Integrand size = 21, antiderivative size = 138 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(2 a+b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(2 a-b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {a^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {\sec ^2(c+d x) \left (\frac {a}{a^2-b^2}-\frac {b \sin (c+d x)}{a^2-b^2}\right )}{2 d} \] Output:
1/4*(2*a+b)*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(2*a-b)*ln(1+sin(d*x+c))/(a-b)^ 2/d-a^3*ln(a+b*sin(d*x+c))/(a^2-b^2)^2/d+1/2*sec(d*x+c)^2*(a/(a^2-b^2)-b*s in(d*x+c)/(a^2-b^2))/d
Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {(2 a+b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {(2 a-b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a^3 \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}-\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
(((2*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^2 + ((2*a - b)*Log[1 + Sin[c + d*x]])/(a - b)^2 - (4*a^3*Log[a + b*Sin[c + d*x]])/((a - b)^2*(a + b)^2) - 1/((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])))/(4*d)
Time = 0.41 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3200, 601, 25, 27, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {\frac {b^2 (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {b^2 \left (\frac {a b^2}{a^2-b^2}-\frac {b \left (2 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {b^2 \left (a b^2-b \left (2 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right ) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a b^2-b \left (2 a^2-b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}+\frac {b^2 (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {\frac {\int \left (-\frac {2 a^3}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {-2 a^2+b a+b^2}{2 (a+b) (b-b \sin (c+d x))}+\frac {(2 a-b) (a+b)}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}+\frac {b^2 (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^2 (a-b \sin (c+d x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {-\frac {2 a^3 \log (a+b \sin (c+d x))}{a^2-b^2}+\frac {(a-b) (2 a+b) \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {(2 a-b) (a+b) \log (b \sin (c+d x)+b)}{2 (a-b)}}{2 \left (a^2-b^2\right )}}{d}\) |
Input:
Int[Tan[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
((((a - b)*(2*a + b)*Log[b - b*Sin[c + d*x]])/(2*(a + b)) - (2*a^3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) + ((2*a - b)*(a + b)*Log[b + b*Sin[c + d*x]] )/(2*(a - b)))/(2*(a^2 - b^2)) + (b^2*(a - b*Sin[c + d*x]))/(2*(a^2 - b^2) *(b^2 - b^2*Sin[c + d*x]^2)))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.55 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
| default | \(\frac {-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(121\) |
| risch | \(\frac {i b x}{2 a^{2}-4 a b +2 b^{2}}+\frac {i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {i a c}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i a^{3} x}{a^{4}-2 b^{2} a^{2}+b^{4}}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )} b \right )}{d \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {i b c}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i a^{3} c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {i a x}{a^{2}+2 a b +b^{2}}-\frac {i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i a x}{a^{2}-2 a b +b^{2}}-\frac {i a c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\) | \(455\) |
Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-a^3/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c))+1/(4*a-4*b)/(1+sin(d*x+c))+1/ 4*(2*a-b)/(a-b)^2*ln(1+sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4*(2*a+b)/ (a+b)^2*ln(sin(d*x+c)-1))
Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {4 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{3} + 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/4*(4*a^3*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - (2*a^3 + 3*a^2*b - b^ 3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^3 - 3*a^2*b + b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*a^3 + 2*a*b^2 + 2*(a^2*b - b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2)
\[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**3/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/4*(4*a^3*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - (2*a - b)*lo g(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - (2*a + b)*log(sin(d*x + c) - 1)/ (a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2))/d
Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{3} b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} + \frac {{\left (2 \, a + b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {{\left (2 \, a - b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} - \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-a^3*b*log(abs(b*sin(d*x + c) + a))/(a^4*b*d - 2*a^2*b^3*d + b^5*d) + 1/4* (2*a + b)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b*d + b^2*d) + 1/4*(2*a - b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) - 1/2*(a^3 - a *b^2 - (a^2*b - b^3)*sin(d*x + c))/((a + b)^2*(a - b)^2*d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1))
Time = 18.07 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.57 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (2\,a-b\right )}{2\,d\,{\left (a-b\right )}^2}-\frac {\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (2\,a+b\right )}{2\,d\,{\left (a+b\right )}^2} \] Input:
int(tan(c + d*x)^3/(a + b*sin(c + d*x)),x)
Output:
(log(tan(c/2 + (d*x)/2) + 1)*(2*a - b))/(2*d*(a - b)^2) - ((b*tan(c/2 + (d *x)/2))/(a^2 - b^2) - (2*a*tan(c/2 + (d*x)/2)^2)/(a^2 - b^2) + (b*tan(c/2 + (d*x)/2)^3)/(a^2 - b^2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2) ^2 + 1)) - (a^3*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/ (d*(a^4 + b^4 - 2*a^2*b^2)) + (log(tan(c/2 + (d*x)/2) - 1)*(2*a + b))/(2*d *(a + b)^2)
Time = 0.18 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.29 \[ \int \frac {\tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{2} a^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{3}+\sin \left (d x +c \right )^{2} a^{3}-\sin \left (d x +c \right )^{2} a \,b^{2}+\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}-2 a^{3}+2 a \,b^{2}}{2 d \left (\sin \left (d x +c \right )^{2} a^{4}-2 \sin \left (d x +c \right )^{2} a^{2} b^{2}+\sin \left (d x +c \right )^{2} b^{4}-a^{4}+2 a^{2} b^{2}-b^{4}\right )} \] Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
(2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b **3 - 2*log(tan((c + d*x)/2) - 1)*a**3 + 3*log(tan((c + d*x)/2) - 1)*a**2* b - log(tan((c + d*x)/2) - 1)*b**3 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d *x)**2*a**3 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b - log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 2*log(tan((c + d*x)/2) + 1)*a**3 - 3*log(tan((c + d*x)/2) + 1)*a**2*b + log(tan((c + d*x)/2) + 1)*b**3 - 2 *log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a** 3 + 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3 + sin(c + d*x)**2*a**3 - sin(c + d*x)**2*a*b**2 + sin(c + d*x)*a**2*b - sin(c + d*x )*b**3 - 2*a**3 + 2*a*b**2)/(2*d*(sin(c + d*x)**2*a**4 - 2*sin(c + d*x)**2 *a**2*b**2 + sin(c + d*x)**2*b**4 - a**4 + 2*a**2*b**2 - b**4))