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

Optimal. Leaf size=75 \[ \frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \sqrt{a^2-b^2}}-\frac{a x}{b^2}-\frac{\cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.106451, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2746, 12, 2735, 2660, 618, 204} \[ \frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \sqrt{a^2-b^2}}-\frac{a x}{b^2}-\frac{\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2746

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

Rule 12

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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{\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac{\cos (c+d x)}{b d}-\frac{\int \frac{a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac{\cos (c+d x)}{b d}-\frac{a \int \frac{\sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac{a x}{b^2}-\frac{\cos (c+d x)}{b d}+\frac{a^2 \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=-\frac{a x}{b^2}-\frac{\cos (c+d x)}{b d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 d}\\ &=-\frac{a x}{b^2}-\frac{\cos (c+d x)}{b d}-\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 d}\\ &=-\frac{a x}{b^2}+\frac{2 a^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2} d}-\frac{\cos (c+d x)}{b d}\\ \end{align*}

Mathematica [A]  time = 0.179816, size = 71, normalized size = 0.95 \[ -\frac{-\frac{2 a^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a (c+d x)+b \cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 96, normalized size = 1.3 \begin{align*} -2\,{\frac{1}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{b}^{2}d}}+2\,{\frac{{a}^{2}}{{b}^{2}d\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b/(1+tan(1/2*d*x+1/2*c)^2)-2/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))+2/d/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*t
an(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*a^2

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87542, size = 609, normalized size = 8.12 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} a^{2} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} d x + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac{\sqrt{a^{2} - b^{2}} a^{2} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) +{\left (a^{3} - a b^{2}\right )} d x +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*a^2*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x
+ c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) +
2*(a^3 - a*b^2)*d*x + 2*(a^2*b - b^3)*cos(d*x + c))/((a^2*b^2 - b^4)*d), -(sqrt(a^2 - b^2)*a^2*arctan(-(a*sin
(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (a^3 - a*b^2)*d*x + (a^2*b - b^3)*cos(d*x + c))/((a^2*b^2 - b
^4)*d)]

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

[Out]

Timed out

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Giac [A]  time = 1.12139, size = 134, normalized size = 1.79 \begin{align*} \frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a^{2}}{\sqrt{a^{2} - b^{2}} b^{2}} - \frac{{\left (d x + c\right )} a}{b^{2}} - \frac{2}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a^2/(sqrt(
a^2 - b^2)*b^2) - (d*x + c)*a/b^2 - 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*b))/d