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

Optimal. Leaf size=86 \[ \frac {2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C x}{b^2}+\frac {C \sin (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3024, 2735, 2659, 205} \[ \frac {2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C x}{b^2}+\frac {C \sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3024

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

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 82, normalized size = 0.95 \[ \frac {-\frac {2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-a C (c+d x)+b C \sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.39, size = 297, normalized size = 3.45 \[ \left [-\frac {2 \, {\left (C a^{3} - C a b^{2}\right )} d x + {\left (C a^{2} + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {{\left (C a^{3} - C a b^{2}\right )} d x - {\left (C a^{2} + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(C*a^3 - C*a*b^2)*d*x + (C*a^2 + A*b^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(
d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*c
os(d*x + c) + a^2)) - 2*(C*a^2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d), -((C*a^3 - C*a*b^2)*d*x - (C*a^2
+ A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (C*a^2*b - C*b^3)*sin(
d*x + c))/((a^2*b^2 - b^4)*d)]

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giac [A]  time = 0.36, size = 136, normalized size = 1.58 \[ -\frac {\frac {{\left (d x + c\right )} C a}{b^{2}} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {2 \, {\left (C a^{2} + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 149, normalized size = 1.73 \[ \frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{2} C}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+2/d/b^2/((a-b)*(a+b))^(1/2)*arc
tan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a^2*C+2/d*C/b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-2/
d/b^2*C*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+C*cos(d*x+c)^2)/(a+b*cos(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 3.55, size = 916, normalized size = 10.65 \[ \frac {C\,\sin \left (c+d\,x\right )}{b\,d}-\frac {2\,C\,a\,\mathrm {atan}\left (\frac {64\,C^3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,C^3\,a^3+128\,A\,C^2\,a^3-\frac {64\,C^3\,a^4}{b}+64\,A^2\,C\,a\,b^2-64\,A^2\,C\,a^2\,b-\frac {128\,A\,C^2\,a^4}{b}}+\frac {64\,C^3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2\,b^2-64\,A^2\,C\,a\,b^3+128\,A\,C^2\,a^4-128\,A\,C^2\,a^3\,b+64\,C^3\,a^4-64\,C^3\,a^3\,b}+\frac {128\,A\,C^2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2\,b^2-64\,A^2\,C\,a\,b^3+128\,A\,C^2\,a^4-128\,A\,C^2\,a^3\,b+64\,C^3\,a^4-64\,C^3\,a^3\,b}+\frac {64\,A^2\,C\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2-\frac {64\,C^3\,a^3}{b}+\frac {64\,C^3\,a^4}{b^2}-64\,A^2\,C\,a\,b-\frac {128\,A\,C^2\,a^3}{b}+\frac {128\,A\,C^2\,a^4}{b^2}}+\frac {128\,A\,C^2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,C^3\,a^3+128\,A\,C^2\,a^3-\frac {64\,C^3\,a^4}{b}+64\,A^2\,C\,a\,b^2-64\,A^2\,C\,a^2\,b-\frac {128\,A\,C^2\,a^4}{b}}-\frac {64\,A^2\,C\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,C\,a^2-\frac {64\,C^3\,a^3}{b}+\frac {64\,C^3\,a^4}{b^2}-64\,A^2\,C\,a\,b-\frac {128\,A\,C^2\,a^3}{b}+\frac {128\,A\,C^2\,a^4}{b^2}}\right )}{b^2\,d}-\frac {\ln \left (\frac {\left (C\,a^2+A\,b^2\right )\,\sqrt {b^2-a^2}\,\left (\frac {32\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a-b\right )\,\left (A^2\,b^4+2\,A\,C\,a^2\,b^2+2\,C^2\,a^4-2\,C^2\,a^3\,b+C^2\,a^2\,b^2\right )}{b^2}-\frac {32\,\left (C\,a^2+A\,b^2\right )\,\sqrt {b^2-a^2}\,\left (a-b\right )\,\left (A\,b^4-A\,a^2\,b^2-C\,a\,b^3+C\,a^3\,b+2\,C\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,A\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\right )}{\left (b^4-a^2\,b^2\right )\,\left (a+b\right )}\right )}{b^4-a^2\,b^2}-\frac {32\,C\,a\,\left (a-b\right )\,\left (A^2\,b^3+A\,C\,a^2\,b+A\,C\,a\,b^2+C^2\,a^3\right )}{b^3}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (C\,a^2+A\,b^2\right )}{d\,\left (b^4-a^2\,b^2\right )}-\frac {\ln \left (b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (A\,b^2\,\sqrt {b^2-a^2}+C\,a^2\,\sqrt {b^2-a^2}\right )}{b^2\,d\,\left (a^2-b^2\right )} \]

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]

