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

Optimal. Leaf size=88 \[ -\frac {2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d \sqrt {a-b} \sqrt {a+b}}+\frac {A x}{a}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4051, 3770, 3919, 3831, 2659, 208} \[ -\frac {2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d \sqrt {a-b} \sqrt {a+b}}+\frac {A x}{a}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4051

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[C/b, I
nt[Csc[e + f*x], x], x] + Dist[1/b, Int[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, A, C}, x]

Rubi steps

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

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Mathematica [C]  time = 0.43, size = 239, normalized size = 2.72 \[ \frac {2 \left (A \cos ^2(c+d x)+C\right ) \left (\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2} \left (-a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+A b d x\right )+2 (\sin (c)+i \cos (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )\right )}{a b d \sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.07, size = 362, normalized size = 4.11 \[ \left [\frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x + {\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}, \frac {2 \, {\left (A a^{2} b - A b^{3}\right )} d x - 2 \, {\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} b - a b^{3}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(A*a^2*b - A*b^3)*d*x + (C*a^2 + A*b^2)*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)) + (C*a^3 - C*a*b^2)*log(sin(d*x + c) + 1) - (C*a^3 - C*a*b^2)*log(-sin(d*x + c) + 1))/((a^3*b
- a*b^3)*d), 1/2*(2*(A*a^2*b - A*b^3)*d*x - 2*(C*a^2 + A*b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*co
s(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (C*a^3 - C*a*b^2)*log(sin(d*x + c) + 1) - (C*a^3 - C*a*b^2)*log(
-sin(d*x + c) + 1))/((a^3*b - a*b^3)*d)]

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giac [A]  time = 0.26, size = 144, normalized size = 1.64 \[ \frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b} - \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*A/a + C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b - 2*(C*a^2 +
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)))/(sqrt(-a^2 + b^2)*a*b))/d

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maple [A]  time = 0.63, size = 158, normalized size = 1.80 \[ -\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d b}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d b}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b/a/((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/b*a/((a-b)*(a+b))^(1/
2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C-1/d/b*ln(tan(1/2*d*x+1/2*c)-1)*C+1/d/b*ln(tan(1/2*d
*x+1/2*c)+1)*C+2/d*A/a*arctan(tan(1/2*d*x+1/2*c))

