Integrand size = 12, antiderivative size = 83 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {b \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \] Output:
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*c os(f*x+e)/(a^2-b^2)/f/(a+b*sin(f*x+e))
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\frac {\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}}{f} \] Input:
Integrate[(a + b*Sin[e + f*x])^(-2),x]
Output:
((2*a*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + (b*Cos[e + f*x])/((a - b)*(a + b)*(a + b*Sin[e + f*x])))/f
Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3143, 25, 27, 3042, 3139, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(a+b \sin (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \sin (e+f x))^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {\int -\frac {a}{a+b \sin (e+f x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a}{a+b \sin (e+f x)}dx}{a^2-b^2}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {1}{a+b \sin (e+f x)}dx}{a^2-b^2}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \int \frac {1}{a+b \sin (e+f x)}dx}{a^2-b^2}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {2 a \int \frac {1}{a \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 b \tan \left (\frac {1}{2} (e+f x)\right )+a}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (a^2-b^2\right )}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (e+f x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
Input:
Int[(a + b*Sin[e + f*x])^(-2),x]
Output:
(2*a*ArcTan[(2*b + 2*a*Tan[(e + f*x)/2])/(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]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
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] + Simp [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] && IntegerQ[2*n]
Time = 0.72 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\) | \(122\) |
default | \(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\) | \(122\) |
risch | \(\frac {2 i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}}{\left (a^{2}-b^{2}\right ) f \left (b \,{\mathrm e}^{2 i \left (f x +e \right )}-b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}\) | \(221\) |
Input:
int(1/(a+b*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.10 (sec) , antiderivative size = 336, normalized size of antiderivative = 4.05 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\left [\frac {{\left (a b \sin \left (f x + e\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}}, -\frac {{\left (a b \sin \left (f x + e\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f}\right ] \] Input:
integrate(1/(a+b*sin(f*x+e))^2,x, algorithm="fricas")
Output:
[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)]
Leaf count of result is larger than twice the leaf count of optimal. 1943 vs. \(2 (66) = 132\).
Time = 43.76 (sec) , antiderivative size = 1943, normalized size of antiderivative = 23.41 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*sin(f*x+e))**2,x)
Output:
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*sq rt(-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*...
Exception generated. \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*sin(f*x+e))^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f} \] Input:
integrate(1/(a+b*sin(f*x+e))^2,x, algorithm="giac")
Output:
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
Time = 17.68 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.10 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\frac {\frac {2\,b}{a^2-b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,\left (a^2-b^2\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\left (a^2\,b-b^3\right )}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{2\,a}\right )}{f\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:
int(1/(a + b*sin(e + f*x))^2,x)
Output:
((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))
Time = 0.23 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (f x +e \right ) a b +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+\cos \left (f x +e \right ) a^{2} b -\cos \left (f x +e \right ) b^{3}}{f \left (\sin \left (f x +e \right ) a^{4} b -2 \sin \left (f x +e \right ) a^{2} b^{3}+\sin \left (f x +e \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:
int(1/(a+b*sin(f*x+e))^2,x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( e + f*x)*a*b + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2 + cos(e + f*x)*a**2*b - cos(e + f*x)*b**3)/(f*(sin(e + f*x) *a**4*b - 2*sin(e + f*x)*a**2*b**3 + sin(e + f*x)*b**5 + a**5 - 2*a**3*b** 2 + a*b**4))