\(\int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx\) [1495]

Optimal result

Integrand size = 29, antiderivative size = 116 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac {(3 a-4 b) b \log (1+\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

[Out]

-1/8*b*(3*a+4*b)*ln(1-sin(d*x+c))/d+1/8*(3*a-4*b)*b*ln(1+sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c))^2/d-1
/4*sec(d*x+c)^2*(a+b*sin(d*x+c))*(2*a+3*b*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2916, 12, 1659, 647, 31} \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac {b (3 a-4 b) \log (\sin (c+d x)+1)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \]

[In]

Int[Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

-1/8*(b*(3*a + 4*b)*Log[1 - Sin[c + d*x]])/d + ((3*a - 4*b)*b*Log[1 + Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(
a + b*Sin[c + d*x])^2)/(4*d) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])*(2*a + 3*b*Sin[c + d*x]))/(4*d)

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 1659

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 2916

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

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^3 (a+x)^2}{b^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^2 \text {Subst}\left (\int \frac {x^3 (a+x)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (-2 b^4-4 a b^2 x-4 b^2 x^2\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\text {Subst}\left (\int \frac {6 a b^4+8 b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d}-\frac {((3 a-4 b) b) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac {(b (3 a+4 b)) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {b (3 a+4 b) \log (1-\sin (c+d x))}{8 d}+\frac {(3 a-4 b) b \log (1+\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (2 a+3 b \sin (c+d x))}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {3 a b \text {arctanh}(\sin (c+d x))}{4 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}-\frac {3 a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {2 a b \sec (c+d x) \tan ^3(c+d x)}{d}+\frac {a^2 \tan ^4(c+d x)}{4 d}-\frac {b^2 \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+2 a b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(136\)
default	\(\frac {\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+2 a b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(136\)
parallelrisch	\(\frac {16 b^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {4 b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {4 b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-4 a^{2}-4 b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (a^{2}+3 b^{2}\right ) \cos \left (4 d x +4 c \right )+6 a b \sin \left (d x +c \right )-10 a b \sin \left (3 d x +3 c \right )+3 a^{2}+b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(225\)
risch	\(i b^{2} x +\frac {2 i b^{2} c}{d}+\frac {i \left (4 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+5 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}\)	\(248\)
norman	\(\frac {\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (2 a^{2}+3 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {5 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {b^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (3 a -4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}-\frac {b \left (3 a +4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\)	\(328\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/4*a^2*sin(d*x+c)^4/cos(d*x+c)^4+2*a*b*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*
sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+b^2*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c
))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {{\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} - 2 \, {\left (5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8*((3*a*b - 4*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a*b + 4*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) +
1) - 4*(a^2 + 2*b^2)*cos(d*x + c)^2 + 2*a^2 + 2*b^2 - 2*(5*a*b*cos(d*x + c)^2 - 2*a*b)*sin(d*x + c))/(d*cos(d
*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a b \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right ) + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} - 3 \, b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*((3*a*b - 4*b^2)*log(sin(d*x + c) + 1) - (3*a*b + 4*b^2)*log(sin(d*x + c) - 1) + 2*(5*a*b*sin(d*x + c)^3 -
3*a*b*sin(d*x + c) + 2*(a^2 + 2*b^2)*sin(d*x + c)^2 - a^2 - 3*b^2)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.12 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, b^{2} \sin \left (d x + c\right )^{4} + 5 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, b^{2} \sin \left (d x + c\right )^{2} - 3 \, a b \sin \left (d x + c\right ) - a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*((3*a*b - 4*b^2)*log(abs(sin(d*x + c) + 1)) - (3*a*b + 4*b^2)*log(abs(sin(d*x + c) - 1)) + 2*(3*b^2*sin(d*
x + c)^4 + 5*a*b*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*b^2*sin(d*x + c)^2 - 3*a*b*sin(d*x + c) - a^2)/(sin
(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.13 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a\,b}{4}-b^2\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (b^2+\frac {3\,a\,b}{4}\right )}{d}+\frac {b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+8\,b^2\right )+2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {11\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((sin(c + d*x)^3*(a + b*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*((3*a*b)/4 - b^2))/d - (log(tan(c/2 + (d*x)/2) - 1)*((3*a*b)/4 + b^2))/d + (b^2*l
og(tan(c/2 + (d*x)/2)^2 + 1))/d - (2*b^2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4*(4*a^2 + 8*b^2) + 2*b^2*t
an(c/2 + (d*x)/2)^6 - (11*a*b*tan(c/2 + (d*x)/2)^3)/2 - (11*a*b*tan(c/2 + (d*x)/2)^5)/2 + (3*a*b*tan(c/2 + (d*
x)/2)^7)/2 + (3*a*b*tan(c/2 + (d*x)/2))/2)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (
d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))