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

Optimal. Leaf size=78 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 \sqrt{a} d (a+b)^{3/2}}-\frac{\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.0883624, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3173, 12, 3181, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 \sqrt{a} d (a+b)^{3/2}}-\frac{\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 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 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)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{2 (a+b) d \left (a+b \sin ^2(c+d x)\right )}+\frac{\int \frac{a}{a+b \sin ^2(c+d x)} \, dx}{2 a (a+b)}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 (a+b) d \left (a+b \sin ^2(c+d x)\right )}+\frac{\int \frac{1}{a+b \sin ^2(c+d x)} \, dx}{2 (a+b)}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 (a+b) d \left (a+b \sin ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 (a+b) d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 \sqrt{a} (a+b)^{3/2} d}-\frac{\cos (c+d x) \sin (c+d x)}{2 (a+b) d \left (a+b \sin ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.517016, size = 74, normalized size = 0.95 \[ \frac{\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{3/2}}-\frac{\sin (2 (c+d x))}{(a+b) (2 a-b \cos (2 (c+d x))+b)}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]/(Sqrt[a]*(a + b)^(3/2)) - Sin[2*(c + d*x)]/((a + b)*(2*a + b - b*C
os[2*(c + d*x)])))/(2*d)

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Maple [A]  time = 0.088, size = 77, normalized size = 1. \begin{align*} -{\frac{\tan \left ( dx+c \right ) }{2\,d \left ( a+b \right ) \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) }}+{\frac{1}{2\,d \left ( a+b \right ) }\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)^2,x)

[Out]

-1/2/d/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+1/2/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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.8503, size = 965, normalized size = 12.37 \begin{align*} \left [\frac{4 \,{\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{-a^{2} - 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 + b\right )} \cos \left (d x + c\right )^{3} -{\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} - 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 )}{8 \,{\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac{2 \,{\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{a^{2} + a b} \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt{a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \,{\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}\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)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.12825, size = 147, normalized size = 1.88 \begin{align*} \frac{\frac{\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 )}{\sqrt{a^{2} + a b}{\left (a + b\right )}} - \frac{\tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}{\left (a + b\right )}}}{2 \, 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)^2,x, algorithm="giac")

[Out]

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