3.458 \(\int \frac {1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=89 \[ -\frac {2 d \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \sqrt {c^2-d^2}}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2747, 2648, 2660, 618, 204} \[ -\frac {2 d \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \sqrt {c^2-d^2}}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])),x]

[Out]

(-2*d*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*Sqrt[c^2 - d^2]*f) - Cos[e + f*x]/((c - d)*
f*(a + a*Sin[e + f*x]))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

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 2648

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

Rule 2660

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 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx &=\frac {\int \frac {1}{a+a \sin (e+f x)} \, dx}{c-d}-\frac {d \int \frac {1}{c+d \sin (e+f x)} \, dx}{a (c-d)}\\ &=-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x))}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a (c-d) f}\\ &=-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x))}+\frac {(4 d) \operatorname {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 )}{a (c-d) f}\\ &=-\frac {2 d \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d) \sqrt {c^2-d^2} f}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.36, size = 114, normalized size = 1.28 \[ \frac {\cos (e+f x) \left (\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {\sin (e+f x)+1}}\right )}{\sqrt {-c-d} \sqrt {c-d} \sqrt {\cos ^2(e+f x)}}+\frac {1}{\sin (e+f x)+1}\right )}{a f (d-c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])),x]

[Out]

(Cos[e + f*x]*((2*d*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/(Sqrt
[-c - d]*Sqrt[c - d]*Sqrt[Cos[e + f*x]^2]) + (1 + Sin[e + f*x])^(-1)))/(a*(-c + d)*f)

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 489, normalized size = 5.49 \[ \left [\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + d \sin \left (f x + e\right ) + d\right )} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 2 \, c^{2} + 2 \, d^{2} - 2 \, {\left (c^{2} - d^{2}\right )} \cos \left (f x + e\right ) + 2 \, {\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \cos \left (f x + e\right ) + {\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \sin \left (f x + e\right ) + {\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f\right )}}, \frac {\sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + d \sin \left (f x + e\right ) + d\right )} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) - c^{2} + d^{2} - {\left (c^{2} - d^{2}\right )} \cos \left (f x + e\right ) + {\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \cos \left (f x + e\right ) + {\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \sin \left (f x + e\right ) + {\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(-c^2 + d^2)*(d*cos(f*x + e) + d*sin(f*x + e) + d)*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)) - 2*c^2 + 2*d^2 - 2*(c^2 - d^2)*cos(f*x + e) + 2*(c^2 - d^2)*sin(f*x + e))/((a
*c^3 - a*c^2*d - a*c*d^2 + a*d^3)*f*cos(f*x + e) + (a*c^3 - a*c^2*d - a*c*d^2 + a*d^3)*f*sin(f*x + e) + (a*c^3
- a*c^2*d - a*c*d^2 + a*d^3)*f), (sqrt(c^2 - d^2)*(d*cos(f*x + e) + d*sin(f*x + e) + d)*arctan(-(c*sin(f*x +
e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) - c^2 + d^2 - (c^2 - d^2)*cos(f*x + e) + (c^2 - d^2)*sin(f*x + e))/((a
*c^3 - a*c^2*d - a*c*d^2 + a*d^3)*f*cos(f*x + e) + (a*c^3 - a*c^2*d - a*c*d^2 + a*d^3)*f*sin(f*x + e) + (a*c^3
- a*c^2*d - a*c*d^2 + a*d^3)*f)]

________________________________________________________________________________________

giac [A]  time = 1.21, size = 100, normalized size = 1.12 \[ -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d}{{\left (a c - a d\right )} \sqrt {c^{2} - d^{2}}} + \frac {1}{{\left (a c - a d\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.29, size = 87, normalized size = 0.98 \[ -\frac {2 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{a f \left (c -d \right ) \sqrt {c^{2}-d^{2}}}-\frac {2}{a f \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*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 details)Is 4*d^2-4*c^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 6.99, size = 121, normalized size = 1.36 \[ \frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (2\,a\,d^2-2\,a\,c\,d\right )}{a\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}-\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c-a\,d\right )}{a\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}}{2\,d}\right )}{a\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}-\frac {2}{f\,\left (a+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (c-d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))*(c + d*sin(e + f*x))),x)

[Out]

(2*d*atan(((d*(2*a*d^2 - 2*a*c*d))/(a*(c + d)^(1/2)*(c - d)^(3/2)) - (2*c*d*tan(e/2 + (f*x)/2)*(a*c - a*d))/(a
*(c + d)^(1/2)*(c - d)^(3/2)))/(2*d)))/(a*f*(c + d)^(1/2)*(c - d)^(3/2)) - 2/(f*(a + a*tan(e/2 + (f*x)/2))*(c
- d))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________