\(\int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [809]

Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {10 a^2 \cot (c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

[Out]

-2*a^2*arctanh(cos(d*x+c))/d-10/3*a^2*cot(d*x+c)/d+2*a^2*cot(d*x+c)/d/(1-sin(d*x+c))+1/3*a^4*cot(d*x+c)/d/(a-a
*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2948, 2845, 3057, 2827, 3852, 8, 3855} \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {10 a^2 \cot (c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))} \]

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(-2*a^2*ArcTanh[Cos[c + d*x]])/d - (10*a^2*Cot[c + d*x])/(3*d) + (2*a^2*Cot[c + d*x])/(d*(1 - Sin[c + d*x])) +
(a^4*Cot[c + d*x])/(3*d*(a - a*Sin[c + d*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2845

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

Rule 2948

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rule 3057

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = a^4 \int \frac {\csc ^2(c+d x)}{(a-a \sin (c+d x))^2} \, dx \\ & = \frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\csc ^2(c+d x) (4 a+2 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx \\ & = \frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} \int \csc ^2(c+d x) \left (10 a^2+6 a^2 \sin (c+d x)\right ) \, dx \\ & = \frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\frac {1}{3} \left (10 a^2\right ) \int \csc ^2(c+d x) \, dx \\ & = -\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\left (10 a^2\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d} \\ & = -\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {10 a^2 \cot (c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.55 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-3 \cot \left (\frac {1}{2} (c+d x)\right )-12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) (-8+7 \sin (c+d x))}{\left (-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(-3*Cot[(c + d*x)/2] - 12*Log[Cos[(c + d*x)/2]] + 12*Log[Sin[(c + d*x)/2]] + 2/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2 + (4*Sin[(c + d*x)/2]*(-8 + 7*Sin[c + d*x]))/(-Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 3*Tan[(c
+ d*x)/2]))/(6*d)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\left (4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-19 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )+31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {50}{3}\right ) a^{2}}{2 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(93\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\)	\(118\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\)	\(118\)
risch	\(\frac {-\frac {44 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-12 i {\mathrm e}^{3 i \left (d x +c \right )} a^{2}+\frac {20 a^{2}}{3}+16 i a^{2} {\mathrm e}^{i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(138\)
norman	\(\frac {\frac {a^{2}}{2 d}-\frac {7 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {35 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {10 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {35 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {7 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {8 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {16 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(278\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(4*(tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))+tan(1/2*d*x+1/2*c)^4-19*tan(1/2*d*x+1/2*c)^2+cot(1/2*d*
x+1/2*c)+31*tan(1/2*d*x+1/2*c)-50/3)*a^2/d/(tan(1/2*d*x+1/2*c)-1)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (82) = 164\).

Time = 0.27 (sec) , antiderivative size = 329, normalized size of antiderivative = 3.78 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {10 \, a^{2} \cos \left (d x + c\right )^{3} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 13 \, a^{2} \cos \left (d x + c\right ) + a^{2} - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (10 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(10*a^2*cos(d*x + c)^3 - 4*a^2*cos(d*x + c)^2 - 13*a^2*cos(d*x + c) + a^2 - 3*(a^2*cos(d*x + c)^3 + 2*a^2*
cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2 - (a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(
1/2*cos(d*x + c) + 1/2) + 3*(a^2*cos(d*x + c)^3 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2 - (a^2*cos(d
*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + (10*a^2*cos(d*x + c)^2 + 14
*a^2*cos(d*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c) - (d*cos(d*x +
c)^2 - d*cos(d*x + c) - 2*d)*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {{\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{2} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{3 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.36 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(12*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^2*tan(1/2*d*x + 1/2*c) - 3*(4*a^2*tan(1/2*d*x + 1/2*c) + a^2)
/tan(1/2*d*x + 1/2*c) - 4*(9*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) + 8*a^2)/(tan(1/2*d*x +
1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 10.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.66 \[ \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+23\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {41\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a^2}{d\,\left (-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^4*sin(c + d*x)^2),x)

[Out]

(2*a^2*log(tan(c/2 + (d*x)/2)))/d - (23*a^2*tan(c/2 + (d*x)/2)^2 - 13*a^2*tan(c/2 + (d*x)/2)^3 + a^2 - (41*a^2
*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c/2 + (d*x)/2) - 6*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2
+ (d*x)/2)^4)) + (a^2*tan(c/2 + (d*x)/2))/(2*d)