Integrand size = 23, antiderivative size = 284 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {2 b^{2/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}-b^{2/3}} d}-\frac {2 \sqrt [3]{-1} b^{2/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}-\frac {2 (-1)^{2/3} b^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {\cot (c+d x)}{a d} \] Output:
2/3*b^(2/3)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^ (1/2))/a^(4/3)/(a^(2/3)-b^(2/3))^(1/2)/d-2/3*(-1)^(1/3)*b^(2/3)*arctan(((- 1)^(2/3)*b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^ (1/2))/a^(4/3)/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)/d-2/3*(-1)^(2/3)*b^(2/3) *arctan((-1)^(1/3)*(b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3 )-(-1)^(2/3)*b^(2/3))^(1/2))/a^(4/3)/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)/d- cot(d*x+c)/a/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.76 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.69 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-3 \cot \left (\frac {1}{2} (c+d x)\right )+2 b \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]+3 \tan \left (\frac {1}{2} (c+d x)\right )}{6 a d} \] Input:
Integrate[Csc[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]
Output:
(-3*Cot[(c + d*x)/2] + 2*b*RootSum[-b + 3*b*#1^2 - (8*I)*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] + I*Log[1 - 2*C os[c + d*x]*#1 + #1^2] + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2)/(b - (4*I)*a*#1 - 2*b*#1^2 + b* #1^4) & ] + 3*Tan[(c + d*x)/2])/(6*a*d)
Time = 0.64 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3699, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \left (a+b \sin (c+d x)^3\right )}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {\csc ^2(c+d x)}{a}-\frac {b \sin (c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{-1} b^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 (-1)^{2/3} b^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {\cot (c+d x)}{a d}\) |
Input:
Int[Csc[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]
Output:
(2*b^(2/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2 /3)]])/(3*a^(4/3)*Sqrt[a^(2/3) - b^(2/3)]*d) - (2*(-1)^(1/3)*b^(2/3)*ArcTa n[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3 )*b^(2/3)]])/(3*a^(4/3)*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) - (2*(-1)^(2 /3)*b^(2/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x) /2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a^(4/3)*Sqrt[a^(2/3) - (-1)^ (2/3)*b^(2/3)]*d) - Cot[c + d*x]/(a*d)
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.40
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 a}}{d}\) | \(114\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 a}}{d}\) | \(114\) |
| risch | \(-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (2985984 a^{10} d^{6}-2985984 a^{8} b^{2} d^{6}\right ) \textit {\_Z}^{6}+62208 a^{6} b^{2} d^{4} \textit {\_Z}^{4}+b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {248832 d^{5} a^{10}}{a^{2} b^{3}+b^{5}}-\frac {248832 d^{5} a^{8} b^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{5}+\left (\frac {20736 i d^{4} a^{9}}{a^{2} b^{3}+b^{5}}-\frac {20736 i d^{4} a^{7} b^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{4}+\left (\frac {3456 d^{3} a^{6} b^{2}}{a^{2} b^{3}+b^{5}}+\frac {1728 d^{3} b^{4} a^{4}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{3}+\left (\frac {288 i d^{2} a^{5} b^{2}}{a^{2} b^{3}+b^{5}}+\frac {144 i d^{2} b^{4} a^{3}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{2}+\left (-\frac {12 d \,a^{4} b^{2}}{a^{2} b^{3}+b^{5}}-\frac {24 d \,b^{4} a^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R} -\frac {i b^{4} a}{a^{2} b^{3}+b^{5}}\right )\right )\) | \(362\) |
Input:
int(csc(d*x+c)^2/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/a*tan(1/2*d*x+1/2*c)-1/2/a/tan(1/2*d*x+1/2*c)-2/3*b/a*sum((_R^3+_ R)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^ 6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a)))
Result contains complex when optimal does not.
Time = 1.09 (sec) , antiderivative size = 21243, normalized size of antiderivative = 74.80 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)**2/(a+b*sin(d*x+c)**3),x)
Output:
Integral(csc(c + d*x)**2/(a + b*sin(c + d*x)**3), x)
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
