3.1344 \(\int \frac{\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=128 \[ -\frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}+\frac{a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.288354, antiderivative size = 150, normalized size of antiderivative = 1.17, number of steps used = 13, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2898, 2622, 321, 207, 2620, 14, 2696, 12, 2660, 618, 204} \[ -\frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac{b \sec (c+d x)}{a^2 d}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \left (-\frac{b \csc (c+d x) \sec ^2(c+d x)}{a^2}+\frac{\csc ^2(c+d x) \sec ^2(c+d x)}{a}+\frac{b^2 \sec ^2(c+d x)}{a^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \csc ^2(c+d x) \sec ^2(c+d x) \, dx}{a}-\frac{b \int \csc (c+d x) \sec ^2(c+d x) \, dx}{a^2}+\frac{b^2 \int \frac{\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2}\\ &=-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}-\frac{b^2 \int \frac{b^2}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{a d}-\frac{b \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac{b \sec (c+d x)}{a^2 d}-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}-\frac{b^4 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{a d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{b \sec (c+d x)}{a^2 d}-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}+\frac{\tan (c+d x)}{a d}-\frac{\left (2 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{b \sec (c+d x)}{a^2 d}-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}+\frac{\tan (c+d x)}{a d}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=-\frac{2 b^4 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{b \sec (c+d x)}{a^2 d}-\frac{b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}+\frac{\tan (c+d x)}{a d}\\ \end{align*}

Mathematica [A]  time = 0.933606, size = 205, normalized size = 1.6 \[ \frac{-\frac{4 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac{2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}+\frac{2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{a}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.107, size = 169, normalized size = 1.3 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{b}^{4}}{d{a}^{2} \left ( a-b \right ) \left ( a+b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.92457, size = 1370, normalized size = 10.7 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right ) \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \,{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}, \frac{2 \, \sqrt{a^{2} - b^{2}} b^{4} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \,{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a^2 + b^2)*b^4*cos(d*x + c)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2
*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a
^2 - b^2))*sin(d*x + c) + 2*a^5 - 2*a^3*b^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/
2)*sin(d*x + c) - (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*a^5
- 3*a^3*b^2 + a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*d*cos(d*x
+ c)*sin(d*x + c)), 1/2*(2*sqrt(a^2 - b^2)*b^4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*c
os(d*x + c)*sin(d*x + c) + 2*a^5 - 2*a^3*b^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)*log(1/2*cos(d*x + c) + 1
/2)*sin(d*x + c) - (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*a^5
- 3*a^3*b^2 + a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*d*cos(d*
x + c)*sin(d*x + c))]

________________________________________________________________________________________

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(csc(d*x+c)**2*sec(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.23745, size = 350, normalized size = 2.73 \begin{align*} -\frac{\frac{12 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{6 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{3} - 3 \, a b^{2}}{{\left (a^{4} - a^{2} b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*b^4/
((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) + 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - (2*a^2
*b*tan(1/2*d*x + 1/2*c)^3 - 2*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x
+ 1/2*c)^2 + 10*a^2*b*tan(1/2*d*x + 1/2*c) + 2*b^3*tan(1/2*d*x + 1/2*c) + 3*a^3 - 3*a*b^2)/((a^4 - a^2*b^2)*(
tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))))/d