3.3.23 \(\int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx\) [223]

Optimal. Leaf size=57 \[ \frac {x}{b}-\frac {2 a \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2814, 2739, 632, 210} \begin {gather*} \frac {x}{b}-\frac {2 a \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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*} \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac {x}{b}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}\\ &=\frac {x}{b}+\frac {(4 a) \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} (c+d x)\right )\right )}{b d}\\ &=\frac {x}{b}-\frac {2 a \tan ^{-1}\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}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 59, normalized size = 1.04 \begin {gather*} \frac {\frac {c}{d}+x-\frac {2 a \tan ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a + b*Sin[c + d*x]),x]

[Out]

(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

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Maple [A]
time = 0.06, size = 68, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(68\)
default	\(\frac {-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(68\)
risch	\(\frac {x}{b}-\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d b}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-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))+2/b*arctan(tan(1/2*d*x+1/
2*c)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.38, size = 237, normalized size = 4.16 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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*arctan(-(a*sin(d*x + c) + b)/(sqrt(a
^2 - b^2)*cos(d*x + c))))/((a^2*b - b^3)*d)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (44) = 88\).
time = 38.05, size = 335, normalized size = 5.88 \begin {gather*} \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 {b d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d \sqrt {b^{2}}} + \frac {2 b}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d \sqrt {b^{2}}} - \frac {d x \sqrt {b^{2}}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - b d \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {b^{2}} \\\frac {b d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d \sqrt {b^{2}}} + \frac {2 b}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d \sqrt {b^{2}}} + \frac {d x \sqrt {b^{2}}}{b^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + b d \sqrt {b^{2}}} & \text {for}\: a = \sqrt {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 {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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)), (b*d*x*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2) - b*d*sqrt(b**2)) + 2*b/(b**2*d*t
an(c/2 + d*x/2) - b*d*sqrt(b**2)) - d*x*sqrt(b**2)/(b**2*d*tan(c/2 + d*x/2) - b*d*sqrt(b**2)), Eq(a, -sqrt(b**
2))), (b*d*x*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2) + b*d*sqrt(b**2)) + 2*b/(b**2*d*tan(c/2 + d*x/2) + b*d*
sqrt(b**2)) + d*x*sqrt(b**2)/(b**2*d*tan(c/2 + d*x/2) + b*d*sqrt(b**2)), Eq(a, sqrt(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)) + 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))

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Giac [A]
time = 5.11, size = 77, normalized size = 1.35 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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

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Mupad [B]
time = 2.00, size = 139, normalized size = 2.44 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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