Integrand size = 21, antiderivative size = 140 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d}+\frac {b \cot ^2(c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac {2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}-\frac {b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))} \] Output:
-(a^2+3*b^2)*cot(d*x+c)/a^4/d+b*cot(d*x+c)^2/a^3/d-1/3*cot(d*x+c)^3/a^2/d- 2*b*(a^2+2*b^2)*ln(tan(d*x+c))/a^5/d+2*b*(a^2+2*b^2)*ln(a+b*tan(d*x+c))/a^ 5/d-b*(a^2+b^2)/a^4/d/(a+b*tan(d*x+c))
Time = 3.48 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.74 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {-\cot ^2(c+d x) \left (2 a^4+9 a^2 b^2+a^4 \csc ^2(c+d x)\right )+3 b^2 \left (a^2+b^2+a^2 \csc ^2(c+d x)-2 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+2 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+4 b^2 \log (a \cos (c+d x)+b \sin (c+d x))\right )+a b \cot (c+d x) \left (-2 a^2-9 b^2+2 a^2 \csc ^2(c+d x)-6 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+6 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+12 b^2 \log (a \cos (c+d x)+b \sin (c+d x))\right )}{3 a^5 d (b+a \cot (c+d x))} \] Input:
Integrate[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]
Output:
(-(Cot[c + d*x]^2*(2*a^4 + 9*a^2*b^2 + a^4*Csc[c + d*x]^2)) + 3*b^2*(a^2 + b^2 + a^2*Csc[c + d*x]^2 - 2*(a^2 + 2*b^2)*Log[Sin[c + d*x]] + 2*a^2*Log[ a*Cos[c + d*x] + b*Sin[c + d*x]] + 4*b^2*Log[a*Cos[c + d*x] + b*Sin[c + d* x]]) + a*b*Cot[c + d*x]*(-2*a^2 - 9*b^2 + 2*a^2*Csc[c + d*x]^2 - 6*(a^2 + 2*b^2)*Log[Sin[c + d*x]] + 6*a^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] + 12 *b^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(3*a^5*d*(b + a*Cot[c + d*x]))
Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3999, 522, 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.
\(\displaystyle \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^4 (a+b \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )}{b^4 (a+b \tan (c+d x))^2}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {b \int \left (\frac {\cot ^4(c+d x)}{a^2 b^2}-\frac {2 \cot ^3(c+d x)}{a^3 b}+\frac {\left (a^2+3 b^2\right ) \cot ^2(c+d x)}{a^4 b^2}-\frac {2 \left (a^2+2 b^2\right ) \cot (c+d x)}{a^5 b}+\frac {2 \left (a^2+2 b^2\right )}{a^5 (a+b \tan (c+d x))}+\frac {a^2+b^2}{a^4 (a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {\cot ^2(c+d x)}{a^3}-\frac {\cot ^3(c+d x)}{3 a^2 b}-\frac {2 \left (a^2+2 b^2\right ) \log (b \tan (c+d x))}{a^5}+\frac {2 \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5}-\frac {a^2+b^2}{a^4 (a+b \tan (c+d x))}-\frac {\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 b}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]
Output:
(b*(-(((a^2 + 3*b^2)*Cot[c + d*x])/(a^4*b)) + Cot[c + d*x]^2/a^3 - Cot[c + d*x]^3/(3*a^2*b) - (2*(a^2 + 2*b^2)*Log[b*Tan[c + d*x]])/a^5 + (2*(a^2 + 2*b^2)*Log[a + b*Tan[c + d*x]])/a^5 - (a^2 + b^2)/(a^4*(a + b*Tan[c + d*x] ))))/d
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Time = 2.63 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+3 b^{2}}{a^{4} \tan \left (d x +c \right )}+\frac {b}{a^{3} \tan \left (d x +c \right )^{2}}-\frac {2 b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{5}}-\frac {\left (a^{2}+b^{2}\right ) b}{a^{4} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}+2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(127\) |
default | \(\frac {-\frac {1}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+3 b^{2}}{a^{4} \tan \left (d x +c \right )}+\frac {b}{a^{3} \tan \left (d x +c \right )^{2}}-\frac {2 b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{5}}-\frac {\left (a^{2}+b^{2}\right ) b}{a^{4} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}+2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(127\) |
risch | \(-\frac {4 i \left (-12 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-18 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-2 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a^{3}+18 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2} b +6 i a \,b^{2}-3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{3}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) a^{4} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{3} d}+\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{5} d}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{5} d}\) | \(353\) |
Input:
int(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/a^2/tan(d*x+c)^3-(a^2+3*b^2)/a^4/tan(d*x+c)+1/a^3*b/tan(d*x+c)^2 -2*b*(a^2+2*b^2)/a^5*ln(tan(d*x+c))-(a^2+b^2)*b/a^4/(a+b*tan(d*x+c))+2*b*( a^2+2*b^2)/a^5*ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (138) = 276\).
