\(\int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\) [795]

Optimal result

Integrand size = 27, antiderivative size = 117 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5 a x}{2}-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d} \]

[Out]

5/2*a*x-3*a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d-3*a*sec(d*x+c)/d+1/3*a*sec(d*x+c)^3/d-5/2*a*tan(d*x+c)/d+5/6*a*t
an(d*x+c)^3/d-1/2*a*sin(d*x+c)^2*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2917, 2671, 294, 308, 209, 2670, 276} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \cos (c+d x)}{d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {5 a x}{2} \]

[In]

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

[Out]

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

Rule 209

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

Rule 276

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

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2670

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

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \sin ^2(c+d x) \tan ^4(c+d x) \, dx+a \int \sin ^3(c+d x) \tan ^4(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}-\frac {a \text {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {(5 a) \text {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {(5 a) \text {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {(5 a) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {5 a x}{2}-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (30 c+30 d x-33 \cos (c+d x)+\cos (3 (c+d x))-36 \sec (c+d x)+4 \sec ^3(c+d x)-3 \sin (2 (c+d x))-28 \tan (c+d x)+4 \sec ^2(c+d x) \tan (c+d x)\right )}{12 d} \]

[In]

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

[Out]

(a*(30*c + 30*d*x - 33*Cos[c + d*x] + Cos[3*(c + d*x)] - 36*Sec[c + d*x] + 4*Sec[c + d*x]^3 - 3*Sin[2*(c + d*x
)] - 28*Tan[c + d*x] + 4*Sec[c + d*x]^2*Tan[c + d*x]))/(12*d)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(-\frac {a \left (-120 d x \cos \left (d x +c \right )+60 d x \sin \left (2 d x +2 c \right )+\sin \left (5 d x +5 c \right )-27 \sin \left (3 d x +3 c \right )+\cos \left (4 d x +4 c \right )+120 \cos \left (2 d x +2 c \right )-60 \sin \left (d x +c \right )+136 \cos \left (d x +c \right )-68 \sin \left (2 d x +2 c \right )+135\right )}{24 d \left (-\sin \left (2 d x +2 c \right )+2 \cos \left (d x +c \right )\right )}\)	\(121\)
derivativedivides	\(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+a \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(164\)
default	\(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+a \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(164\)
risch	\(\frac {5 a x}{2}+\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {11 a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {11 a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {2 \left (5 a \,{\mathrm e}^{i \left (d x +c \right )}-9 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a \,{\mathrm e}^{3 i \left (d x +c \right )}-7 i a \right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(173\)
norman	\(\frac {-\frac {5 a x}{2}+\frac {32 a}{3 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {14 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {14 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {5 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {15 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {32 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(205\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)^6*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/24/d*a*(-120*d*x*cos(d*x+c)+60*d*x*sin(2*d*x+2*c)+sin(5*d*x+5*c)-27*sin(3*d*x+3*c)+cos(4*d*x+4*c)+120*cos(2
*d*x+2*c)-60*sin(d*x+c)+136*cos(d*x+c)-68*sin(2*d*x+2*c)+135)/(-sin(2*d*x+2*c)+2*cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 29 \, a \cos \left (d x + c\right )^{2} + {\left (2 \, a \cos \left (d x + c\right )^{4} + 15 \, a d x \cos \left (d x + c\right ) - 15 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(a*cos(d*x + c)^4 - 15*a*d*x*cos(d*x + c) + 29*a*cos(d*x + c)^2 + (2*a*cos(d*x + c)^4 + 15*a*d*x*cos(d*x +
c) - 15*a*cos(d*x + c)^2 - 4*a)*sin(d*x + c) + 2*a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**6*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.82 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a + {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(2*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a + (2*tan(d*x + c)^3 + 15*d*
x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.56 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {15 \, {\left (d x + c\right )} a - \frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {33 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 102 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 330 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 402 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 410 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 264 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 150 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 61 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)*a - 3*a/(tan(1/2*d*x + 1/2*c) + 1) + (33*a*tan(1/2*d*x + 1/2*c)^8 - 102*a*tan(1/2*d*x + 1/2*
c)^7 + 200*a*tan(1/2*d*x + 1/2*c)^6 - 330*a*tan(1/2*d*x + 1/2*c)^5 + 402*a*tan(1/2*d*x + 1/2*c)^4 - 410*a*tan(
1/2*d*x + 1/2*c)^3 + 264*a*tan(1/2*d*x + 1/2*c)^2 - 150*a*tan(1/2*d*x + 1/2*c) + 61*a)/(tan(1/2*d*x + 1/2*c)^3
- tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 16.94 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.62 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5\,a\,x}{2}+\frac {\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-30\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {a\,\left (45\,d\,x-60\right )}{6}-\frac {15\,a\,d\,x}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (10\,a\,d\,x-\frac {a\,\left (60\,d\,x-80\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {a\,\left (30\,d\,x-100\right )}{6}-5\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-28\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {a\,\left (60\,d\,x-176\right )}{6}-10\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {15\,a\,d\,x}{2}-\frac {a\,\left (45\,d\,x-132\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {a\,\left (30\,d\,x-98\right )}{6}-5\,a\,d\,x\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\left (15\,d\,x-64\right )}{6}+\frac {5\,a\,d\,x}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

[In]

int((sin(c + d*x)^6*(a + a*sin(c + d*x)))/cos(c + d*x)^4,x)

[Out]

(5*a*x)/2 + (tan(c/2 + (d*x)/2)*((a*(30*d*x - 98))/6 - 5*a*d*x) - (a*(15*d*x - 64))/6 - tan(c/2 + (d*x)/2)^4*(
(a*(30*d*x - 28))/6 - 5*a*d*x) - tan(c/2 + (d*x)/2)^9*((a*(30*d*x - 30))/6 - 5*a*d*x) + tan(c/2 + (d*x)/2)^8*(
(a*(45*d*x - 60))/6 - (15*a*d*x)/2) + tan(c/2 + (d*x)/2)^6*((a*(30*d*x - 100))/6 - 5*a*d*x) - tan(c/2 + (d*x)/
2)^7*((a*(60*d*x - 80))/6 - 10*a*d*x) - tan(c/2 + (d*x)/2)^2*((a*(45*d*x - 132))/6 - (15*a*d*x)/2) + tan(c/2 +
(d*x)/2)^3*((a*(60*d*x - 176))/6 - 10*a*d*x) + 6*a*tan(c/2 + (d*x)/2)^5 + (5*a*d*x)/2)/(d*(tan(c/2 + (d*x)/2)
+ 1)*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1)^3)