3.1474 \(\int \frac{\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=204 \[ \frac{3 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

[Out]

(3*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) - (a*Sec[c + d*
x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - ((3*a^2 + 2*b^2)*Sec[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c
+ d*x])) + (Sec[c + d*x]*(3*a*(2*a^2 + 3*b^2) - b*(11*a^2 + 4*b^2)*Sin[c + d*x]))/(2*(a^2 - b^2)^3*d)

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Rubi [A]  time = 0.363555, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2864, 2866, 12, 2660, 618, 204} \[ \frac{3 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac{a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*Tan[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(3*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) - (a*Sec[c + d*
x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - ((3*a^2 + 2*b^2)*Sec[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c
+ d*x])) + (Sec[c + d*x]*(3*a*(2*a^2 + 3*b^2) - b*(11*a^2 + 4*b^2)*Sin[c + d*x]))/(2*(a^2 - b^2)^3*d)

Rule 2864

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

Rule 2866

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

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{\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) (2 b-3 a \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\int \frac{\sec ^2(c+d x) \left (-5 a b+2 \left (3 a^2+2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}-\frac{\int -\frac{3 a b \left (2 a^2+3 b^2\right )}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac{\left (3 a b \left (2 a^2+3 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac{\left (3 a b \left (2 a^2+3 b^2\right )\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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}-\frac{\left (6 a b \left (2 a^2+3 b^2\right )\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 )}{\left (a^2-b^2\right )^3 d}\\ &=\frac{3 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}-\frac{a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 2.97236, size = 206, normalized size = 1.01 \[ \frac{\frac{6 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{b^2 \cos (c+d x) \left (b \left (5 a^2+2 b^2\right ) \sin (c+d x)+a \left (6 a^2+b^2\right )\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}+\sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{2}{(a+b)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2}{(a-b)^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*Tan[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

((6*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]
*(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) +
(b^2*Cos[c + d*x]*(a*(6*a^2 + b^2) + b*(5*a^2 + 2*b^2)*Sin[c + d*x]))/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x
])^2))/(2*d)

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Maple [B]  time = 0.132, size = 643, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)+7/d*a^2/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^
2*tan(1/2*d*x+1/2*c)^3*b^3+6/d*a^3/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2
*d*x+1/2*c)^2*b^2+13/d*a/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c
)^2*b^4+2/d*b^6/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^2+17/
d*a^2/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)*b^3+4/d*b^5/(a-b)
^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+6/d*a^3/(a-b)^3/(a+b)^3/(tan
(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^2+1/d*b^4/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*
d*x+1/2*c)*b+a)^2*a+6/d*a^3/(a-b)^3/(a+b)^3*b/(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))+9/d*a/(a-b)^3/(a+b)^3*b^3/(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)^3/(tan(1/2*d*x+1/2*c)+1)

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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(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.12024, size = 1979, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*a^7 - 12*a^5*b^2 + 12*a^3*b^4 - 4*a*b^6 + 2*(16*a^5*b^2 - 17*a^3*b^4 + a*b^6)*cos(d*x + c)^2 - 3*((2*
a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)*sin(d*x + c) - (2*a^5*b + 5*a^3*b^3
+ 3*a*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*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)) - 2*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2)
*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 +
6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^
2*b^8 + b^10)*d*cos(d*x + c)), -1/2*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (16*a^5*b^2 - 17*a^3*b^4 + a*b^
6)*cos(d*x + c)^2 + 3*((2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)*sin(d*x +
c) - (2*a^5*b + 5*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b
^2)*cos(d*x + c))) - (2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^
2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3
+ 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*
a^2*b^8 + b^10)*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(sin(c + d*x)*sec(c + d*x)**2/(a + b*sin(c + d*x))**3, x)

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Giac [A]  time = 1.25063, size = 493, normalized size = 2.42 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )}{\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 )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}} + \frac{7 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 13 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 17 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{4} b^{2} + a^{2} b^{4}}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(3*(2*a^3*b + 3*a*b^3)*(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)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/
2*d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 - 1)) + (7*a^3*b^
3*tan(1/2*d*x + 1/2*c)^3 + 6*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 + 13*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/
2*d*x + 1/2*c)^2 + 17*a^3*b^3*tan(1/2*d*x + 1/2*c) + 4*a*b^5*tan(1/2*d*x + 1/2*c) + 6*a^4*b^2 + a^2*b^4)/((a^7
- 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2))/d