\(\int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx\) [675]

Optimal result

Integrand size = 23, antiderivative size = 64 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\frac {b x}{d}-\frac {2 (b c-3 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d \sqrt {c^2-d^2} f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02, 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} \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\frac {b x}{d}-\frac {2 (b c-a d) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}} \]

[In]

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

[Out]

(b*x)/d - (2*(b*c - a*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d*Sqrt[c^2 - d^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 {b x}{d}-\frac {(b c-a d) \int \frac {1}{c+d \sin (e+f x)} \, dx}{d} \\ & = \frac {b x}{d}-\frac {(2 (b c-a d)) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f} \\ & = \frac {b x}{d}+\frac {(4 (b c-a d)) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f} \\ & = \frac {b x}{d}-\frac {2 (b c-a d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d \sqrt {c^2-d^2} f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\frac {b (e+f x)+\frac {(-2 b c+6 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{d f} \]

[In]

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

[Out]

(b*(e + f*x) + ((-2*b*c + 6*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2])/(d*f)

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.19

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.98 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\left [\frac {2 \, {\left (b c^{2} - b d^{2}\right )} f x + {\left (b c - a d\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right )}{2 \, {\left (c^{2} d - d^{3}\right )} f}, \frac {{\left (b c^{2} - b d^{2}\right )} f x + {\left (b c - a d\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right )}{{\left (c^{2} d - d^{3}\right )} f}\right ] \]

[In]

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

[Out]

[1/2*(2*(b*c^2 - b*d^2)*f*x + (b*c - a*d)*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x +
e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c
*d*sin(f*x + e) - c^2 - d^2)))/((c^2*d - d^3)*f), ((b*c^2 - b*d^2)*f*x + (b*c - a*d)*sqrt(c^2 - d^2)*arctan(-(
c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))))/((c^2*d - d^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 = 13.18 (sec) , antiderivative size = 425, normalized size of antiderivative = 6.64 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \left (a + b \sin {\left (e \right )}\right )}{\sin {\left (e \right )}} & \text {for}\: c = 0 \wedge d = 0 \wedge f = 0 \\\frac {\frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} \right )}}{f} + b x}{d} & \text {for}\: c = 0 \\\frac {a x - \frac {b \cos {\left (e + f x \right )}}{f}}{c} & \text {for}\: d = 0 \\\frac {x \left (a + b \sin {\left (e \right )}\right )}{c + d \sin {\left (e \right )}} & \text {for}\: f = 0 \\\frac {2 a}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - d f} + \frac {b f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - d f} - \frac {b f x}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - d f} + \frac {2 b}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - d f} & \text {for}\: c = - d \\- \frac {2 a}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + d f} + \frac {b f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + d f} + \frac {b f x}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + d f} + \frac {2 b}{d f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + d f} & \text {for}\: c = d \\\frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} - \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{f \sqrt {- c^{2} + d^{2}}} - \frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} + \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{f \sqrt {- c^{2} + d^{2}}} - \frac {b c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} - \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{d f \sqrt {- c^{2} + d^{2}}} + \frac {b c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} + \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{d f \sqrt {- c^{2} + d^{2}}} + \frac {b x}{d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((a+b*sin(f*x+e))/(c+d*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*d^2-4*c^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\frac {\frac {{\left (f x + e\right )} b}{d} - \frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (b c - a d\right )}}{\sqrt {c^{2} - d^{2}} d}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.63 (sec) , antiderivative size = 342, normalized size of antiderivative = 5.34 \[ \int \frac {3+b \sin (e+f x)}{c+d \sin (e+f x)} \, dx=\frac {2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{d\,f}+\frac {c\,\left (b\,\ln \left (\frac {d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {-\left (c+d\right )\,\left (c-d\right )}-b\,\ln \left (\frac {d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {d^2-c^2}\right )-a\,d\,\ln \left (\frac {d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {-\left (c+d\right )\,\left (c-d\right )}+a\,d\,\ln \left (\frac {d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\,\sqrt {d^2-c^2}}{d\,f\,\left (c^2-d^2\right )} \]

[In]

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

[Out]

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