3.206 \(\int \frac {a+b \sin ^2(x)}{c+d \cos (x)} \, dx\)

Optimal. Leaf size=105 \[ \frac {2 a \tan ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} \sqrt {c+d}}+\frac {b c x}{d^2}-\frac {2 b \sqrt {c-d} \sqrt {c+d} \tan ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{d^2}-\frac {b \sin (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {4401, 2659, 205, 2695, 2735} \[ \frac {2 a \tan ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} \sqrt {c+d}}+\frac {b c x}{d^2}-\frac {2 b \sqrt {c-d} \sqrt {c+d} \tan ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {x}{2}\right )}{\sqrt {c+d}}\right )}{d^2}-\frac {b \sin (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*x)/d^2 + (2*a*ArcTan[(Sqrt[c - d]*Tan[x/2])/Sqrt[c + d]])/(Sqrt[c - d]*Sqrt[c + d]) - (2*b*Sqrt[c - d]*Sq
rt[c + d]*ArcTan[(Sqrt[c - d]*Tan[x/2])/Sqrt[c + d]])/d^2 - (b*Sin[x])/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 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

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 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 73, normalized size = 0.70 \[ \frac {-\frac {2 \left (a d^2+b \left (d^2-c^2\right )\right ) \tanh ^{-1}\left (\frac {(c-d) \tan \left (\frac {x}{2}\right )}{\sqrt {d^2-c^2}}\right )}{\sqrt {d^2-c^2}}+b c x-b d \sin (x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 254, normalized size = 2.42 \[ \left [\frac {{\left (b c^{2} - {\left (a + b\right )} d^{2}\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {2 \, c d \cos \relax (x) + {\left (2 \, c^{2} - d^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-c^{2} + d^{2}} {\left (c \cos \relax (x) + d\right )} \sin \relax (x) - c^{2} + 2 \, d^{2}}{d^{2} \cos \relax (x)^{2} + 2 \, c d \cos \relax (x) + c^{2}}\right ) + 2 \, {\left (b c^{3} - b c d^{2}\right )} x - 2 \, {\left (b c^{2} d - b d^{3}\right )} \sin \relax (x)}{2 \, {\left (c^{2} d^{2} - d^{4}\right )}}, -\frac {{\left (b c^{2} - {\left (a + b\right )} d^{2}\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \cos \relax (x) + d}{\sqrt {c^{2} - d^{2}} \sin \relax (x)}\right ) - {\left (b c^{3} - b c d^{2}\right )} x + {\left (b c^{2} d - b d^{3}\right )} \sin \relax (x)}{c^{2} d^{2} - d^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 110, normalized size = 1.05 \[ \frac {b c x}{d^{2}} - \frac {2 \, b \tan \left (\frac {1}{2} \, x\right )}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} d} + \frac {2 \, {\left (b c^{2} - a d^{2} - b d^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac {c \tan \left (\frac {1}{2} \, x\right ) - d \tan \left (\frac {1}{2} \, x\right )}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c*x/d^2 - 2*b*tan(1/2*x)/((tan(1/2*x)^2 + 1)*d) + 2*(b*c^2 - a*d^2 - b*d^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2
*c + 2*d) + arctan(-(c*tan(1/2*x) - d*tan(1/2*x))/sqrt(c^2 - d^2)))/(sqrt(c^2 - d^2)*d^2)

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maple [A]  time = 0.11, size = 148, normalized size = 1.41 \[ \frac {2 \arctan \left (\frac {\left (c -d \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) a}{\sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {2 \arctan \left (\frac {\left (c -d \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) c^{2} b}{d^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 \arctan \left (\frac {\left (c -d \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) b}{\sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {2 b \tan \left (\frac {x}{2}\right )}{d \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}+\frac {2 b c \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((c+d)*(c-d))^(1/2)*arctan((c-d)*tan(1/2*x)/((c+d)*(c-d))^(1/2))*a-2/d^2/((c+d)*(c-d))^(1/2)*arctan((c-d)*ta
n(1/2*x)/((c+d)*(c-d))^(1/2))*c^2*b+2/((c+d)*(c-d))^(1/2)*arctan((c-d)*tan(1/2*x)/((c+d)*(c-d))^(1/2))*b-2*b/d
*tan(1/2*x)/(1+tan(1/2*x)^2)+2*b/d^2*c*arctan(tan(1/2*x))

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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*sin(x)^2)/(c+d*cos(x)),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*d^2-4*c^2>0)', see `assume?`
for more details)Is 4*d^2-4*c^2 positive or negative?