Time = 0.11 (sec) , antiderivative size = 442, normalized size of antiderivative = 3.16 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, {\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} - 3 \, {\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 3 \, {\left ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) - 2 \, {\left (6 \, a b^{3} \cos \left (d x + c\right ) - {\left (a^{3} b + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{5} b d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b d \cos \left (d x + c\right )^{2} + a^{5} b d - {\left (a^{6} d \cos \left (d x + c\right )^{3} - a^{6} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/3*(2*(a^4 + 6*a^2*b^2)*cos(d*x + c)^4 + 6*a^2*b^2 - 3*(a^4 + 6*a^2*b^2)* cos(d*x + c)^2 + 3*((a^2*b^2 + 2*b^4)*cos(d*x + c)^4 + a^2*b^2 + 2*b^4 - 2 *(a^2*b^2 + 2*b^4)*cos(d*x + c)^2 - ((a^3*b + 2*a*b^3)*cos(d*x + c)^3 - (a ^3*b + 2*a*b^3)*cos(d*x + c))*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 3*((a^2*b^2 + 2*b^4)*cos(d*x + c)^4 + a^2*b^2 + 2*b^4 - 2*(a^2*b^2 + 2*b^4)*cos(d*x + c)^2 - ((a^3*b + 2 *a*b^3)*cos(d*x + c)^3 - (a^3*b + 2*a*b^3)*cos(d*x + c))*sin(d*x + c))*log (-1/4*cos(d*x + c)^2 + 1/4) - 2*(6*a*b^3*cos(d*x + c) - (a^3*b + 6*a*b^3)* cos(d*x + c)^3)*sin(d*x + c))/(a^5*b*d*cos(d*x + c)^4 - 2*a^5*b*d*cos(d*x + c)^2 + a^5*b*d - (a^6*d*cos(d*x + c)^3 - a^6*d*cos(d*x + c))*sin(d*x + c ))
\[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(csc(d*x+c)**4/(a+b*tan(d*x+c))**2,x)
Output:
Integral(csc(c + d*x)**4/(a + b*tan(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {2 \, a^{2} b \tan \left (d x + c\right ) - 6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{3} - a^{3} - 3 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{4} b \tan \left (d x + c\right )^{4} + a^{5} \tan \left (d x + c\right )^{3}} + \frac {6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5}} - \frac {6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{5}}}{3 \, d} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/3*((2*a^2*b*tan(d*x + c) - 6*(a^2*b + 2*b^3)*tan(d*x + c)^3 - a^3 - 3*(a ^3 + 2*a*b^2)*tan(d*x + c)^2)/(a^4*b*tan(d*x + c)^4 + a^5*tan(d*x + c)^3) + 6*(a^2*b + 2*b^3)*log(b*tan(d*x + c) + a)/a^5 - 6*(a^2*b + 2*b^3)*log(ta n(d*x + c))/a^5)/d
Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{5} d} + \frac {2 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b d} + \frac {2 \, a^{3} b \tan \left (d x + c\right ) - a^{4} - 6 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \tan \left (d x + c\right )^{3} - 3 \, {\left (a^{4} + 2 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{3 \, {\left (b \tan \left (d x + c\right ) + a\right )} a^{5} d \tan \left (d x + c\right )^{3}} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-2*(a^2*b + 2*b^3)*log(abs(tan(d*x + c)))/(a^5*d) + 2*(a^2*b^2 + 2*b^4)*lo g(abs(b*tan(d*x + c) + a))/(a^5*b*d) + 1/3*(2*a^3*b*tan(d*x + c) - a^4 - 6 *(a^3*b + 2*a*b^3)*tan(d*x + c)^3 - 3*(a^4 + 2*a^2*b^2)*tan(d*x + c)^2)/(( b*tan(d*x + c) + a)*a^5*d*tan(d*x + c)^3)
