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]

(-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]))

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Rubi [A]  time = 0.12999, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]

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 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 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])

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 = 0.471396, size = 115, normalized size = 1.29 \[ \frac{\cos (e+f x) \left (\frac{1}{(d-c) (\sin (e+f x)+1)}+\frac{2 d \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{\sqrt{-c-d} (d-c)^{3/2} \sqrt{\cos ^2(e+f x)}}\right )}{a f} \]

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*ArcTan[(Sqrt[-c + d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/(Sqrt
[-c - d]*(-c + d)^(3/2)*Sqrt[Cos[e + f*x]^2]) + 1/((-c + d)*(1 + Sin[e + f*x]))))/(a*f)

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Maple [A]  time = 0.073, size = 87, normalized size = 1. \begin{align*} -2\,{\frac{d}{af \left ( c-d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{1}{af \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

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)

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

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

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Fricas [B]  time = 1.70122, size = 1089, normalized size = 12.24 \begin{align*} \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 ] \end{align*}

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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [A]  time = 1.54384, size = 135, normalized size = 1.52 \begin{align*} -\frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \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} \end{align*}

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