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mupad [B]  time = 4.04, size = 2429, normalized size = 23.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^2*d*sin(x))/(d^4 - c^2*d^2) - (b*d^3*sin(x))/(d^4 - c^2*d^2) - (2*b*c^3*atan(sin(x/2)/cos(x/2)))/(d^4 - c
^2*d^2) - (a*d^2*atan((a^2*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c^5*sin(x/2)*(d^2 - c^2)^(3/2)*2i - b^2*c^7
*sin(x/2)*(d^2 - c^2)^(1/2)*2i + b^2*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*
2i + a^2*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*2i + b^2*c*d^6*sin(x/2)*(d
^2 - c^2)^(1/2)*1i - a^2*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b
^2*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*2i + b^2*c^3*d^2*sin(x/2)*(d^2 - c^2)^(3/2)*4i - b^2*c^3*d^4*sin(x/2)*(d
^2 - c^2)^(1/2)*4i + b^2*c^4*d^3*sin(x/2)*(d^2 - c^2)^(1/2)*1i + b^2*c^5*d^2*sin(x/2)*(d^2 - c^2)^(1/2)*5i + a
*b*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*2i - a*b*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*4i + a*b*c*d^6*sin(x/2)*(d^2 - c^2
)^(1/2)*2i - a*b*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*4i + a*b*c^3*d^2*sin(x/2)*(d^2 - c^2)^(3/2)*4i - a*b*c^3*d
^4*sin(x/2)*(d^2 - c^2)^(1/2)*4i + a*b*c^4*d^3*sin(x/2)*(d^2 - c^2)^(1/2)*2i + a*b*c^5*d^2*sin(x/2)*(d^2 - c^2
)^(1/2)*2i)/(a^2*d^8*cos(x/2) + b^2*d^8*cos(x/2) + 2*a*b*d^8*cos(x/2) - 2*a^2*c^2*d^6*cos(x/2) + a^2*c^4*d^4*c
os(x/2) - 3*b^2*c^2*d^6*cos(x/2) + 3*b^2*c^4*d^4*cos(x/2) - b^2*c^6*d^2*cos(x/2) - 6*a*b*c^2*d^6*cos(x/2) + 6*
a*b*c^4*d^4*cos(x/2) - 2*a*b*c^6*d^2*cos(x/2)))*(d^2 - c^2)^(1/2)*2i)/(d^4 - c^2*d^2) + (b*c^2*atan((a^2*d^7*s
in(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c^5*sin(x/2)*(d^2 - c^2)^(3/2)*2i - b^2*c^7*sin(x/2)*(d^2 - c^2)^(1/2)*2i +
b^2*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*2i + a^2*c*d^6*sin(x/2)*(d^2 - c
^2)^(1/2)*1i - b^2*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*2i + b^2*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c^2*d^5
*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c^2*d^5*sin(x/2)*(d^2 - c^2)^
(1/2)*2i + b^2*c^3*d^2*sin(x/2)*(d^2 - c^2)^(3/2)*4i - b^2*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*4i + b^2*c^4*d^3
