3.108 \(\int \frac{\sin ^2(c+d x)}{(a+b \sin ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=131 \[ \frac{(4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{3/2} d (a+b)^{5/2}}-\frac{(2 a-b) \sin (c+d x) \cos (c+d x)}{8 a d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}-\frac{\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]

[Out]

((4*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*a^(3/2)*(a + b)^(5/2)*d) - (Cos[c + d*x]*Sin[c + d*x
])/(4*(a + b)*d*(a + b*Sin[c + d*x]^2)^2) - ((2*a - b)*Cos[c + d*x]*Sin[c + d*x])/(8*a*(a + b)^2*d*(a + b*Sin[
c + d*x]^2))

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Rubi [A]  time = 0.14898, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3173, 12, 3181, 205} \[ \frac{(4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{3/2} d (a+b)^{5/2}}-\frac{(2 a-b) \sin (c+d x) \cos (c+d x)}{8 a d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )}-\frac{\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

((4*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(8*a^(3/2)*(a + b)^(5/2)*d) - (Cos[c + d*x]*Sin[c + d*x
])/(4*(a + b)*d*(a + b*Sin[c + d*x]^2)^2) - ((2*a - b)*Cos[c + d*x]*Sin[c + d*x])/(8*a*(a + b)^2*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 )^3} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{\int \frac{a+2 a \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx}{4 a (a+b)}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{(2 a-b) \cos (c+d x) \sin (c+d x)}{8 a (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac{\int \frac{a (4 a+b)}{a+b \sin ^2(c+d x)} \, dx}{8 a^2 (a+b)^2}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{(2 a-b) \cos (c+d x) \sin (c+d x)}{8 a (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac{(4 a+b) \int \frac{1}{a+b \sin ^2(c+d x)} \, dx}{8 a (a+b)^2}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{(2 a-b) \cos (c+d x) \sin (c+d x)}{8 a (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}+\frac{(4 a+b) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 a (a+b)^2 d}\\ &=\frac{(4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{8 a^{3/2} (a+b)^{5/2} d}-\frac{\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{(2 a-b) \cos (c+d x) \sin (c+d x)}{8 a (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.37209, size = 112, normalized size = 0.85 \[ \frac{\frac{(4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2} (a+b)^{5/2}}-\frac{\sin (2 (c+d x)) \left (8 a^2+b (b-2 a) \cos (2 (c+d x))+4 a b-b^2\right )}{a (a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2/(a + b*Sin[c + d*x]^2)^3,x]

[Out]

(((4*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(3/2)*(a + b)^(5/2)) - ((8*a^2 + 4*a*b - b^2 + b*(-
2*a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(a*(a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2))/(8*d)

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Maple [B]  time = 0.088, size = 278, normalized size = 2.1 \begin{align*} -{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ( a+b \right ) }}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}b}{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2}a \left ( a+b \right ) }}-{\frac{a\tan \left ( dx+c \right ) }{2\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{\tan \left ( dx+c \right ) b}{8\,d \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{1}{2\,d \left ({a}^{2}+2\,ab+{b}^{2} \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 ) }}}}+{\frac{b}{8\,d \left ({a}^{2}+2\,ab+{b}^{2} \right ) a}\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)^3,x)

[Out]

-1/2/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/(a+b)*tan(d*x+c)^3+1/8/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2/a/(a+b
)*tan(d*x+c)^3*b-1/2/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2*a/(a^2+2*a*b+b^2)*tan(d*x+c)-1/8/d/(a*tan(d*x+c)^2+
tan(d*x+c)^2*b+a)^2/(a^2+2*a*b+b^2)*tan(d*x+c)*b+1/2/d/(a^2+2*a*b+b^2)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)
/(a*(a+b))^(1/2))+1/8/d/(a^2+2*a*b+b^2)/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.18566, size = 1719, normalized size = 13.12 \begin{align*} \text{result too large to display} \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)^3,x, algorithm="fricas")

[Out]

[-1/32*(((4*a*b^2 + b^3)*cos(d*x + c)^4 + 4*a^3 + 9*a^2*b + 6*a*b^2 + b^3 - 2*(4*a^2*b + 5*a*b^2 + b^3)*cos(d*
x + c)^2)*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)) - 4*((2*a^3*b + a^2*b^2 - a*b^3)*cos(d*x +
c)^3 - (4*a^4 + 7*a^3*b + 2*a^2*b^2 - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 +
a^2*b^5)*d*cos(d*x + c)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*d*cos(d*x + c)^2 + (a^7 +
5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*d), -1/16*(((4*a*b^2 + b^3)*cos(d*x + c)^4 + 4*a^3 +
9*a^2*b + 6*a*b^2 + b^3 - 2*(4*a^2*b + 5*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*co
s(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c))) - 2*((2*a^3*b + a^2*b^2 - a*b^3)*cos(d*x +
c)^3 - (4*a^4 + 7*a^3*b + 2*a^2*b^2 - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a
^2*b^5)*d*cos(d*x + c)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*d*cos(d*x + c)^2 + (a^7 + 5
*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20015, size = 258, normalized size = 1.97 \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 )}{\left (4 \, a + b\right )}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{a^{2} + a b}} - \frac{4 \, a^{2} \tan \left (d x + c\right )^{3} + 3 \, a b \tan \left (d x + c\right )^{3} - b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{2} \tan \left (d x + c\right ) + a b \tan \left (d x + c\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, 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)^3,x, algorithm="giac")

[Out]

1/8*((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)))
*(4*a + b)/((a^3 + 2*a^2*b + a*b^2)*sqrt(a^2 + a*b)) - (4*a^2*tan(d*x + c)^3 + 3*a*b*tan(d*x + c)^3 - b^2*tan(
d*x + c)^3 + 4*a^2*tan(d*x + c) + a*b*tan(d*x + c))/((a^3 + 2*a^2*b + a*b^2)*(a*tan(d*x + c)^2 + b*tan(d*x + c
)^2 + a)^2))/d