Integrand size = 23, antiderivative size = 281 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 (-1)^{2/3} b^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \sqrt [3]{b}-\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}+\frac {2 b^{2/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}-b^{2/3}} d}-\frac {2 \sqrt [3]{-1} b^{2/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{4/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}-\frac {\cot (c+d x)}{a d} \]
-cot(d*x+c)/a/d+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)/d/(a^(2/3)-b^(2/3))^(1/2)-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)/d/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)-2/3*(- 1)^(2/3)*b^(2/3)*arctan(((-1)^(1/3)*b^(1/3)-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)/d/(a^(2/3)-(-1)^(2/3)*b^(2/3))^( 1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.70 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {-3 \cot \left (\frac {1}{2} (c+d x)\right )+2 b \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]+3 \tan \left (\frac {1}{2} (c+d x)\right )}{6 a d} \]
(-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 = 0.63 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.01, 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 |
| \(\displaystyle \frac {2 b^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{-1} b^{2/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 (-1)^{2/3} b^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{4/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {\cot (c+d x)}{a d}\) |
(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)
3.2.93.3.1 Defintions of rubi rules used
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 = 1.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.41
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {2 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(114\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {2 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(114\) |
| risch | \(-\frac {2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (2985984 a^{10} d^{6}-2985984 a^{8} b^{2} d^{6}\right ) \textit {\_Z}^{6}+62208 a^{6} b^{2} d^{4} \textit {\_Z}^{4}+b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {248832 d^{5} a^{10}}{a^{2} b^{3}+b^{5}}-\frac {248832 d^{5} a^{8} b^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{5}+\left (\frac {20736 i d^{4} a^{9}}{a^{2} b^{3}+b^{5}}-\frac {20736 i d^{4} a^{7} b^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{4}+\left (\frac {3456 d^{3} a^{6} b^{2}}{a^{2} b^{3}+b^{5}}+\frac {1728 d^{3} b^{4} a^{4}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{3}+\left (\frac {288 i d^{2} a^{5} b^{2}}{a^{2} b^{3}+b^{5}}+\frac {144 i d^{2} b^{4} a^{3}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R}^{2}+\left (-\frac {12 d \,a^{4} b^{2}}{a^{2} b^{3}+b^{5}}-\frac {24 d \,b^{4} a^{2}}{a^{2} b^{3}+b^{5}}\right ) \textit {\_R} -\frac {i b^{4} a}{a^{2} b^{3}+b^{5}}\right )\right )\) | \(362\) |
1/d*(1/2*tan(1/2*d*x+1/2*c)/a-2/3/a*b*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))-1/2/a/tan(1/2*d*x+1/2*c))
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 21243, normalized size of antiderivative = 75.60 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
((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)
\[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 14.46 (sec) , antiderivative size = 697, normalized size of antiderivative = 2.48 \[ \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} \]
(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