(C*sin(c + d*x))/(b*d) - (2*C*a*atan((64*C^3*a^3*tan(c/2 + (d*x)/2))/(64*C^3*a^3 + 128*A*C^2*a^3 - (64*C^3*a^4
)/b + 64*A^2*C*a*b^2 - 64*A^2*C*a^2*b - (128*A*C^2*a^4)/b) + (64*C^3*a^4*tan(c/2 + (d*x)/2))/(64*C^3*a^4 + 128
*A*C^2*a^4 - 64*C^3*a^3*b - 128*A*C^2*a^3*b - 64*A^2*C*a*b^3 + 64*A^2*C*a^2*b^2) + (128*A*C^2*a^4*tan(c/2 + (d
*x)/2))/(64*C^3*a^4 + 128*A*C^2*a^4 - 64*C^3*a^3*b - 128*A*C^2*a^3*b - 64*A^2*C*a*b^3 + 64*A^2*C*a^2*b^2) + (6
4*A^2*C*a^2*tan(c/2 + (d*x)/2))/(64*A^2*C*a^2 - (64*C^3*a^3)/b + (64*C^3*a^4)/b^2 - 64*A^2*C*a*b - (128*A*C^2*
a^3)/b + (128*A*C^2*a^4)/b^2) + (128*A*C^2*a^3*tan(c/2 + (d*x)/2))/(64*C^3*a^3 + 128*A*C^2*a^3 - (64*C^3*a^4)/
b + 64*A^2*C*a*b^2 - 64*A^2*C*a^2*b - (128*A*C^2*a^4)/b) - (64*A^2*C*a*b*tan(c/2 + (d*x)/2))/(64*A^2*C*a^2 - (
64*C^3*a^3)/b + (64*C^3*a^4)/b^2 - 64*A^2*C*a*b - (128*A*C^2*a^3)/b + (128*A*C^2*a^4)/b^2)))/(b^2*d) - (log(((
A*b^2 + C*a^2)*(b^2 - a^2)^(1/2)*((32*tan(c/2 + (d*x)/2)*(a - b)*(A^2*b^4 + 2*C^2*a^4 - 2*C^2*a^3*b + C^2*a^2*
b^2 + 2*A*C*a^2*b^2))/b^2 - (32*(A*b^2 + C*a^2)*(b^2 - a^2)^(1/2)*(a - b)*(A*b^4 - A*a^2*b^2 - C*a*b^3 + C*a^3
*b + 2*C*a^3*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A*a*b^2*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))/((b^4 - a
^2*b^2)*(a + b))))/(b^4 - a^2*b^2) - (32*C*a*(a - b)*(A^2*b^3 + C^2*a^3 + A*C*a*b^2 + A*C*a^2*b))/b^3)*(-(a +
b)*(a - b))^(1/2)*(A*b^2 + C*a^2))/(d*(b^4 - a^2*b^2)) - (log(b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2) + (b
^2 - a^2)^(1/2))*(A*b^2*(b^2 - a^2)^(1/2) + C*a^2*(b^2 - a^2)^(1/2)))/(b^2*d*(a^2 - b^2))

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sympy [A]  time = 129.16, size = 2518, normalized size = 29.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(A + C*cos(c)**2)/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (A*tan(c/2 + d*x/2)**2/(b*d*tan(c/
2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + A/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + C*d*x*tan(c/2 + d
*x/2)**3/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + C*d*x*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**3 +
b*d*tan(c/2 + d*x/2)) + 3*C*tan(c/2 + d*x/2)**2/(b*d*tan(c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)) + C/(b*d*tan(
c/2 + d*x/2)**3 + b*d*tan(c/2 + d*x/2)), Eq(a, -b)), ((A*x + C*x*sin(c + d*x)**2/2 + C*x*cos(c + d*x)**2/2 + C
*sin(c + d*x)*cos(c + d*x)/(2*d))/a, Eq(b, 0)), (x*(A + C*cos(c)**2)/(a + b*cos(c)), Eq(d, 0)), (A*tan(c/2 + d
*x/2)**3/(b*d*tan(c/2 + d*x/2)**2 + b*d) + A*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**2 + b*d) - C*d*x*tan(c/2
+ d*x/2)**2/(b*d*tan(c/2 + d*x/2)**2 + b*d) - C*d*x/(b*d*tan(c/2 + d*x/2)**2 + b*d) + C*tan(c/2 + d*x/2)**3/(b
*d*tan(c/2 + d*x/2)**2 + b*d) + 3*C*tan(c/2 + d*x/2)/(b*d*tan(c/2 + d*x/2)**2 + b*d), Eq(a, b)), (A*b**2*log(-
sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*ta
n(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/
2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + A*b**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a
*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt
(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - A*b**2*log(sqrt(-a/(a -
b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2
)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*
d*sqrt(-a/(a - b) - b/(a - b))) - A*b**2*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b**2*d*sqrt(-
a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) -
b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - C*a**2*d*x*sqrt(-a/(a - b) - b/(a - b)
)*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) -
b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) -
C*a**2*d*x*sqrt(-a/(a - b) - b/(a - b))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*
sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b
) - b/(a - b))) + C*a**2*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*s
qrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a -
b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + C*a**2*log(-sqrt(-a/(a - b) - b/(
a - b)) + tan(c/2 + d*x/2))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a -
b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b
))) - C*a**2*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b
) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b
))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - C*a**2*log(sqrt(-a/(a - b) - b/(a - b)) + tan(
c/2 + d*x/2))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b
)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + C*a*b*d*
x*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2
+ a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqr
t(-a/(a - b) - b/(a - b))) + C*a*b*d*x*sqrt(-a/(a - b) - b/(a - b))/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan
(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2
)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) + 2*C*a*b*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)/(a*b**2*d*
sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) - b/(a - b)) - b**3*d*sqrt(-a/(a -
b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))) - 2*C*b**2*sqrt(-a/(a - b) - b/(a
- b))*tan(c/2 + d*x/2)/(a*b**2*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 + a*b**2*d*sqrt(-a/(a - b) -
b/(a - b)) - b**3*d*sqrt(-a/(a - b) - b/(a - b))*tan(c/2 + d*x/2)**2 - b**3*d*sqrt(-a/(a - b) - b/(a - b))),
True))

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