3.603 \(\int \frac{\cos ^2(c+d x) (1-\cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=154 \[ \frac{2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (6 a^2-b^2\right )}{2 b^4}+\frac{3 a \sin (c+d x)}{b^3 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{b d (a+b \cos (c+d x))}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

[Out]

-((6*a^2 - b^2)*x)/(2*b^4) + (2*a*(3*a^2 - 2*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a
- b]*b^4*Sqrt[a + b]*d) + (3*a*Sin[c + d*x])/(b^3*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*d) + (Cos[c + d*x]
^2*Sin[c + d*x])/(b*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.397235, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3050, 3023, 2735, 2659, 205} \[ \frac{2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (6 a^2-b^2\right )}{2 b^4}+\frac{3 a \sin (c+d x)}{b^3 d}+\frac{\sin (c+d x) \cos ^2(c+d x)}{b d (a+b \cos (c+d x))}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]

[Out]

-((6*a^2 - b^2)*x)/(2*b^4) + (2*a*(3*a^2 - 2*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a
- b]*b^4*Sqrt[a + b]*d) + (3*a*Sin[c + d*x])/(b^3*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*d) + (Cos[c + d*x]
^2*Sin[c + d*x])/(b*d*(a + b*Cos[c + d*x]))

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 3023

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 \left (a^2-b^2\right )+3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{3 a \left (a^2-b^2\right )-b \left (a^2-b^2\right ) \cos (c+d x)-6 a \left (a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{3 a \sin (c+d x)}{b^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}-\frac{\int \frac{3 a b \left (a^2-b^2\right )+\left (a^2-b^2\right ) \left (6 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}+\frac{3 a \sin (c+d x)}{b^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}+\frac{\left (a \left (3 a^2-2 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}+\frac{3 a \sin (c+d x)}{b^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}+\frac{\left (2 a \left (3 a^2-2 b^2\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^4 d}\\ &=-\frac{\left (6 a^2-b^2\right ) x}{2 b^4}+\frac{2 a \left (3 a^2-2 b^2\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^4 \sqrt{a+b} d}+\frac{3 a \sin (c+d x)}{b^3 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{b d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.29238, size = 131, normalized size = 0.85 \[ \frac{2 \left (b^2-6 a^2\right ) (c+d x)-\frac{8 a \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+\frac{4 a^2 b \sin (c+d x)}{a+b \cos (c+d x)}+8 a b \sin (c+d x)-b^2 \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(1 - Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^2,x]

[Out]

(2*(-6*a^2 + b^2)*(c + d*x) - (8*a*(3*a^2 - 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[
-a^2 + b^2] + 8*a*b*Sin[c + d*x] + (4*a^2*b*Sin[c + d*x])/(a + b*Cos[c + d*x]) - b^2*Sin[2*(c + d*x)])/(4*b^4*
d)

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Maple [B]  time = 0.034, size = 321, normalized size = 2.1 \begin{align*} 4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}a}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{1}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+4\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) a}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{1}{d{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{4}}}+{\frac{1}{d{b}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{3} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+6\,{\frac{{a}^{3}}{d{b}^{4}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{a}{d{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x)

[Out]

4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*a+1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c
)^3+4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)*a-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2
*c)-6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^2+1/d/b^2*arctan(tan(1/2*d*x+1/2*c))+2/d*a^2/b^3*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)+6/d*a^3/b^4/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/
2*c)/((a+b)*(a-b))^(1/2))-4/d*a/b^2/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62666, size = 1191, normalized size = 7.73 \begin{align*} \left [-\frac{{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x \cos \left (d x + c\right ) +{\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x -{\left (3 \, a^{4} - 2 \, a^{2} b^{2} +{\left (3 \, a^{3} b - 2 \, 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 (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 ) -{\left (6 \, a^{4} b - 6 \, a^{2} b^{3} -{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) +{\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}, -\frac{{\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x \cos \left (d x + c\right ) +{\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x - 2 \,{\left (3 \, a^{4} - 2 \, a^{2} b^{2} +{\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\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 (6 \, a^{4} b - 6 \, a^{2} b^{3} -{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) +{\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*((6*a^4*b - 7*a^2*b^3 + b^5)*d*x*cos(d*x + c) + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - (3*a^4 - 2*a^2*b^2 + (
3*a^3*b - 2*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*s
qrt(-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*cos(d*x + c) + a^
2)) - (6*a^4*b - 6*a^2*b^3 - (a^2*b^3 - b^5)*cos(d*x + c)^2 + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/
((a^2*b^5 - b^7)*d*cos(d*x + c) + (a^3*b^4 - a*b^6)*d), -1/2*((6*a^4*b - 7*a^2*b^3 + b^5)*d*x*cos(d*x + c) + (
6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - 2*(3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arct
an(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*a^4*b - 6*a^2*b^3 - (a^2*b^3 - b^5)*cos(d*x + c)
^2 + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*cos(d*x + c) + (a^3*b^4 - a*b^6)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(1-cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.67666, size = 321, normalized size = 2.08 \begin{align*} \frac{\frac{4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} b^{3}} - \frac{{\left (6 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{4 \,{\left (3 \, a^{3} - 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^{4}} + \frac{2 \,{\left (4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, 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 )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(1-cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(4*a^2*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*b^3) - (6*a^2 -
b^2)*(d*x + c)/b^4 - 4*(3*a^3 - 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^4) + 2*(4*a*tan(1/2*d*x + 1/2*c
)^3 + b*tan(1/2*d*x + 1/2*c)^3 + 4*a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 +
1)^2*b^3))/d