\(\int \frac {1}{(a+b \sin (x))^2} \, dx\) [189]

Optimal result

Integrand size = 8, antiderivative size = 65 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

2*a*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)+b*cos(x)/(a^2-b^2)/(a+b*sin(x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2743, 12, 2739, 632, 210} \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\frac {2 a \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[In]

Int[(a + b*Sin[x])^(-2),x]

[Out]

(2*a*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + (b*Cos[x])/((a^2 - b^2)*(a + b*Sin[x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {a}{a+b \sin (x)} \, dx}{-a^2+b^2} \\ & = \frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac {a \int \frac {1}{a+b \sin (x)} \, dx}{a^2-b^2} \\ & = \frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^2-b^2} \\ & = \frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\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 {x}{2}\right )\right )}{a^2-b^2} \\ & = \frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \]

[In]

Integrate[(a + b*Sin[x])^(-2),x]

[Out]

(2*a*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + (b*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x]))

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {\frac {2 b^{2} \tan \left (\frac {x}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\)	\(98\)
risch	\(\frac {2 i b +2 a \,{\mathrm e}^{i x}}{\left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}\)	\(192\)

[In]

int(1/(a+b*sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

2*(b^2/a/(a^2-b^2)*tan(1/2*x)+b/(a^2-b^2))/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)+2*a/(a^2-b^2)^(3/2)*arctan(1/2*(2
*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 268, normalized size of antiderivative = 4.12 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\left [\frac {{\left (a b \sin \left (x\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}}, -\frac {{\left (a b \sin \left (x\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}\right ] \]

[In]

integrate(1/(a+b*sin(x))^2,x, algorithm="fricas")

[Out]

[1/2*((a*b*sin(x) + a^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 - 2*(a*cos(x
)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*(a^2*b - b^3)*cos(x))/(a
^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x)), -((a*b*sin(x) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*si
n(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - (a^2*b - b^3)*cos(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^
5)*sin(x))]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1474 vs. \(2 (51) = 102\).

Time = 35.75 (sec) , antiderivative size = 1474, normalized size of antiderivative = 22.68 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*sin(x))**2,x)

[Out]

Piecewise((zoo*(tan(x/2)/2 - 1/(2*tan(x/2))), Eq(a, 0) & Eq(b, 0)), ((tan(x/2)/2 - 1/(2*tan(x/2)))/b**2, Eq(a,
0)), (-6*tan(x/2)**2/(3*b**2*tan(x/2)**3 - 9*b**2*tan(x/2)**2 + 9*b**2*tan(x/2) - 3*b**2) + 6*tan(x/2)/(3*b**
2*tan(x/2)**3 - 9*b**2*tan(x/2)**2 + 9*b**2*tan(x/2) - 3*b**2) - 4/(3*b**2*tan(x/2)**3 - 9*b**2*tan(x/2)**2 +
9*b**2*tan(x/2) - 3*b**2), Eq(a, -b)), (-6*tan(x/2)**2/(3*b**2*tan(x/2)**3 + 9*b**2*tan(x/2)**2 + 9*b**2*tan(x
/2) + 3*b**2) - 6*tan(x/2)/(3*b**2*tan(x/2)**3 + 9*b**2*tan(x/2)**2 + 9*b**2*tan(x/2) + 3*b**2) - 4/(3*b**2*ta
n(x/2)**3 + 9*b**2*tan(x/2)**2 + 9*b**2*tan(x/2) + 3*b**2), Eq(a, b)), (a**3*log(tan(x/2) + b/a - sqrt(-a**2 +
b**2)/a)*tan(x/2)**2/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b
**2)*tan(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2
+ b**2)*tan(x/2)) + a**3*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**
4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*tan(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*
b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2)*tan(x/2)) - a**3*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)
/a)*tan(x/2)**2/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*t
an(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2
)*tan(x/2)) - a**3*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**4*sqrt
(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*tan(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*b**2*s
qrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2)*tan(x/2)) + 2*a**2*b*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a
)*tan(x/2)/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*tan(x/
2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2)*tan
(x/2)) - 2*a**2*b*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(x/2)/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a
**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*tan(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**
2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2)*tan(x/2)) + 2*a*b*sqrt(-a**2 + b**2)/(a**4*sqrt(-a**2
+ b**2)*tan(x/2)**2 + a**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*tan(x/2) - a**2*b**2*sqrt(-a**2 +
b**2)*tan(x/2)**2 - a**2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2)*tan(x/2)) + 2*b**2*sqrt(-a**2 +
b**2)*tan(x/2)/(a**4*sqrt(-a**2 + b**2)*tan(x/2)**2 + a**4*sqrt(-a**2 + b**2) + 2*a**3*b*sqrt(-a**2 + b**2)*t
an(x/2) - a**2*b**2*sqrt(-a**2 + b**2)*tan(x/2)**2 - a**2*b**2*sqrt(-a**2 + b**2) - 2*a*b**3*sqrt(-a**2 + b**2
)*tan(x/2)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*sin(x))^2,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 = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, x\right ) + a b\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} \]

[In]

integrate(1/(a+b*sin(x))^2,x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*a/(a^2 - b^2)^(3/2) + 2*(b^2*
tan(1/2*x) + a*b)/((a^3 - a*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a))

Mupad [B] (verification not implemented)

Time = 6.71 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(a+b \sin (x))^2} \, dx=\frac {\frac {2\,b}{a^2-b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{a\,\left (a^2-b^2\right )}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}+\frac {2\,a\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\left (a^2\,b-b^3\right )}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{2\,a}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

[In]

int(1/(a + b*sin(x))^2,x)

[Out]

((2*b)/(a^2 - b^2) + (2*b^2*tan(x/2))/(a*(a^2 - b^2)))/(a + 2*b*tan(x/2) + a*tan(x/2)^2) + (2*a*atan(((a^2 - b
^2)*((2*a^2*tan(x/2))/((a + b)^(3/2)*(a - b)^(3/2)) + (2*a*(a^2*b - b^3))/((a + b)^(3/2)*(a^2 - b^2)*(a - b)^(
3/2))))/(2*a)))/((a + b)^(3/2)*(a - b)^(3/2))