Integrand size = 21, antiderivative size = 270 \[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}-b^{2/3}} \sqrt [3]{b} d}+\frac {2 \sqrt [3]{-1} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} \sqrt [3]{b} d}+\frac {2 (-1)^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} \sqrt [3]{b} d} \] Output:
-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^(1/3)/(a^(2/3)-b^(2/3))^(1/2)/b^(1/3)/d+2/3*(-1)^(1/3)*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^(1/3)/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)/b^(1/3)/d+2/3*(-1)^(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^(1/3)/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)/b^(1/3)/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 11.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.64 \[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {\text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 d} \] Input:
Integrate[Sin[c + d*x]/(a + b*Sin[c + d*x]^3),x]
Output:
-1/3*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] + 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) & ]/d
Time = 0.54 (sec) , antiderivative size = 270, 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.
\(\displaystyle \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{a+b \sin (c+d x)^3}dx\) |
\(\Big \downarrow \) 3699 |
\(\displaystyle \int \left (-\frac {1}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}-\frac {(-1)^{2/3}}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)\right )}+\frac {\sqrt [3]{-1}}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \sqrt [3]{-1} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 (-1)^{2/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt [3]{a} \sqrt [3]{b} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\) |
Input:
Int[Sin[c + d*x]/(a + b*Sin[c + d*x]^3),x]
Output:
(-2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/ (3*a^(1/3)*Sqrt[a^(2/3) - b^(2/3)]*b^(1/3)*d) + (2*(-1)^(1/3)*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^(1/3)*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b^(1/3)*d) + (2*(-1)^( 2/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2]))/S qrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*a^(1/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b ^(2/3)]*b^(1/3)*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 = 0.75 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.29
method | result | size |
derivativedivides | \(\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 d}\) | \(78\) |
default | \(\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R} \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 d}\) | \(78\) |
risch | \(-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (-64+\left (729 a^{4} b^{2} d^{6}-729 a^{2} b^{4} d^{6}\right ) \textit {\_Z}^{6}-972 a^{2} b^{2} d^{4} \textit {\_Z}^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {243 i d^{5} b^{2} a^{5}}{32 a^{2}+32 b^{2}}+\frac {243 i d^{5} b^{4} a^{3}}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{5}+\left (\frac {162 i d^{4} b \,a^{5}}{32 a^{2}+32 b^{2}}-\frac {162 i d^{4} b^{3} a^{3}}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{4}+\left (\frac {216 i d^{3} b^{2} a^{3}}{32 a^{2}+32 b^{2}}+\frac {108 i d^{3} b^{4} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{3}+\left (-\frac {144 i d^{2} b \,a^{3}}{32 a^{2}+32 b^{2}}-\frac {72 i d^{2} b^{3} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R}^{2}+\left (\frac {48 i d \,a^{3}}{32 a^{2}+32 b^{2}}+\frac {96 i d \,b^{2} a}{32 a^{2}+32 b^{2}}\right ) \textit {\_R} -\frac {32 i a b}{32 a^{2}+32 b^{2}}\right )\right )}{2}\) | \(338\) |
Input:
int(sin(d*x+c)/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
2/3/d*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 = 1.11 (sec) , antiderivative size = 18879, normalized size of antiderivative = 69.92 \[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)**3),x)
Output:
Integral(sin(c + d*x)/(a + b*sin(c + d*x)**3), x)
\[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
integrate(sin(d*x + c)/(b*sin(d*x + c)^3 + a), x)
\[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
integrate(sin(d*x + c)/(b*sin(d*x + c)^3 + a), x)
Time = 39.62 (sec) , antiderivative size = 652, normalized size of antiderivative = 2.41 \[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {\sum _{k=1}^6\ln \left (-8192\,a^3\,b+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^3\,b^3\,294912+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^4\,b^3\,1548288+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^5\,b^3\,1990656-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^4\,b^5\,7962624+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^6\,b^3\,5971968+65536\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^2\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,221184+{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^4\,a^4\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,2654208-{\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}^5\,a^5\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^2\,b^4\,d^6+243\,a^2\,b^2\,d^4+1,d,k\right )}{d} \] Input:
int(sin(c + d*x)/(a + b*sin(c + d*x)^3),x)
Output:
symsum(log(294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^2*a^3*b^3 - 8192*a^3*b + 1548288*root(729*a^4*b^2*d^6 - 729*a^ 2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^3*a^4*b^3 + 1990656*root(729*a^4*b^ 2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^4*a^5*b^3 - 7962624*r oot(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^5*a^4*b ^5 + 5971968*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^5*a^6*b^3 + 65536*a^2*b^2*tan(c/2 + (d*x)/2) + 196608*root(729*a^4* b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)*a^3*b^2*tan(c/2 + ( d*x)/2) + 294912*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^2*a^4*b^2*tan(c/2 + (d*x)/2) - 1769472*root(729*a^4*b^2*d^6 - 7 29*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^3*a^3*b^4*tan(c/2 + (d*x)/2) + 221184*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k )^3*a^5*b^2*tan(c/2 + (d*x)/2) + 2654208*root(729*a^4*b^2*d^6 - 729*a^2*b^ 4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^4*a^4*b^4*tan(c/2 + (d*x)/2) - 1990656* root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b^2*d^4 + 1, d, k)^5*a^5* b^4*tan(c/2 + (d*x)/2))*root(729*a^4*b^2*d^6 - 729*a^2*b^4*d^6 + 243*a^2*b ^2*d^4 + 1, d, k), k, 1, 6)/d
\[ \int \frac {\sin (c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin \left (d x +c \right )}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(sin(d*x+c)/(a+b*sin(d*x+c)^3),x)
Output:
int(sin(c + d*x)/(sin(c + d*x)**3*b + a),x)