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

Optimal. Leaf size=46 \[ \frac{x}{b}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b d \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0747304, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3171, 3181, 205} \[ \frac{x}{b}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3171

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(B*x
)/b, x] + Dist[(A*b - a*B)/b, Int[1/(a + b*Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{x}{b}-\frac{a \int \frac{1}{a+b \sin ^2(c+d x)} \, dx}{b}\\ &=\frac{x}{b}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{x}{b}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b \sqrt{a+b} d}\\ \end{align*}

Mathematica [A]  time = 0.145377, size = 46, normalized size = 1. \[ \frac{-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a+b}}+c+d x}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.075, size = 50, normalized size = 1.1 \begin{align*}{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{bd}}-{\frac{a}{bd}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85861, size = 617, normalized size = 13.41 \begin{align*} \left [\frac{4 \, d x + \sqrt{-\frac{a}{a + b}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{4 \, b d}, \frac{2 \, d x + \sqrt{\frac{a}{a + b}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, b 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)^2),x, algorithm="fricas")

[Out]

[1/4*(4*d*x + sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c
)^2 + 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c
) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)))/(b*d), 1/2*(2
*d*x + sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x +
c))))/(b*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)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.19646, size = 109, normalized size = 2.37 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} a}{\sqrt{a^{2} + a b} b} - \frac{d x + c}{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)^2),x, algorithm="giac")

[Out]

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