)/2)*(A^2*b^6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4
*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^2*
b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a
^2*b^4 - a^3*b^3) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*(2*a*b^9 -
2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^
6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b
^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*
b^2))/((64*(B^3*a^5 - A*B^2*b^5 + A^2*B*b^5 + 2*B^3*a*b^4 - B^3*a^4*b + 2*B^3*a^2*b^3 - 3*B^3*a^3*b^2 - 3*A*B^
2*a*b^4 + A*B^2*a^2*b^3 + A*B^2*a^3*b^2))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (((32*tan(c/2 + (d*x)/2)*(A^2*b^
6 + 2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*
b^5 + 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2
*b^7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)
/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a
^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(
a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1
/2)*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) + (((32*tan(c/2 + (d*x)/2)*(A^2*b^6 +
2*B^2*a^6 + B^2*b^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b + 3*B^2*a^2*b^4 + 4*B^2*a^3*b^3 - 5*B^2*a^4*b^2 - 4*A*B*a*b^5
+ 2*A*B*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(A*a^2*b^7 - B*b^9 - A*b^9 - A*a^3*b^6 + B*a^2*b^
7 - 3*B*a^3*b^6 + B*a^5*b^4 + A*a*b^8 + 2*B*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*tan(c/2 + (d*x)/2)
*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*
b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a -
b)^3)^(1/2)*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)
*(A*b^3 + B*a^3 - 2*B*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^3 + B
*a^3 - 2*B*a*b^2)*2i)/(d*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)) - (2*tan(c/2 + (d*x)/2)*(B*a^2 - A*a*b))/(d*
(a + b)*(a*b - b^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 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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