Time = 1.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,\left (a^2+2\,b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (2\,a^2\,b+4\,b^3\right )}\right )\,\left (a^2+2\,b^2\right )}{a^5\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+2\,b^2\right )}{a^3}-\frac {2\,b\,\mathrm {tan}\left (c+d\,x\right )}{3\,a^2}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2+2\,b^2\right )}{a^4}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )} \] Input:
int(1/(sin(c + d*x)^4*(a + b*tan(c + d*x))^2),x)
Output:
(4*b*atanh((2*b*(a^2 + 2*b^2)*(a + 2*b*tan(c + d*x)))/(a*(2*a^2*b + 4*b^3) ))*(a^2 + 2*b^2))/(a^5*d) - (1/(3*a) + (tan(c + d*x)^2*(a^2 + 2*b^2))/a^3 - (2*b*tan(c + d*x))/(3*a^2) + (2*b*tan(c + d*x)^3*(a^2 + 2*b^2))/a^4)/(d* (a*tan(c + d*x)^3 + b*tan(c + d*x)^4))
Time = 0.16 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.49 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {48 \cos \left (d x +c \right ) \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 )^{3} a^{3} b^{2}+96 \cos \left (d x +c \right ) \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 )^{3} a \,b^{4}-48 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}-96 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a \,b^{4}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}-88 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}-96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{4}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+48 \,\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 )^{4} a^{2} b^{3}+96 \,\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 )^{4} b^{5}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2} b^{3}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{5}+7 \sin \left (d x +c \right )^{4} a^{4} b +24 \sin \left (d x +c \right )^{4} a^{2} b^{3}-8 \sin \left (d x +c \right )^{2} a^{4} b -48 \sin \left (d x +c \right )^{2} a^{2} b^{3}-8 a^{4} b}{24 \sin \left (d x +c \right )^{3} a^{5} b d \left (\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b \right )} \] Input:
int(csc(d*x+c)^4/(a+b*tan(d*x+c))^2,x)
Output:
(48*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin (c + d*x)**3*a**3*b**2 + 96*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan ((c + d*x)/2)*b - a)*sin(c + d*x)**3*a*b**4 - 48*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**3*b**2 - 96*cos(c + d*x)*log(tan((c + d*x)/2) )*sin(c + d*x)**3*a*b**4 - 9*cos(c + d*x)*sin(c + d*x)**3*a**5 - 88*cos(c + d*x)*sin(c + d*x)**3*a**3*b**2 - 96*cos(c + d*x)*sin(c + d*x)**3*a*b**4 + 16*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 48*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**4*a**2*b**3 + 96*log(tan((c + d*x) /2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**4*b**5 - 48*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**2*b**3 - 96*log(tan((c + d*x)/2))*sin(c + d* x)**4*b**5 + 7*sin(c + d*x)**4*a**4*b + 24*sin(c + d*x)**4*a**2*b**3 - 8*s in(c + d*x)**2*a**4*b - 48*sin(c + d*x)**2*a**2*b**3 - 8*a**4*b)/(24*sin(c + d*x)**3*a**5*b*d*(cos(c + d*x)*a + sin(c + d*x)*b))