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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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),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 = 9.81, size = 3656, normalized size = 41.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*atan((16384*A^5*b^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b
^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^
2*a*b^3 - (32768*A^4*C*b^5)/a) + (16384*A^5*b^5*tan(c/2 + (d*x)/2))/(16384*A^5*b^5 - 16384*A*C^4*a^5 + 32768*A
^4*C*b^5 - 16384*A^5*a*b^4 + 32768*A^2*C^3*a^2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^3
*C^2*a^3*b^2 + 16384*A*C^4*a^4*b - 32768*A^4*C*a*b^4) + (16384*A*C^4*a^4*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 +
16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*
A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (32768*A^4*C*b^4*tan(c/2 + (d
*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768
*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (327
68*A^4*C*b^5*tan(c/2 + (d*x)/2))/(16384*A^5*b^5 - 16384*A*C^4*a^5 + 32768*A^4*C*b^5 - 16384*A^5*a*b^4 + 32768*
A^2*C^3*a^2*b^3 - 32768*A^2*C^3*a^3*b^2 + 32768*A^3*C^2*a^2*b^3 - 32768*A^3*C^2*a^3*b^2 + 16384*A*C^4*a^4*b -
32768*A^4*C*a*b^4) - (16384*A*C^4*a^3*b*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4
- (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3
- 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) - (32768*A^2*C^3*a*b^3*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 1638
4*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^
4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) - (32768*A^3*C^2*a*b^3*tan(c/2 + (d
*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768
*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a) + (327
68*A^2*C^3*a^2*b^2*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768*A^4*C*b^4 - (16384*A^5*b^5)/a
+ 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2*C^3*a*b^3 - 32768*A^3*C^2*a*b^
3 - (32768*A^4*C*b^5)/a) + (32768*A^3*C^2*a^2*b^2*tan(c/2 + (d*x)/2))/(16384*A^5*b^4 + 16384*A*C^4*a^4 + 32768
*A^4*C*b^4 - (16384*A^5*b^5)/a + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A*C^4*a^3*b - 32768*A^2
*C^3*a*b^3 - 32768*A^3*C^2*a*b^3 - (32768*A^4*C*b^5)/a)))/(a*d) + (2*C*atanh((16384*C^5*a^4*tan(c/2 + (d*x)/2)
)/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C
^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (16384*C^5
*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4*a^5 - 16384*A^4*C*b^5 - 16384*C^5*a^4*b - 32768*A^2*C^3*
a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b^3 + 32768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^
4*C*a*b^4) + (32768*A*C^4*a^4*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*
C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A
^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A*C^4*a^5*tan(c/2 + (d*x)/2))/(16384*C^5*a^5 + 32768*A*C^4*a^5 -
16384*A^4*C*b^5 - 16384*C^5*a^4*b - 32768*A^2*C^3*a^2*b^3 + 32768*A^2*C^3*a^3*b^2 - 32768*A^3*C^2*a^2*b^3 + 32
768*A^3*C^2*a^3*b^2 - 32768*A*C^4*a^4*b + 16384*A^4*C*a*b^4) + (16384*A^4*C*b^4*tan(c/2 + (d*x)/2))/(16384*C^5
*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 -
16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) - (16384*A^4*C*a*b^3*tan
(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^
2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/
b) - (32768*A^2*C^3*a^3*b*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*
a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C
^2*a^3*b - (32768*A*C^4*a^5)/b) - (32768*A^3*C^2*a^3*b*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 +
16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 3276
8*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A^2*C^3*a^2*b^2*tan(c/2 + (d*x)/2))/(163
84*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*C^3*a^2*b^2 + 32768*A^3*C^2*a^2
*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A*C^4*a^5)/b) + (32768*A^3*C^2*a
^2*b^2*tan(c/2 + (d*x)/2))/(16384*C^5*a^4 + 32768*A*C^4*a^4 + 16384*A^4*C*b^4 - (16384*C^5*a^5)/b + 32768*A^2*
C^3*a^2*b^2 + 32768*A^3*C^2*a^2*b^2 - 16384*A^4*C*a*b^3 - 32768*A^2*C^3*a^3*b - 32768*A^3*C^2*a^3*b - (32768*A
*C^4*a^5)/b)))/(b*d) - (log(24576*A^2*C^3*a^4 - 8192*A^4*C*b^4 - 8192*A*C^4*a^4 + 24576*A^3*C^2*b^4 + 16384*A^
2*C^3*a^2*b^2 + 16384*A^3*C^2*a^2*b^2 + 8192*A*C^4*a^3*b + 8192*A^4*C*a*b^3 + 8192*A^2*C^3*a*b^3 - 49152*A^2*C
^3*a^3*b - 49152*A^3*C^2*a*b^3 + 8192*A^3*C^2*a^3*b + (((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8192*tan(c/2 +
(d*x)/2)*(a - b)*(A^4*b^4 + C^4*a^4 + 2*A^2*C^2*a^4 + 2*A^2*C^2*b^4 + 6*A^2*C^2*a^2*b^2 + 2*A*C^3*a^2*b^2 - 4
*A^2*C^2*a*b^3 - 4*A^2*C^2*a^3*b + 2*A^3*C*a^2*b^2) + (((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(24576*C^3*a^6
- 24576*A^3*b^6 + 49152*A^3*a*b^5 - 49152*C^3*a^5*b - 32768*A^3*a^2*b^4 + 8192*A^3*a^3*b^3 - 8192*C^3*a^3*b^3
+ 32768*C^3*a^4*b^2 + 8192*A*C^2*a^5*b - 8192*A^2*C*a*b^5 + 24576*A*C^2*a^2*b^4 - 65536*A*C^2*a^3*b^3 + 32768*
A*C^2*a^4*b^2 - 32768*A^2*C*a^2*b^4 + 65536*A^2*C*a^3*b^3 - 24576*A^2*C*a^4*b^2 + (((a + b)*(a - b))^(1/2)*(A*
b^2 + C*a^2)*(16384*tan(c/2 + (d*x)/2)*(a - b)^2*(A^2*b^5 + C^2*a^5 - A^2*a*b^4 - C^2*a^4*b + A^2*a^2*b^3 - A^
2*a^3*b^2 - C^2*a^2*b^3 + C^2*a^3*b^2 + A*C*a^2*b^3 + A*C*a^3*b^2) - (((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*
(24576*A*a^2*b^6 - 57344*A*a^3*b^5 + 40960*A*a^4*b^4 - 8192*A*a^5*b^3 - 8192*C*a^3*b^5 + 40960*C*a^4*b^4 - 573
44*C*a^5*b^3 + 24576*C*a^6*b^2 + (16384*a*b*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(a - b)
^3*(a^2 + b^2))/(a^2 - b^2)))/(a*b*(a^2 - b^2))))/(a*b*(a^2 - b^2))))/(a*b*(a^2 - b^2))))/(a*b*(a^2 - b^2)))*(
(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2))/(d*(a*b^3 - a^3*b)) - (log(a*tan(c/2 + (d*x)/2) - b*tan(c/2 + (d*x)/2)
+ (a^2 - b^2)^(1/2))*(A*b^2*(a^2 - b^2)^(1/2) + C*a^2*(a^2 - b^2)^(1/2)))/(a*b*d*(a^2 - b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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