Integrand size = 19, antiderivative size = 57 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {x}{b}-\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d} \] Output:
x/b-2*a*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(1/2) /d
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {c}{d}+x-\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}}{b} \] Input:
Integrate[Sin[c + d*x]/(a + b*Sin[c + d*x]),x]
Output:
(c/d + x - (2*a*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(Sqrt[a^ 2 - b^2]*d))/b
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3214, 3042, 3139, 1083, 217}
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.
\(\displaystyle \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {x}{b}-\frac {2 a \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x}{b}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\) |
Input:
Int[Sin[c + d*x]/(a + b*Sin[c + d*x]),x]
Output:
x/b - (2*a*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*Sq rt[a^2 - b^2]*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.46 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19
method | result | size |
derivativedivides | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}}{d}\) | \(68\) |
default | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}}{d}\) | \(68\) |
risch | \(\frac {x}{b}-\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d b}\) | \(149\) |
Input:
int(sin(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(2/b*arctan(tan(1/2*d*x+1/2*c))-2/b*a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a* tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 237, normalized size of antiderivative = 4.16 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d x - \sqrt {-a^{2} + b^{2}} a \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x + \sqrt {a^{2} - b^{2}} a \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
[1/2*(2*(a^2 - b^2)*d*x - sqrt(-a^2 + b^2)*a*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)))/((a^2*b - b^3)*d), ((a^2 - b^2)*d*x + sqrt(a^2 - b^2)*a*ar ctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))))/((a^2*b - b^3) *d)]
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (44) = 88\).
Time = 11.83 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.44 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\- \frac {\cos {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {x \sin {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\\frac {d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d} - \frac {d x}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d} + \frac {2}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d} & \text {for}\: a = - b \\\frac {d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d} + \frac {d x}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d} + \frac {2}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d} & \text {for}\: a = b \\- \frac {a \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {- a^{2} + b^{2}}} + \frac {a \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b d \sqrt {- a^{2} + b^{2}}} + \frac {x}{b} & \text {otherwise} \end {cases} \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/b, Eq(a, 0)), (-cos( c + d*x)/(a*d), Eq(b, 0)), (x*sin(c)/(a + b*sin(c)), Eq(d, 0)), (d*x*tan(c /2 + d*x/2)/(b*d*tan(c/2 + d*x/2) - b*d) - d*x/(b*d*tan(c/2 + d*x/2) - b*d ) + 2/(b*d*tan(c/2 + d*x/2) - b*d), Eq(a, -b)), (d*x*tan(c/2 + d*x/2)/(b*d *tan(c/2 + d*x/2) + b*d) + d*x/(b*d*tan(c/2 + d*x/2) + b*d) + 2/(b*d*tan(c /2 + d*x/2) + b*d), Eq(a, b)), (-a*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(b*d*sqrt(-a**2 + b**2)) + a*log(tan(c/2 + d*x/2) + b/a + sqrt (-a**2 + b**2)/a)/(b*d*sqrt(-a**2 + b**2)) + x/b, True))
Exception generated. \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.60 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b} - \frac {d x + c}{b}}{d} \] Input:
integrate(sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2 *c) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^2 - b^2)*b) - (d*x + c)/b)/d
Time = 36.43 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.44 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {x}{b}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b-3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^4}{{\left (b^2-a^2\right )}^{3/2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{b\,d\,\sqrt {b^2-a^2}} \] Input:
int(sin(c + d*x)/(a + b*sin(c + d*x)),x)
Output:
x/b - (2*a*atanh((a^4*sin(c/2 + (d*x)/2) + 2*b^4*sin(c/2 + (d*x)/2) + a*b^ 3*cos(c/2 + (d*x)/2) - a^3*b*cos(c/2 + (d*x)/2) - 3*a^2*b^2*sin(c/2 + (d*x )/2))/((b^2 - a^2)^(3/2)*(a*cos(c/2 + (d*x)/2) + 2*b*sin(c/2 + (d*x)/2)))) )/(b*d*(b^2 - a^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a +a^{2} d x -b^{2} d x}{b d \left (a^{2}-b^{2}\right )} \] Input:
int(sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a + a**2*d*x - b**2*d*x)/(b*d*(a**2 - b**2))