*sin(x/2)*(d^2 - c^2)^(1/2)*1i + b^2*c^5*d^2*sin(x/2)*(d^2 - c^2)^(1/2)*5i + a*b*d^7*sin(x/2)*(d^2 - c^2)^(1/2
)*2i - a*b*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*4i + a*b*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*2i - a*b*c^2*d^5*sin(x/2
)*(d^2 - c^2)^(1/2)*4i + a*b*c^3*d^2*sin(x/2)*(d^2 - c^2)^(3/2)*4i - a*b*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*4i
+ a*b*c^4*d^3*sin(x/2)*(d^2 - c^2)^(1/2)*2i + a*b*c^5*d^2*sin(x/2)*(d^2 - c^2)^(1/2)*2i)/(a^2*d^8*cos(x/2) +
b^2*d^8*cos(x/2) + 2*a*b*d^8*cos(x/2) - 2*a^2*c^2*d^6*cos(x/2) + a^2*c^4*d^4*cos(x/2) - 3*b^2*c^2*d^6*cos(x/2)
+ 3*b^2*c^4*d^4*cos(x/2) - b^2*c^6*d^2*cos(x/2) - 6*a*b*c^2*d^6*cos(x/2) + 6*a*b*c^4*d^4*cos(x/2) - 2*a*b*c^6
*d^2*cos(x/2)))*(d^2 - c^2)^(1/2)*2i)/(d^4 - c^2*d^2) - (b*d^2*atan((a^2*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b
^2*c^5*sin(x/2)*(d^2 - c^2)^(3/2)*2i - b^2*c^7*sin(x/2)*(d^2 - c^2)^(1/2)*2i + b^2*d^7*sin(x/2)*(d^2 - c^2)^(1
/2)*1i - a^2*c*d^4*sin(x/2)*(d^2 - c^2)^(3/2)*2i + a^2*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c*d^4*sin(x/2
)*(d^2 - c^2)^(3/2)*2i + b^2*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*1i - a^2*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*1i -
a^2*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*1i - b^2*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*2i + b^2*c^3*d^2*sin(x/2)*
(d^2 - c^2)^(3/2)*4i - b^2*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*4i + b^2*c^4*d^3*sin(x/2)*(d^2 - c^2)^(1/2)*1i +
b^2*c^5*d^2*sin(x/2)*(d^2 - c^2)^(1/2)*5i + a*b*d^7*sin(x/2)*(d^2 - c^2)^(1/2)*2i - a*b*c*d^4*sin(x/2)*(d^2 -
c^2)^(3/2)*4i + a*b*c*d^6*sin(x/2)*(d^2 - c^2)^(1/2)*2i - a*b*c^2*d^5*sin(x/2)*(d^2 - c^2)^(1/2)*4i + a*b*c^3
*d^2*sin(x/2)*(d^2 - c^2)^(3/2)*4i - a*b*c^3*d^4*sin(x/2)*(d^2 - c^2)^(1/2)*4i + a*b*c^4*d^3*sin(x/2)*(d^2 - c
^2)^(1/2)*2i + a*b*c^5*d^2*sin(x/2)*(d^2 - c^2)^(1/2)*2i)/(a^2*d^8*cos(x/2) + b^2*d^8*cos(x/2) + 2*a*b*d^8*cos
(x/2) - 2*a^2*c^2*d^6*cos(x/2) + a^2*c^4*d^4*cos(x/2) - 3*b^2*c^2*d^6*cos(x/2) + 3*b^2*c^4*d^4*cos(x/2) - b^2*
c^6*d^2*cos(x/2) - 6*a*b*c^2*d^6*cos(x/2) + 6*a*b*c^4*d^4*cos(x/2) - 2*a*b*c^6*d^2*cos(x/2)))*(d^2 - c^2)^(1/2
)*2i)/(d^4 - c^2*d^2) + (2*b*c*d^2*atan(sin(x/2)/cos(x/2)))/(d^4 - c^2*d^2)

