\(\int \frac {1}{(3+b \sin (e+f x))^2} \, dx\) [710]
Optimal result
Integrand size = 12, antiderivative size = 76 \[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\frac {6 \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{3/2} f}+\frac {b \cos (e+f x)}{\left (9-b^2\right ) f (3+b \sin (e+f x))}
\]
[Out]
2*a*arctan((b+a*tan(1/2*f*x+1/2*e))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/f+b*cos(f*x+e)/(a^2-b^2)/f/(a+b*sin(f*x+e
))
Rubi [A] (verified)
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2743, 12, 2739, 632, 210}
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\frac {2 a \arctan \left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}
\]
[In]
Int[(a + b*Sin[e + f*x])^(-2),x]
[Out]
(2*a*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(3/2)*f) + (b*Cos[e + f*x])/((a^2 - b^2)*f
*(a + b*Sin[e + f*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 (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {\int \frac {a}{a+b \sin (e+f x)} \, dx}{-a^2+b^2} \\ & = \frac {b \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {a \int \frac {1}{a+b \sin (e+f x)} \, dx}{a^2-b^2} \\ & = \frac {b \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(2 a) \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 )}{\left (a^2-b^2\right ) f} \\ & = \frac {b \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f 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 {1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right ) f} \\ & = \frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {b \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=-\frac {\frac {6 \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\sqrt {9-b^2}}+\frac {b \cos (e+f x)}{3+b \sin (e+f x)}}{(-3+b) (3+b) f}
\]
[In]
Integrate[(3 + b*Sin[e + f*x])^(-2),x]
[Out]
-(((6*ArcTan[(b + 3*Tan[(e + f*x)/2])/Sqrt[9 - b^2]])/Sqrt[9 - b^2] + (b*Cos[e + f*x])/(3 + b*Sin[e + f*x]))/(
(-3 + b)*(3 + b)*f))
Maple [A] (verified)
Time = 0.76 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61
| | |
| method | result | size |
| | |
| derivativedivides |
\(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\) |
\(122\) |
| default |
\(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\) |
\(122\) |
| risch |
\(\frac {2 i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}}{\left (a^{2}-b^{2}\right ) f \left (b \,{\mathrm e}^{2 i \left (f x +e \right )}-b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}\) |
\(221\) |
| | |
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[In]
int(1/(a+b*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
[Out]
1/f*(2*(b^2/a/(a^2-b^2)*tan(1/2*f*x+1/2*e)+b/(a^2-b^2))/(tan(1/2*f*x+1/2*e)^2*a+2*b*tan(1/2*f*x+1/2*e)+a)+2*a/
(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2)))
Fricas [A] (verification not implemented)
none
Time = 0.35 (sec) , antiderivative size = 336, normalized size of antiderivative = 4.42
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\left [\frac {{\left (a b \sin \left (f x + e\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}}, -\frac {{\left (a b \sin \left (f x + e\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f}\right ]
\]
[In]
integrate(1/(a+b*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
[1/2*((a*b*sin(f*x + e) + a^2)*sqrt(-a^2 + b^2)*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)) + 2*(a^2*b - b^3)*cos(f*x + e))/((a^4*b - 2*a^2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b
^2 + a*b^4)*f), -((a*b*sin(f*x + e) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f
*x + e))) - (a^2*b - b^3)*cos(f*x + e))/((a^4*b - 2*a^2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*
f)]
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1943 vs. \(2 (66) = 132\).
