3.103 \(\int \frac {1}{(a+b \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=87 \[ \frac {(2 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a+b)^{3/2}}+\frac {b \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3184, 12, 3181, 205} \[ \frac {(2 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} d (a+b)^{3/2}}+\frac {b \sin (c+d x) \cos (c+d x)}{2 a d (a+b) \left (a+b \sin ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^2)^(-2),x]

[Out]

((2*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(3/2)*d) + (b*Cos[c + d*x]*Sin[c + d
*x])/(2*a*(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 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 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=\frac {b \cos (c+d x) \sin (c+d x)}{2 a (a+b) d \left (a+b \sin ^2(c+d x)\right )}-\frac {\int \frac {-2 a-b}{a+b \sin ^2(c+d x)} \, dx}{2 a (a+b)}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{2 a (a+b) d \left (a+b \sin ^2(c+d x)\right )}+\frac {(2 a+b) \int \frac {1}{a+b \sin ^2(c+d x)} \, dx}{2 a (a+b)}\\ &=\frac {b \cos (c+d x) \sin (c+d x)}{2 a (a+b) d \left (a+b \sin ^2(c+d x)\right )}+\frac {(2 a+b) \operatorname {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 a (a+b) d}\\ &=\frac {(2 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2} d}+\frac {b \cos (c+d x) \sin (c+d x)}{2 a (a+b) d \left (a+b \sin ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 84, normalized size = 0.97 \[ \frac {\frac {(2 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2}}+\frac {\sqrt {a} b \sin (2 (c+d x))}{(a+b) (2 a-b \cos (2 (c+d x))+b)}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 463, normalized size = 5.32 \[ \left [-\frac {4 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 3 \, a b - b^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{8 \, {\left ({\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} d\right )}}, -\frac {2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 3 \, a b - b^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left ({\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 113, normalized size = 1.30 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (2 \, a + b\right )}}{{\left (a^{2} + a b\right )}^{\frac {3}{2}}} + \frac {b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a^{2} + a b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)))
*(2*a + b)/(a^2 + a*b)^(3/2) + b*tan(d*x + c)/((a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a^2 + a*b)))/d

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maple [A]  time = 0.36, size = 119, normalized size = 1.37 \[ \frac {b \tan \left (d x +c \right )}{2 d a \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \left (a +b \right ) \sqrt {a \left (a +b \right )}}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right ) b}{2 d a \left (a +b \right ) \sqrt {a \left (a +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.48, size = 89, normalized size = 1.02 \[ \frac {\frac {b \tan \left (d x + c\right )}{a^{3} + a^{2} b + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{2}} + \frac {{\left (2 \, a + b\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + a b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.43, size = 79, normalized size = 0.91 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )}{2\,\sqrt {a}\,\sqrt {a+b}}\right )\,\left (2\,a+b\right )}{2\,a^{3/2}\,d\,{\left (a+b\right )}^{3/2}}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2\,a\,d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )\,\left (a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d*x)^2)^2,x)

[Out]

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

[Out]

Timed out

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