Integrand size = 23, antiderivative size = 259 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {x}{b}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b d}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b d}+\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b d} \] Output:
x/b-2/3*a^(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^(2/3)-b^(2/3))^(1/2)/b/d-2/3*a^(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^(2/ 3)+(-1)^(1/3)*b^(2/3))^(1/2)/b/d+2/3*a^(1/3)*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^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)/b/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 11.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.54 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {3 c+3 d x+2 i a \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 b d} \] Input:
Integrate[Sin[c + d*x]^3/(a + b*Sin[c + d*x]^3),x]
Output:
(3*c + 3*d*x + (2*I)*a*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)]*#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*b*d)
Time = 0.64 (sec) , antiderivative size = 259, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {1}{b}-\frac {a}{b \left (a+b \sin ^3(c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {x}{b}\) |
Input:
Int[Sin[c + d*x]^3/(a + b*Sin[c + d*x]^3),x]
Output:
x/b - (2*a^(1/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b*d) - (2*a^(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*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b*d) + (2*a^(1/3)*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*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b*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 = 0.78 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.40
| method | result | size |
| derivativedivides | \(\frac {-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 b}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(104\) |
| default | \(\frac {-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 b}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\) | \(104\) |
| risch | \(\frac {x}{b}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{6} d^{6}-729 b^{8} d^{6}\right ) \textit {\_Z}^{6}-15552 a^{2} b^{4} d^{4} \textit {\_Z}^{4}+110592 a^{2} b^{2} d^{2} \textit {\_Z}^{2}-262144 a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 i a \,b^{4} d^{5}}{16384}-\frac {243 i b^{6} d^{5}}{16384 a}\right ) \textit {\_R}^{5}+\left (-\frac {81 i a \,b^{3} d^{4}}{4096}+\frac {81 i b^{5} d^{4}}{4096 a}\right ) \textit {\_R}^{4}+\left (-\frac {135 i a \,b^{2} d^{3}}{512}-\frac {27 i b^{4} d^{3}}{512 a}\right ) \textit {\_R}^{3}+\frac {27 i a b \,d^{2} \textit {\_R}^{2}}{64}+\frac {9 i a d \textit {\_R}}{8}-\frac {2 i a}{b}\right )\right )}{8}\) | \(190\) |
Input:
int(sin(d*x+c)^3/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/b*a*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))+2/b*arct an(tan(1/2*d*x+1/2*c)))
Result contains complex when optimal does not.
Time = 1.05 (sec) , antiderivative size = 29221, normalized size of antiderivative = 112.82 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3/(a+b*sin(d*x+c)**3),x)
Output:
Timed out
\[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
(8*a*b*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))/(b^3*cos(6*d*x + 6*c)^2 + 9*b^3*cos(4*d*x + 4*c)^2 + 64*a^2* b*cos(3*d*x + 3*c)^2 + 9*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(6*d*x + 6*c)^2 + 9*b^3*sin(4*d*x + 4*c)^2 + 64*a^2*b*sin(3*d*x + 3*c)^2 - 48*a*b^2*cos(3*d *x + 3*c)*sin(2*d*x + 2*c) + 9*b^3*sin(2*d*x + 2*c)^2 - 6*b^3*cos(2*d*x + 2*c) + b^3 - 2*(3*b^3*cos(4*d*x + 4*c) - 3*b^3*cos(2*d*x + 2*c) - 8*a*b^2* sin(3*d*x + 3*c) + b^3)*cos(6*d*x + 6*c) - 6*(3*b^3*cos(2*d*x + 2*c) + 8*a *b^2*sin(3*d*x + 3*c) - b^3)*cos(4*d*x + 4*c) - 2*(8*a*b^2*cos(3*d*x + 3*c ) + 3*b^3*sin(4*d*x + 4*c) - 3*b^3*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6* (8*a*b^2*cos(3*d*x + 3*c) - 3*b^3*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16* (3*a*b^2*cos(2*d*x + 2*c) - a*b^2)*sin(3*d*x + 3*c)), x) + x)/b
\[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 37.71 (sec) , antiderivative size = 1672, normalized size of antiderivative = 6.46 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sin(c + d*x)^3/(a + b*sin(c + d*x)^3),x)
Output:
symsum(log(134217728*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^7*tan(c/2 + (d*x)/2) - 268435456*root(729* a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^ 2*a^7*b - 1073741824*a^6*tan(c/2 + (d*x)/2) + 4831838208*root(729*a^2*b^6* z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^2*a^5*b^ 3 + 33722204160*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27* a^2*b^2*z^2 + a^2, z, k)^3*a^6*b^3 + 15703474176*root(729*a^2*b^6*z^6 - 72 9*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^4*a^5*b^5 - 4831 838208*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z ^2 + a^2, z, k)^4*a^7*b^3 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^ 6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^5*a^4*b^7 + 154014842880 *root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a ^2, z, k)^5*a^6*b^5 + 35332816896*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243 *a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^6*a^5*b^7 - 21743271936*root(72 9*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k )^6*a^7*b^5 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^ 4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^4*b^9 + 122305904640*root(729*a^2* b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^ 6*b^7 + 2013265920*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^6*b - 3221225472*root(729*a^2*b^6*z^6 - 7...
\[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(sin(d*x+c)^3/(a+b*sin(d*x+c)^3),x)
Output:
int(sin(c + d*x)**3/(sin(c + d*x)**3*b + a),x)