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

Optimal. Leaf size=17 \[ \frac {\log \left (a^2+b^2 \sin ^2(x)\right )}{b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 260} \[ \frac {\log \left (a^2+b^2 \sin ^2(x)\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/(a^2 + b^2*Sin[x]^2),x]

[Out]

Log[a^2 + b^2*Sin[x]^2]/b^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin {align*} \int \frac {\sin (2 x)}{a^2+b^2 \sin ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {2 x}{a^2+b^2 x^2} \, dx,x,\sin (x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x}{a^2+b^2 x^2} \, dx,x,\sin (x)\right )\\ &=\frac {\log \left (a^2+b^2 \sin ^2(x)\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ \frac {\log \left (a^2+b^2 \sin ^2(x)\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a^2 + b^2*Sin[x]^2]/b^2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin (2 x)}{a^2+b^2 \sin ^2(x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sin[2*x]/(a^2 + b^2*Sin[x]^2),x]

[Out]

Could not integrate

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fricas [A]  time = 1.04, size = 21, normalized size = 1.24 \[ \frac {\log \left (-b^{2} \cos \relax (x)^{2} + a^{2} + b^{2}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-b^2*cos(x)^2 + a^2 + b^2)/b^2

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giac [A]  time = 1.04, size = 17, normalized size = 1.00 \[ \frac {\log \left (b^{2} \sin \relax (x)^{2} + a^{2}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b^2*sin(x)^2 + a^2)/b^2

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maple [A]  time = 0.18, size = 18, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (a^{2}+b^{2} \left (\sin ^{2}\relax (x )\right )\right )}{b^{2}}\)	\(18\)
default	\(\frac {\ln \left (a^{2}+b^{2} \left (\sin ^{2}\relax (x )\right )\right )}{b^{2}}\)	\(18\)
risch	\(-\frac {2 i x}{b^{2}}+\frac {\ln \left ({\mathrm e}^{4 i x}-\frac {2 \left (2 a^{2}+b^{2}\right ) {\mathrm e}^{2 i x}}{b^{2}}+1\right )}{b^{2}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(a^2+b^2*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

ln(a^2+b^2*sin(x)^2)/b^2

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maxima [A]  time = 0.44, size = 17, normalized size = 1.00 \[ \frac {\log \left (b^{2} \sin \relax (x)^{2} + a^{2}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b^2*sin(x)^2 + a^2)/b^2

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mupad [B]  time = 0.61, size = 48, normalized size = 2.82 \[ \frac {\mathrm {atan}\left (\frac {b^2\,{\sin \relax (x)}^2}{a^2\,{\cos \relax (x)}^2\,2{}\mathrm {i}+a^2\,{\sin \relax (x)}^2\,2{}\mathrm {i}+b^2\,{\sin \relax (x)}^2\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(b^2*sin(x)^2 + a^2),x)

[Out]

(atan((b^2*sin(x)^2)/(a^2*cos(x)^2*2i + a^2*sin(x)^2*2i + b^2*sin(x)^2*1i))*2i)/b^2

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sympy [A]  time = 2.98, size = 32, normalized size = 1.88 \[ 2 \left (\begin {cases} - \frac {\cos ^{2}{\relax (x )}}{2 a^{2}} & \text {for}\: b^{2} = 0 \\\frac {\log {\left (a^{2} + b^{2} \sin ^{2}{\relax (x )} \right )}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a**2+b**2*sin(x)**2),x)

[Out]

2*Piecewise((-cos(x)**2/(2*a**2), Eq(b**2, 0)), (log(a**2 + b**2*sin(x)**2)/(2*b**2), True))

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