\(\int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx\) [701]

Optimal result

Integrand size = 23, antiderivative size = 60 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\frac {d x}{b}+\frac {2 (b c-3 d) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{b \sqrt {9-b^2} f} \]

[Out]

d*x/b+2*(-a*d+b*c)*arctan((b+a*tan(1/2*f*x+1/2*e))/(a^2-b^2)^(1/2))/b/f/(a^2-b^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2814, 2739, 632, 210} \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\frac {2 (b c-a d) \arctan \left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b f \sqrt {a^2-b^2}}+\frac {d x}{b} \]

[In]

Int[(c + d*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]

[Out]

(d*x)/b + (2*(b*c - a*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2]*f)

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {d x}{b}-\frac {(-b c+a d) \int \frac {1}{a+b \sin (e+f x)} \, dx}{b} \\ & = \frac {d x}{b}+\frac {(2 (b c-a d)) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{b f} \\ & = \frac {d x}{b}-\frac {(4 (b c-a d)) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )}{b f} \\ & = \frac {d x}{b}+\frac {2 (b c-a d) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\frac {d (e+f x)+\frac {2 (b c-3 d) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\sqrt {9-b^2}}}{b f} \]

[In]

Integrate[(c + d*Sin[e + f*x])/(3 + b*Sin[e + f*x]),x]

[Out]

(d*(e + f*x) + (2*(b*c - 3*d)*ArcTan[(b + 3*Tan[(e + f*x)/2])/Sqrt[9 - b^2]])/Sqrt[9 - b^2])/(b*f)

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {\frac {2 \left (-d a +c b \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}+\frac {2 d \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b}}{f}\)	\(76\)
default	\(\frac {\frac {2 \left (-d a +c b \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}+\frac {2 d \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b}}{f}\)	\(76\)
risch	\(\frac {d x}{b}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) d a}{\sqrt {-a^{2}+b^{2}}\, f b}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) c}{\sqrt {-a^{2}+b^{2}}\, f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) d a}{\sqrt {-a^{2}+b^{2}}\, f b}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) c}{\sqrt {-a^{2}+b^{2}}\, f}\)	\(282\)

[In]

int((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*(2*(-a*d+b*c)/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))+2*d/b*arctan(tan(
1/2*f*x+1/2*e)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.22 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d f x + \sqrt {-a^{2} + b^{2}} {\left (b c - a d\right )} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} f}, \frac {{\left (a^{2} - b^{2}\right )} d f x - \sqrt {a^{2} - b^{2}} {\left (b c - a d\right )} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right )}{{\left (a^{2} b - b^{3}\right )} f}\right ] \]

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*(2*(a^2 - b^2)*d*f*x + sqrt(-a^2 + b^2)*(b*c - a*d)*log(-((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x +
e) - a^2 - b^2 - 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*
b*sin(f*x + e) - a^2 - b^2)))/((a^2*b - b^3)*f), ((a^2 - b^2)*d*f*x - sqrt(a^2 - b^2)*(b*c - a*d)*arctan(-(a*s
in(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))))/((a^2*b - b^3)*f)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (53) = 106\).

Time = 13.67 (sec) , antiderivative size = 425, normalized size of antiderivative = 7.08 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \left (c + d \sin {\left (e \right )}\right )}{\sin {\left (e \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {\frac {c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} \right )}}{f} + d x}{b} & \text {for}\: a = 0 \\\frac {c x - \frac {d \cos {\left (e + f x \right )}}{f}}{a} & \text {for}\: b = 0 \\\frac {x \left (c + d \sin {\left (e \right )}\right )}{a + b \sin {\left (e \right )}} & \text {for}\: f = 0 \\\frac {2 c}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - b f} + \frac {d f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - b f} - \frac {d f x}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - b f} + \frac {2 d}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - b f} & \text {for}\: a = - b \\- \frac {2 c}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + b f} + \frac {d f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + b f} + \frac {d f x}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + b f} + \frac {2 d}{b f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + b f} & \text {for}\: a = b \\- \frac {a d \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b f \sqrt {- a^{2} + b^{2}}} + \frac {a d \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{b f \sqrt {- a^{2} + b^{2}}} + \frac {c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{f \sqrt {- a^{2} + b^{2}}} - \frac {c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{f \sqrt {- a^{2} + b^{2}}} + \frac {d x}{b} & \text {otherwise} \end {cases} \]