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sympy [A]  time = 111.37, size = 2608, normalized size = 24.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-a*log(tan(x/2) - 1)*tan(x/2)**2/(tan(x/2)**2 + 1) - a*log(tan(x/2) - 1)/(tan(x/2)**2 + 1) + a
*log(tan(x/2) + 1)*tan(x/2)**2/(tan(x/2)**2 + 1) + a*log(tan(x/2) + 1)/(tan(x/2)**2 + 1) - b*log(tan(x/2) - 1)
*tan(x/2)**2/(tan(x/2)**2 + 1) - b*log(tan(x/2) - 1)/(tan(x/2)**2 + 1) + b*log(tan(x/2) + 1)*tan(x/2)**2/(tan(
x/2)**2 + 1) + b*log(tan(x/2) + 1)/(tan(x/2)**2 + 1) - 2*b*tan(x/2)/(tan(x/2)**2 + 1)), Eq(c, 0) & Eq(d, 0)),
(a*tan(x/2)**3/(d*tan(x/2)**2 + d) + a*tan(x/2)/(d*tan(x/2)**2 + d) + b*x*tan(x/2)**2/(d*tan(x/2)**2 + d) + b*
x/(d*tan(x/2)**2 + d) - 2*b*tan(x/2)/(d*tan(x/2)**2 + d), Eq(c, d)), (a*tan(x/2)**2/(d*tan(x/2)**3 + d*tan(x/2
)) + a/(d*tan(x/2)**3 + d*tan(x/2)) - b*x*tan(x/2)**3/(d*tan(x/2)**3 + d*tan(x/2)) - b*x*tan(x/2)/(d*tan(x/2)*
*3 + d*tan(x/2)) - 2*b*tan(x/2)**2/(d*tan(x/2)**3 + d*tan(x/2)), Eq(c, -d)), ((a*x + b*x*sin(x)**2/2 + b*x*cos
(x)**2/2 - b*sin(x)*cos(x)/2)/c, Eq(d, 0)), (a*d**2*log(-sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))*tan(x/2)**2/
(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d)
- d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) + a*d**2*log(-sqrt(-c/(c - d) - d/(c - d)) + tan
(x/2))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(
c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) - a*d**2*log(sqrt(-c/(c - d) - d/(c - d))
+ tan(x/2))*tan(x/2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)
) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) - a*d**2*log(sqrt(-c/(c
- d) - d/(c - d)) + tan(x/2))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(
c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) + b*c**2*x*sqrt(-
c/(c - d) - d/(c - d))*tan(x/2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) -
d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) + b*c**2*x*sq
rt(-c/(c - d) - d/(c - d))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c -
d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) - b*c**2*log(-sqrt(-c
/(c - d) - d/(c - d)) + tan(x/2))*tan(x/2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-
c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) -
b*c**2*log(-sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2
*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c -
d))) + b*c**2*log(sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))*tan(x/2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*ta
n(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c
/(c - d) - d/(c - d))) + b*c**2*log(sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))/(c*d**2*sqrt(-c/(c - d) - d/(c -
d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*s
qrt(-c/(c - d) - d/(c - d))) - b*c*d*x*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c
- d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**
3*sqrt(-c/(c - d) - d/(c - d))) - b*c*d*x*sqrt(-c/(c - d) - d/(c - d))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*ta
n(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c
/(c - d) - d/(c - d))) - 2*b*c*d*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*ta
n(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c
/(c - d) - d/(c - d))) + 2*b*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*t
an(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-
c/(c - d) - d/(c - d))) + b*d**2*log(-sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))*tan(x/2)**2/(c*d**2*sqrt(-c/(c
- d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))*tan(x/
2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) + b*d**2*log(-sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))/(c*d**2*sqrt
(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c - d) - d/(c - d))
*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) - b*d**2*log(sqrt(-c/(c - d) - d/(c - d)) + tan(x/2))*tan(x/
2)**2/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt(-c/(c
- d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))) - b*d**2*log(sqrt(-c/(c - d) - d/(c - d))
+ tan(x/2))/(c*d**2*sqrt(-c/(c - d) - d/(c - d))*tan(x/2)**2 + c*d**2*sqrt(-c/(c - d) - d/(c - d)) - d**3*sqrt
(-c/(c - d) - d/(c - d))*tan(x/2)**2 - d**3*sqrt(-c/(c - d) - d/(c - d))), True))

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