3.459 \(\int \frac {1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=150 \[ -\frac {2 d (2 c+d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}-\frac {d (c+2 d) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2768, 2754, 12, 2660, 618, 204} \[ -\frac {2 d (2 c+d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}-\frac {d (c+2 d) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2),x]

[Out]

(-2*d*(2*c + d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 - d^2)^(3/2)*f) - (d*(c + 2*
d)*Cos[e + f*x])/(a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])) - Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])*(
c + d*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 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 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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2768

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

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 162, normalized size = 1.08 \[ \frac {\cos (e+f x) \left (-\frac {d}{(\sin (e+f x)+1) (c+d \sin (e+f x))}+\frac {c+2 d}{(c-d) (\sin (e+f x)+1)}+\frac {2 d (2 c+d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {\sin (e+f x)+1}}\right )}{\sqrt {-c-d} (c-d)^{3/2} \sqrt {\cos ^2(e+f x)}}\right )}{a f (d-c) (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2),x]

[Out]

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

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fricas [B]  time = 0.49, size = 1120, normalized size = 7.47 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

[1/2*(2*c^4 - 4*c^2*d^2 + 2*d^4 + 2*(c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e)^2 + (2*c^2*d + 3*c*d^2 +
d^3 - (2*c*d^2 + d^3)*cos(f*x + e)^2 + (2*c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 +
d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-c^2 + d^2)*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^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e) - 2*(c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2
*d^2 - c*d^3 - 2*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5
- a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e)
- (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5
- a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e)), (c^4 - 2*c^2*d^2 + d^
4 + (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e)^2 - (2*c^2*d + 3*c*d^2 + d^3 - (2*c*d^2 + d^3)*cos(f*x +
e)^2 + (2*c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))
*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (c^4 + c^3*d - c*d^3 - d^4)*co
s(f*x + e) - (c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*
d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2
+ 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5
*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*
d^4 - a*d^6)*f)*sin(f*x + e))]

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giac [B]  time = 0.34, size = 311, normalized size = 2.07 \[ -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (2 \, c d + d^{2}\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c^{3} + c^{2} d + c d^{2}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \]

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))^2,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)))*(2*c*d +
d^2)/((a*c^3 - a*c^2*d - a*c*d^2 + a*d^3)*sqrt(c^2 - d^2)) + (c^3*tan(1/2*f*x + 1/2*e)^2 + c^2*d*tan(1/2*f*x +
1/2*e)^2 + d^3*tan(1/2*f*x + 1/2*e)^2 + 2*c^2*d*tan(1/2*f*x + 1/2*e) + 3*c*d^2*tan(1/2*f*x + 1/2*e) + d^3*tan
(1/2*f*x + 1/2*e) + c^3 + c^2*d + c*d^2)/((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2*e)^3 +
c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 1/2*e) + 2*d*tan(1/2*f*x + 1/2*e) + c)
))/f

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maple [A]  time = 0.34, size = 273, normalized size = 1.82 \[ -\frac {2 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right ) c}-\frac {2 d^{2}}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}-\frac {4 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) c}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {2}{a f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

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))^2,x)

[Out]

-2/a/f*d^3/(c-d)^2/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)/c*tan(1/2*f*x+1/2*e)-2/a/f*d^2/(c-d
)^2/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)-4/a/f*d/(c-d)^2/(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))*c-2/a/f*d^2/(c-d)^2/(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)^2/(tan(1/2*f*x+1/2*e)+1)

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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))^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*d^2-4*c^2>0)', see `assume?`
for more details)Is 4*d^2-4*c^2 positive or negative?

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mupad [B]  time = 8.57, size = 309, normalized size = 2.06 \[ -\frac {\frac {2\,\left (c^2+c\,d+d^2\right )}{\left (c+d\right )\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (d^2+2\,c\,d\right )}{c\,{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^3+c^2\,d+d^3\right )}{c\,\left (c+d\right )\,{\left (c-d\right )}^2}}{f\,\left (a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (a\,c+2\,a\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,c\right )}-\frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (2\,c+d\right )\,\left (2\,a\,c^3\,d-2\,a\,c^2\,d^2-2\,a\,c\,d^3+2\,a\,d^4\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c+d\right )\,\left (a\,c^3-a\,c^2\,d-a\,c\,d^2+a\,d^3\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}}{2\,d^2+4\,c\,d}\right )\,\left (2\,c+d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}} \]

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))^2),x)

[Out]

- ((2*(c*d + c^2 + d^2))/((c + d)*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(2*c*d + d^2))/(c*(c - d)^2) + (2*tan(e/2
+ (f*x)/2)^2*(c^2*d + c^3 + d^3))/(c*(c + d)*(c - d)^2))/(f*(a*c + tan(e/2 + (f*x)/2)^2*(a*c + 2*a*d) + tan(e
/2 + (f*x)/2)*(a*c + 2*a*d) + a*c*tan(e/2 + (f*x)/2)^3)) - (2*d*atan(((d*(2*c + d)*(2*a*d^4 - 2*a*c^2*d^2 - 2*
a*c*d^3 + 2*a*c^3*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)) + (2*c*d*tan(e/2 + (f*x)/2)*(2*c + d)*(a*c^3 + a*d^3 - a
*c*d^2 - a*c^2*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)))/(4*c*d + 2*d^2))*(2*c + d))/(a*f*(c + d)^(3/2)*(c - d)^(5/
2))

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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))**2,x)

[Out]

Timed out

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