Time = 42.50 (sec) , antiderivative size = 1943, normalized size of antiderivative = 25.57
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\text {Too large to display}
\]
[In]
integrate(1/(a+b*sin(f*x+e))**2,x)
[Out]
Piecewise((zoo*x/sin(e)**2, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((tan(e/2 + f*x/2)/(2*f) - 1/(2*f*tan(e/2 + f*x/2
)))/b**2, Eq(a, 0)), (-6*tan(e/2 + f*x/2)**2/(3*b**2*f*tan(e/2 + f*x/2)**3 - 9*b**2*f*tan(e/2 + f*x/2)**2 + 9*
b**2*f*tan(e/2 + f*x/2) - 3*b**2*f) + 6*tan(e/2 + f*x/2)/(3*b**2*f*tan(e/2 + f*x/2)**3 - 9*b**2*f*tan(e/2 + f*
x/2)**2 + 9*b**2*f*tan(e/2 + f*x/2) - 3*b**2*f) - 4/(3*b**2*f*tan(e/2 + f*x/2)**3 - 9*b**2*f*tan(e/2 + f*x/2)*
*2 + 9*b**2*f*tan(e/2 + f*x/2) - 3*b**2*f), Eq(a, -b)), (-6*tan(e/2 + f*x/2)**2/(3*b**2*f*tan(e/2 + f*x/2)**3
+ 9*b**2*f*tan(e/2 + f*x/2)**2 + 9*b**2*f*tan(e/2 + f*x/2) + 3*b**2*f) - 6*tan(e/2 + f*x/2)/(3*b**2*f*tan(e/2
+ f*x/2)**3 + 9*b**2*f*tan(e/2 + f*x/2)**2 + 9*b**2*f*tan(e/2 + f*x/2) + 3*b**2*f) - 4/(3*b**2*f*tan(e/2 + f*x
/2)**3 + 9*b**2*f*tan(e/2 + f*x/2)**2 + 9*b**2*f*tan(e/2 + f*x/2) + 3*b**2*f), Eq(a, b)), (x/(a + b*sin(e))**2
, Eq(f, 0)), (a**3*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(e/2 + f*x/2)**2/(a**4*f*sqrt(-a**2 +
b**2)*tan(e/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2
*b**2*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2
)*tan(e/2 + f*x/2)) + a**3*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**4*f*sqrt(-a**2 + b**2)*tan(e
/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**2*f*sqrt
(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*tan(e/2 +
f*x/2)) - a**3*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(e/2 + f*x/2)**2/(a**4*f*sqrt(-a**2 + b**
2)*tan(e/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**
2*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*ta
n(e/2 + f*x/2)) - a**3*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**4*f*sqrt(-a**2 + b**2)*tan(e/2 +
f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**2*f*sqrt(-a*
*2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/
2)) + 2*a**2*b*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(e/2 + f*x/2)/(a**4*f*sqrt(-a**2 + b**2)*
tan(e/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**2*f
*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*tan(e
/2 + f*x/2)) - 2*a**2*b*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(e/2 + f*x/2)/(a**4*f*sqrt(-a**2
+ b**2)*tan(e/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a*
*2*b**2*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b*
*2)*tan(e/2 + f*x/2)) + 2*a*b*sqrt(-a**2 + b**2)/(a**4*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 + a**4*f*sqrt(
-a**2 + b**2) + 2*a**3*b*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**2*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/
2)**2 - a**2*b**2*f*sqrt(-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)) + 2*b**2*sqrt(-a**2 +
b**2)*tan(e/2 + f*x/2)/(a**4*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 + a**4*f*sqrt(-a**2 + b**2) + 2*a**3*b*
f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2) - a**2*b**2*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)**2 - a**2*b**2*f*sqrt(
-a**2 + b**2) - 2*a*b**3*f*sqrt(-a**2 + b**2)*tan(e/2 + f*x/2)), True))
Maxima [F(-2)]
Exception generated. \[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(1/(a+b*sin(f*x+e))^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 = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f}
\]
[In]
integrate(1/(a+b*sin(f*x+e))^2,x, algorithm="giac")
[Out]
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)))*a/(a^2 - b
^2)^(3/2) + (b^2*tan(1/2*f*x + 1/2*e) + a*b)/((a^3 - a*b^2)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*
e) + a)))/f
Mupad [B] (verification not implemented)
Time = 8.20 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.29
\[
\int \frac {1}{(3+b \sin (e+f x))^2} \, dx=\frac {\frac {2\,b}{a^2-b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,\left (a^2-b^2\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\left (a^2\,b-b^3\right )}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{2\,a}\right )}{f\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}
\]
[In]
int(1/(a + b*sin(e + f*x))^2,x)
[Out]
((2*b)/(a^2 - b^2) + (2*b^2*tan(e/2 + (f*x)/2))/(a*(a^2 - b^2)))/(f*(a + 2*b*tan(e/2 + (f*x)/2) + a*tan(e/2 +
(f*x)/2)^2)) + (2*a*atan(((a^2 - b^2)*((2*a^2*tan(e/2 + (f*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)))/(f*(a + b)^(3/2)*(a - b)^(3/2))