Integrand size = 23, antiderivative size = 287 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 b \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-b^{2/3}} d}-\frac {2 b \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}+\frac {2 b \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{5/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]
-1/2*arctanh(cos(d*x+c))/a/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d-2/3*b*arctan((b ^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/a^(5/3)/d/(a^( 2/3)-b^(2/3))^(1/2)-2/3*b*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^(5/3)/d/(a^(2/3)+(-1)^(1/3)*b ^(2/3))^(1/2)+2/3*b*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^(5/3)/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 = 0.40 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.63 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {16 i b \text {RootSum}\left [-b+3 b \text {$\#$1}^2-8 i a \text {$\#$1}^3-3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]-3 \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )+4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )}{24 a d} \]
((16*I)*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)]*#1 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ] - 3*(Csc[(c + d*x)/2] ^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] - Sec[(c + d*x)/2]^ 2))/(24*a*d)
Time = 0.62 (sec) , antiderivative size = 287, 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.
| \(\displaystyle \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^3 \left (a+b \sin (c+d x)^3\right )}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {\csc ^3(c+d x)}{a}-\frac {b}{a \left (a+b \sin ^3(c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 b \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 b \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{5/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {\text {arctanh}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
(-2*b*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]] )/(3*a^(5/3)*Sqrt[a^(2/3) - b^(2/3)]*d) - (2*b*ArcTan[((-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^(5/3 )*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) + (2*b*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^(5/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*d) - ArcTanh[Cos[c + d*x ]]/(2*a*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)
3.2.87.3.1 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 = 1.49 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.47
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{d}\) | \(136\) |
| default | \(\frac {\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 a}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{d}\) | \(136\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-8 i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (191102976 a^{12} d^{6}-191102976 a^{10} b^{2} d^{6}\right ) \textit {\_Z}^{6}-995328 a^{8} b^{2} d^{4} \textit {\_Z}^{4}+1728 a^{4} b^{4} d^{2} \textit {\_Z}^{2}-b^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {15925248 i d^{5} a^{11}}{b^{6}}+\frac {15925248 i d^{5} a^{9}}{b^{4}}\right ) \textit {\_R}^{5}+\left (-\frac {331776 i d^{4} a^{9}}{b^{5}}+\frac {331776 i d^{4} a^{7}}{b^{3}}\right ) \textit {\_R}^{4}+\left (\frac {69120 i d^{3} a^{7}}{b^{4}}+\frac {13824 i d^{3} a^{5}}{b^{2}}\right ) \textit {\_R}^{3}+\frac {1728 i d^{2} a^{5} \textit {\_R}^{2}}{b^{3}}-\frac {72 i d \,a^{3} \textit {\_R}}{b^{2}}-\frac {2 i a}{b}\right )\right )+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}\) | \(279\) |
1/d*(1/8*tan(1/2*d*x+1/2*c)^2/a-1/3/a*b*sum((_R^4+2*_R^2+1)/(_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))-1/8/a/tan(1/2*d*x+1/2*c)^2+1/2/a*ln(tan(1/2*d*x+1/2*c)))
Result contains complex when optimal does not.
Time = 14.83 (sec) , antiderivative size = 29431, normalized size of antiderivative = 102.55 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\csc ^{3}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
1/4*(4*(cos(3*d*x + 3*c) + cos(d*x + c))*cos(4*d*x + 4*c) - 4*(2*cos(2*d*x + 2*c) - 1)*cos(3*d*x + 3*c) - 8*cos(2*d*x + 2*c)*cos(d*x + c) + 32*(a*b* d*cos(4*d*x + 4*c)^2 + 4*a*b*d*cos(2*d*x + 2*c)^2 + a*b*d*sin(4*d*x + 4*c) ^2 - 4*a*b*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*a*b*d*sin(2*d*x + 2*c)^ 2 - 4*a*b*d*cos(2*d*x + 2*c) + a*b*d - 2*(2*a*b*d*cos(2*d*x + 2*c) - a*b*d )*cos(4*d*x + 4*c))*integrate(-(8*a*cos(3*d*x + 3*c)^2 - b*cos(3*d*x + 3*c )*sin(6*d*x + 6*c) + 3*b*cos(3*d*x + 3*c)*sin(4*d*x + 4*c) + b*cos(6*d*x + 6*c)*sin(3*d*x + 3*c) - 3*b*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) + 8*a*sin(3 *d*x + 3*c)^2 - 3*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + (3*b*cos(2*d*x + 2 *c) - b)*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*s in(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*(2*cos(2*d*x + 2*c) - 1)*cos(4*d*x + 4*c) ...
\[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\csc \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 15.06 (sec) , antiderivative size = 1573, normalized size of antiderivative = 5.48 \[ \int \frac {\csc ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
symsum(log(-(65536*a*b^9 - 262144*b^10*tan(c/2 + (d*x)/2) - 131072*root(72 9*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)*a^2*b^9 - 61440*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)*a^4*b^7 + 860160*root(729*a^10*b^2*z^6 - 72 9*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^2*a^5*b^7 - 324 4032*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z ^2 - b^6, z, k)^3*a^6*b^7 - 1105920*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^3*a^8*b^5 + 3538944*root(72 9*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^4*a^7*b^7 + 3870720*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2 *z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^4*a^9*b^5 + 663552*root(729*a^10*b^2*z^ 6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^5*a^10*b^ 5 - 4976640*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^ 4*b^4*z^2 - b^6, z, k)^5*a^12*b^3 - 7962624*root(729*a^10*b^2*z^6 - 729*a^ 12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^11*b^5 + 995328 0*root(729*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^6*a^13*b^3 + 24576*a^2*b^8*tan(c/2 + (d*x)/2) + 540672*root(7 29*a^10*b^2*z^6 - 729*a^12*z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z , k)*a^3*b^8*tan(c/2 + (d*x)/2) - 7077888*root(729*a^10*b^2*z^6 - 729*a^12 *z^6 - 243*a^8*b^2*z^4 - 27*a^4*b^4*z^2 - b^6, z, k)^2*a^4*b^8*tan(c/2 ...