Integrand size = 21, antiderivative size = 264 \[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a \sqrt {a^{2/3}-b^{2/3}} d}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {2 \sqrt [3]{b} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 a \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} d}+\frac {2 \sqrt [3]{b} \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} d} \]
-arctanh(cos(d*x+c))/a/d-2/3*b^(1/3)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1 /2*c))/(a^(2/3)-b^(2/3))^(1/2))/a/d/(a^(2/3)-b^(2/3))^(1/2)+2/3*b^(1/3)*ar ctanh((b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)+ b^(2/3))^(1/2))/a/d/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)+2/3*b^(1/3)*arctanh ((b^(1/3)-(-1)^(1/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/3)+b^(2 /3))^(1/2))/a/d/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.50 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+i b \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]}{6 a d} \]
-1/6*(6*Log[Cos[(c + d*x)/2]] - 6*Log[Sin[(c + d*x)/2]] + I*b*RootSum[(-I) *b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 4*Arc Tan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]* #1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[ 1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1 ^5) & ])/(a*d)
Time = 0.69 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, 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.
| \(\displaystyle \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \left (a+b \sin (c+d x)^3\right )}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {\csc (c+d x)}{a}-\frac {b \sin ^2(c+d x)}{a \left (a+b \sin ^3(c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{b} \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 a d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 \sqrt [3]{b} \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\text {arctanh}(\cos (c+d x))}{a d}\) |
(-2*b^(1/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^( 2/3)]])/(3*a*Sqrt[a^(2/3) - b^(2/3)]*d) - ArcTanh[Cos[c + d*x]]/(a*d) + (2 *b^(1/3)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[-((- 1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*a*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]* d) + (2*b^(1/3)*ArcTanh[(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sq rt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*a*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)] *d)
3.2.86.3.1 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 = 1.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.36
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {4 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}}{d}\) | \(96\) |
| default | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {4 b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}}{d}\) | \(96\) |
| risch | \(2 i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (46656 a^{8} d^{6}-46656 b^{2} a^{6} d^{6}\right ) \textit {\_Z}^{6}-3888 b^{2} a^{4} d^{4} \textit {\_Z}^{4}-108 a^{2} b^{2} d^{2} \textit {\_Z}^{2}-b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {7776 i a^{7} d^{5}}{b^{2}}+7776 i a^{5} d^{5}\right ) \textit {\_R}^{5}+\left (-\frac {1296 i a^{5} d^{4}}{b}+1296 i a^{3} b \,d^{4}\right ) \textit {\_R}^{4}+648 i a^{3} d^{3} \textit {\_R}^{3}+\left (\frac {36 i a^{3} d^{2}}{b}+72 i a b \,d^{2}\right ) \textit {\_R}^{2}+18 i a d \textit {\_R} +\frac {i b}{a}\right )\right )-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}\) | \(222\) |
1/d*(1/a*ln(tan(1/2*d*x+1/2*c))-4/3/a*b*sum(_R^2/(_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 = 3.77 (sec) , antiderivative size = 29139, normalized size of antiderivative = 110.38 \[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc {\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
-1/2*(2*a*d*integrate(2*(16*a*b*cos(3*d*x + 3*c)^2 + 16*a*b*sin(3*d*x + 3* c)^2 + 3*b^2*cos(d*x + c)*sin(2*d*x + 2*c) - 3*b^2*cos(2*d*x + 2*c)*sin(d* x + c) + b^2*sin(d*x + c) - (b^2*sin(5*d*x + 5*c) - 2*b^2*sin(3*d*x + 3*c) + b^2*sin(d*x + c))*cos(6*d*x + 6*c) - (8*a*b*cos(3*d*x + 3*c) + 3*b^2*si n(4*d*x + 4*c) - 3*b^2*sin(2*d*x + 2*c))*cos(5*d*x + 5*c) - 3*(2*b^2*sin(3 *d*x + 3*c) - b^2*sin(d*x + c))*cos(4*d*x + 4*c) - 2*(4*a*b*cos(d*x + c) + 3*b^2*sin(2*d*x + 2*c))*cos(3*d*x + 3*c) + (b^2*cos(5*d*x + 5*c) - 2*b^2* cos(3*d*x + 3*c) + b^2*cos(d*x + c))*sin(6*d*x + 6*c) + (3*b^2*cos(4*d*x + 4*c) - 3*b^2*cos(2*d*x + 2*c) - 8*a*b*sin(3*d*x + 3*c) + b^2)*sin(5*d*x + 5*c) + 3*(2*b^2*cos(3*d*x + 3*c) - b^2*cos(d*x + c))*sin(4*d*x + 4*c) + 2 *(3*b^2*cos(2*d*x + 2*c) - 4*a*b*sin(d*x + c) - b^2)*sin(3*d*x + 3*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*si n(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)*s in(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*s in(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*...
\[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 15.20 (sec) , antiderivative size = 1439, normalized size of antiderivative = 5.45 \[ \int \frac {\csc (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
symsum(log(98304*b^5 + 1048576*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^ 4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)*b^6*tan(c/2 + (d*x)/2) - 98304*roo t(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^2*a^2*b^5 + 5898240*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2 *z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^3*a^3*b^5 - 7962624*root(729*a^6*b^2*z^ 6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^4*a^4*b^5 - 663552*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2 *z^2 - b^2, z, k)^4*a^6*b^3 - 5308416*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^5*a^5*b^5 + 10616832*root(7 29*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^5*a^7*b^3 + 7962624*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^ 4 + 27*a^2*b^2*z^2 - b^2, z, k)^6*a^6*b^5 - 9953280*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^6*a^8*b^3 - 5 89824*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^ 2 - b^2, z, k)*a*b^5 - 24576*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4* b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)*a^2*b^4*tan(c/2 + (d*x)/2) - 3145728 *root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b ^2, z, k)^2*a*b^6*tan(c/2 + (d*x)/2) + 466944*root(729*a^6*b^2*z^6 - 729*a ^8*z^6 - 243*a^4*b^2*z^4 + 27*a^2*b^2*z^2 - b^2, z, k)^2*a^3*b^4*tan(c/2 + (d*x)/2) - 18874368*root(729*a^6*b^2*z^6 - 729*a^8*z^6 - 243*a^4*b^2*z...