((a*d*cos(2*d*x + 2*c)^2 + a*d*sin(2*d*x + 2*c)^2 - 2*a*d*cos(2*d*x + 2*c) + a*d)*integrate(-4*(3*b^2*cos(4*d*x + 4*c)^2 + 3*b^2*cos(2*d*x + 2*c)^2 + 3*b^2*sin(4*d*x + 4*c)^2 + 8*a*b*cos(2*d*x + 2*c)*sin(3*d*x + 3*c) - 8*a *b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 3*b^2*sin(2*d*x + 2*c)^2 - b^2*cos( 2*d*x + 2*c) - (b^2*cos(4*d*x + 4*c) - b^2*cos(2*d*x + 2*c))*cos(6*d*x + 6 *c) - (6*b^2*cos(2*d*x + 2*c) + 8*a*b*sin(3*d*x + 3*c) - b^2)*cos(4*d*x + 4*c) - (b^2*sin(4*d*x + 4*c) - b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 2* (4*a*b*cos(3*d*x + 3*c) - 3*b^2*sin(2*d*x + 2*c))*sin(4*d*x + 4*c))/(a*b^2 *cos(6*d*x + 6*c)^2 + 9*a*b^2*cos(4*d*x + 4*c)^2 + 64*a^3*cos(3*d*x + 3*c) ^2 + 9*a*b^2*cos(2*d*x + 2*c)^2 + a*b^2*sin(6*d*x + 6*c)^2 + 9*a*b^2*sin(4 *d*x + 4*c)^2 + 64*a^3*sin(3*d*x + 3*c)^2 - 48*a^2*b*cos(3*d*x + 3*c)*sin( 2*d*x + 2*c) + 9*a*b^2*sin(2*d*x + 2*c)^2 - 6*a*b^2*cos(2*d*x + 2*c) + a*b ^2 - 2*(3*a*b^2*cos(4*d*x + 4*c) - 3*a*b^2*cos(2*d*x + 2*c) - 8*a^2*b*sin( 3*d*x + 3*c) + a*b^2)*cos(6*d*x + 6*c) - 6*(3*a*b^2*cos(2*d*x + 2*c) + 8*a ^2*b*sin(3*d*x + 3*c) - a*b^2)*cos(4*d*x + 4*c) - 2*(8*a^2*b*cos(3*d*x + 3 *c) + 3*a*b^2*sin(4*d*x + 4*c) - 3*a*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c ) + 6*(8*a^2*b*cos(3*d*x + 3*c) - 3*a*b^2*sin(2*d*x + 2*c))*sin(4*d*x + 4* c) + 16*(3*a^2*b*cos(2*d*x + 2*c) - a^2*b)*sin(3*d*x + 3*c)), x) - 2*sin(2 *d*x + 2*c))/(a*d*cos(2*d*x + 2*c)^2 + a*d*sin(2*d*x + 2*c)^2 - 2*a*d*cos( 2*d*x + 2*c) + a*d)
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
integrate(csc(d*x + c)^2/(b*sin(d*x + c)^3 + a), x)
Time = 38.76 (sec) , antiderivative size = 697, normalized size of antiderivative = 2.45 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\left (\sum _{k=1}^6\ln \left (8192\,a^7\,b^6-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^2\,a^9\,b^6\,294912+{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^3\,a^{11}\,b^5\,1548288-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^4\,a^{13}\,b^4\,1990656-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^5\,a^{13}\,b^5\,7962624+{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^5\,a^{15}\,b^3\,5971968-65536\,a^6\,b^7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )\,a^8\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^2\,a^{10}\,b^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^3\,a^{10}\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472+{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^3\,a^{12}\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,221184-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^4\,a^{12}\,b^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,2654208-{\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )}^5\,a^{14}\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656\right )\,\mathrm {root}\left (729\,a^8\,b^2\,d^6-729\,a^{10}\,d^6-243\,a^6\,b^2\,d^4-b^4,d,k\right )\right )-\frac {1}{2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a}}{d} \] Input:
int(1/(sin(c + d*x)^2*(a + b*sin(c + d*x)^3)),x)
Output:
(symsum(log(8192*a^7*b^6 - 294912*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 24 3*a^6*b^2*d^4 - b^4, d, k)^2*a^9*b^6 + 1548288*root(729*a^8*b^2*d^6 - 729* a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^3*a^11*b^5 - 1990656*root(729*a^8* b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^4*a^13*b^4 - 7962624 *root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^5*a^13 *b^5 + 5971968*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4 , d, k)^5*a^15*b^3 - 65536*a^6*b^7*tan(c/2 + (d*x)/2) + 196608*root(729*a^ 8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)*a^8*b^6*tan(c/2 + (d*x)/2) - 294912*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^2*a^10*b^5*tan(c/2 + (d*x)/2) - 1769472*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^3*a^10*b^6*tan(c/2 + (d*x)/2) + 221184*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k )^3*a^12*b^4*tan(c/2 + (d*x)/2) - 2654208*root(729*a^8*b^2*d^6 - 729*a^10* d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^4*a^12*b^5*tan(c/2 + (d*x)/2) - 1990656 *root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2*d^4 - b^4, d, k)^5*a^14 *b^4*tan(c/2 + (d*x)/2))*root(729*a^8*b^2*d^6 - 729*a^10*d^6 - 243*a^6*b^2 *d^4 - b^4, d, k), k, 1, 6) - 1/(2*a*tan(c/2 + (d*x)/2)) + tan(c/2 + (d*x) /2)/(2*a))/d
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^2/(a+b*sin(d*x+c)^3),x)
Output:
( - 3*cos(c + d*x)*a**2 - 24*cos(c + d*x)*b**2 + 6*int(tan((c + d*x)/2)**3 /(tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/2)**4*a + 8*tan((c + d*x)/2)**3* b + 3*tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)*a**2*b*d + 12*int(tan((c + d*x)/2)/(tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/2)**4*a + 8*tan((c + d* x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)*a**2*b*d - 96*in t(tan((c + d*x)/2)/(tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/2)**4*a + 8*ta n((c + d*x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)*b**3*d - 12*int(1/(tan((c + d*x)/2)**8*a + 3*tan((c + d*x)/2)**6*a + 8*tan((c + d *x)/2)**5*b + 3*tan((c + d*x)/2)**4*a + tan((c + d*x)/2)**2*a),x)*sin(c + d*x)*a*b**2*d + 6*int(1/(tan((c + d*x)/2)**7*a + 3*tan((c + d*x)/2)**5*a + 8*tan((c + d*x)/2)**4*b + 3*tan((c + d*x)/2)**3*a + tan((c + d*x)/2)*a),x )*sin(c + d*x)*a**2*b*d - 36*int(1/(tan((c + d*x)/2)**6*a + 3*tan((c + d*x )/2)**4*a + 8*tan((c + d*x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + a),x)*sin( c + d*x)*a*b**2*d + 3*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a*b + log( tan((c + d*x)/2)**6*a + 3*tan((c + d*x)/2)**4*a + 8*tan((c + d*x)/2)**3*b + 3*tan((c + d*x)/2)**2*a + a)*sin(c + d*x)*a*b - 12*log(tan((c + d*x)/2)) *sin(c + d*x)*a*b - 12*sin(c + d*x)*b**2*d*x - 24*b**2)/(3*sin(c + d*x)*a* *3*d)