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]

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

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Rubi [A]  time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 12

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

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 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]

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*}

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Mathematica [A]  time = 0.53, 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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fricas [B]  time = 0.46, size = 419, normalized size = 5.37 \[ \left [\frac {4 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{8 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac {2 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}\right ] \]

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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giac [A]  time = 0.18, size = 109, normalized size = 1.40 \[ \frac {\frac {\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )}{\sqrt {a^{2} + a b} {\left (a + b\right )}} - \frac {\tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a + b\right )}}}{2 \, d} \]

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

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maple [A]  time = 0.25, size = 77, normalized size = 0.99 \[ -\frac {\tan \left (d x +c \right )}{2 d \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 d \left (a +b \right ) \sqrt {a \left (a +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2/(a+b*sin(d*x+c)^2)^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 [A]  time = 0.47, size = 74, normalized size = 0.95 \[ -\frac {\frac {\tan \left (d x + c\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} + a b} - \frac {\arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a + b\right )}}}{2 \, d} \]

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]

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

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mupad [B]  time = 13.43, size = 72, normalized size = 0.92 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (2\,a+2\,b\right )}^2}{4\,\sqrt {a}\,{\left (a+b\right )}^{3/2}}\right )}{2\,\sqrt {a}\,d\,{\left (a+b\right )}^{3/2}}-\frac {\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )\,\left (a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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