3.501 \(\int \frac {a+b \sin (d+e x)}{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx\)

Optimal. Leaf size=23 \[ -\frac {\cos (d+e x)}{e (a \sin (d+e x)+b)} \]

[Out]

-cos(e*x+d)/e/(b+a*sin(e*x+d))

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Rubi [A]  time = 0.09, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3288, 2754, 8} \[ -\frac {\cos (d+e x)}{e (a \sin (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[d + e*x]/(e*(b + a*Sin[d + e*x])))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3288

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sin[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {a+b \sin (d+e x)}{b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)} \, dx &=\left (4 a^2\right ) \int \frac {a+b \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^2} \, dx\\ &=-\frac {\cos (d+e x)}{e (b+a \sin (d+e x))}+\frac {\int 0 \, dx}{a^2-b^2}\\ &=-\frac {\cos (d+e x)}{e (b+a \sin (d+e x))}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 1.00 \[ -\frac {\cos (d+e x)}{e (a \sin (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[d + e*x]/(e*(b + a*Sin[d + e*x])))

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fricas [A]  time = 0.83, size = 23, normalized size = 1.00 \[ -\frac {\cos \left (e x + d\right )}{a e \sin \left (e x + d\right ) + b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2),x, algorithm="fricas")

[Out]

-cos(e*x + d)/(a*e*sin(e*x + d) + b*e)

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giac [B]  time = 0.18, size = 52, normalized size = 2.26 \[ -\frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + b\right )} e^{\left (-1\right )}}{{\left (b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + b\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2),x, algorithm="giac")

[Out]

-2*(a*tan(1/2*x*e + 1/2*d) + b)*e^(-1)/((b*tan(1/2*x*e + 1/2*d)^2 + 2*a*tan(1/2*x*e + 1/2*d) + b)*b)

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maple [B]  time = 0.34, size = 52, normalized size = 2.26 \[ \frac {-\frac {2 a \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{b}-2}{e \left (b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 a \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2),x)

[Out]

2/e*(-a*tan(1/2*d+1/2*e*x)/b-1)/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)

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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((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 2.84, size = 39, normalized size = 1.70 \[ -\frac {a\,\sin \left (d+e\,x\right )+b\,\left (\cos \left (d+e\,x\right )+1\right )}{b\,e\,\left (b+a\,\sin \left (d+e\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(d + e*x))/(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x)),x)

[Out]

-(a*sin(d + e*x) + b*(cos(d + e*x) + 1))/(b*e*(b + a*sin(d + e*x)))

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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((a+b*sin(e*x+d))/(b**2+2*a*b*sin(e*x+d)+a**2*sin(e*x+d)**2),x)

[Out]

Timed out

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