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x)

[Out]

Piecewise((zoo*x*(c + d*sin(e))/sin(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((c*log(tan(e/2 + f*x/2))/f + d*x)/b,
Eq(a, 0)), ((c*x - d*cos(e + f*x)/f)/a, Eq(b, 0)), (x*(c + d*sin(e))/(a + b*sin(e)), Eq(f, 0)), (2*c/(b*f*tan
(e/2 + f*x/2) - b*f) + d*f*x*tan(e/2 + f*x/2)/(b*f*tan(e/2 + f*x/2) - b*f) - d*f*x/(b*f*tan(e/2 + f*x/2) - b*f
) + 2*d/(b*f*tan(e/2 + f*x/2) - b*f), Eq(a, -b)), (-2*c/(b*f*tan(e/2 + f*x/2) + b*f) + d*f*x*tan(e/2 + f*x/2)/
(b*f*tan(e/2 + f*x/2) + b*f) + d*f*x/(b*f*tan(e/2 + f*x/2) + b*f) + 2*d/(b*f*tan(e/2 + f*x/2) + b*f), Eq(a, b)
), (-a*d*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(b*f*sqrt(-a**2 + b**2)) + a*d*log(tan(e/2 + f*x/2
) + b/a + sqrt(-a**2 + b**2)/a)/(b*f*sqrt(-a**2 + b**2)) + c*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a
)/(f*sqrt(-a**2 + b**2)) - c*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(f*sqrt(-a**2 + b**2)) + d*x/b
, True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\frac {\frac {{\left (f x + e\right )} d}{b} + \frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (b c - a d\right )}}{\sqrt {a^{2} - b^{2}} b}}{f} \]

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

((f*x + e)*d/b + 2*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b
^2)))*(b*c - a*d)/(sqrt(a^2 - b^2)*b))/f

Mupad [B] (verification not implemented)

Time = 10.39 (sec) , antiderivative size = 343, normalized size of antiderivative = 5.72 \[ \int \frac {c+d \sin (e+f x)}{3+b \sin (e+f x)} \, dx=\frac {2\,d\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{b\,f}-\frac {a\,\left (d\,\ln \left (\frac {b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}-d\,\ln \left (\frac {b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}\right )-b\,c\,\ln \left (\frac {b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}+b\,c\,\ln \left (\frac {b\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b\,f\,\left (a^2-b^2\right )} \]

[In]

int((c + d*sin(e + f*x))/(a + b*sin(e + f*x)),x)

[Out]

(2*d*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(b*f) - (a*(d*log((b*cos(e/2 + (f*x)/2) + a*sin(e/2 + (f*x)/
2) + cos(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2))/cos(e/2 + (f*x)/2))*(-(a + b)*(a - b))^(1/2) - d*log((b*cos(e/2 + (
f*x)/2) + a*sin(e/2 + (f*x)/2) - cos(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2))/cos(e/2 + (f*x)/2))*(b^2 - a^2)^(1/2))
- b*c*log((b*cos(e/2 + (f*x)/2) + a*sin(e/2 + (f*x)/2) + cos(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2))/cos(e/2 + (f*x)
/2))*(-(a + b)*(a - b))^(1/2) + b*c*log((b*cos(e/2 + (f*x)/2) + a*sin(e/2 + (f*x)/2) - cos(e/2 + (f*x)/2)*(b^2
- a^2)^(1/2))/cos(e/2 + (f*x)/2))*(b^2 - a^2)^(1/2))/(b*f*(a^2 - b^2))