3.321 \(\int \frac{\sec ^2(x)}{(a+b \sin ^2(x))^2} \, dx\)

Optimal. Leaf size=76 \[ \frac{b (4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{5/2}}+\frac{b^2 \tan (x)}{2 a (a+b)^2 \left ((a+b) \tan ^2(x)+a\right )}+\frac{\tan (x)}{(a+b)^2} \]

[Out]

(b*(4*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(5/2)) + Tan[x]/(a + b)^2 + (b^2*Tan[x])
/(2*a*(a + b)^2*(a + (a + b)*Tan[x]^2))

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Rubi [A]  time = 0.123771, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3191, 390, 385, 205} \[ \frac{b (4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{5/2}}+\frac{b^2 \tan (x)}{2 a (a+b)^2 \left ((a+b) \tan ^2(x)+a\right )}+\frac{\tan (x)}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(a + b*Sin[x]^2)^2,x]

[Out]

(b*(4*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(5/2)) + Tan[x]/(a + b)^2 + (b^2*Tan[x])
/(2*a*(a + b)^2*(a + (a + b)*Tan[x]^2))

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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{\sec ^2(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{(a+b)^2}+\frac{b (2 a+b)+2 b (a+b) x^2}{(a+b)^2 \left (a+(a+b) x^2\right )^2}\right ) \, dx,x,\tan (x)\right )\\ &=\frac{\tan (x)}{(a+b)^2}+\frac{\operatorname{Subst}\left (\int \frac{b (2 a+b)+2 b (a+b) x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )}{(a+b)^2}\\ &=\frac{\tan (x)}{(a+b)^2}+\frac{b^2 \tan (x)}{2 a (a+b)^2 \left (a+(a+b) \tan ^2(x)\right )}+\frac{(b (4 a+b)) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{2 a (a+b)^2}\\ &=\frac{b (4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{5/2}}+\frac{\tan (x)}{(a+b)^2}+\frac{b^2 \tan (x)}{2 a (a+b)^2 \left (a+(a+b) \tan ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.518547, size = 76, normalized size = 1. \[ \frac{1}{2} \left (\frac{b (4 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{a^{3/2} (a+b)^{5/2}}+\frac{\frac{b^2 \sin (2 x)}{a (2 a-b \cos (2 x)+b)}+2 \tan (x)}{(a+b)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.091, size = 112, normalized size = 1.5 \begin{align*}{\frac{\tan \left ( x \right ) }{{a}^{2}+2\,ab+{b}^{2}}}+{\frac{{b}^{2}\tan \left ( x \right ) }{2\, \left ( a+b \right ) ^{2}a \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}+2\,{\frac{b}{ \left ( a+b \right ) ^{2}\sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( x \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }+{\frac{{b}^{2}}{2\, \left ( a+b \right ) ^{2}a}\arctan \left ({ \left ( a+b \right ) \tan \left ( x \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(sec(x)^2/(a+b*sin(x)^2)^2,x)

[Out]

1/(a^2+2*a*b+b^2)*tan(x)+1/2*b^2/(a+b)^2/a*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)+2*b/(a+b)^2/(a*(a+b))^(1/2)*arctan
((a+b)*tan(x)/(a*(a+b))^(1/2))+1/2*b^2/(a+b)^2/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(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(sec(x)^2/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31892, size = 1152, normalized size = 15.16 \begin{align*} \left [-\frac{{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{3} -{\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sqrt{-a^{2} - a b} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \,{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} -{\left (a + b\right )} \cos \left (x\right )\right )} \sqrt{-a^{2} - a b} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \,{\left (2 \, a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} -{\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{8 \,{\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )^{3} -{\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )\right )}}, -\frac{{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{3} -{\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sqrt{a^{2} + a b} \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b}{2 \, \sqrt{a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) + 2 \,{\left (2 \, a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} -{\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{4 \,{\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )^{3} -{\left (a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \cos \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(((4*a*b^2 + b^3)*cos(x)^3 - (4*a^2*b + 5*a*b^2 + b^3)*cos(x))*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^
2)*cos(x)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(x)^2 + 4*((2*a + b)*cos(x)^3 - (a + b)*cos(x))*sqrt(-a^2 - a*b)*sin(
x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*b + b^2)*cos(x)^2 + a^2 + 2*a*b + b^2)) + 4*(2*a^4 + 4*a^3*b + 2*
a^2*b^2 - (2*a^3*b + a^2*b^2 - a*b^3)*cos(x)^2)*sin(x))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*cos(x)^3 -
(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(x)), -1/4*(((4*a*b^2 + b^3)*cos(x)^3 - (4*a^2*b + 5*a*b^
2 + b^3)*cos(x))*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x))) + 2*
(2*a^4 + 4*a^3*b + 2*a^2*b^2 - (2*a^3*b + a^2*b^2 - a*b^3)*cos(x)^2)*sin(x))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 +
a^2*b^4)*cos(x)^3 - (a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(x))]

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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(sec(x)**2/(a+b*sin(x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.14809, size = 153, normalized size = 2.01 \begin{align*} \frac{b^{2} \tan \left (x\right )}{2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}{\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )}} + \frac{{\left (4 \, a b + b^{2}\right )} \arctan \left (\frac{a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt{a^{2} + a b}}\right )}{2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{a^{2} + a b}} + \frac{\tan \left (x\right )}{a^{2} + 2 \, a b + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+b*sin(x)^2)^2,x, algorithm="giac")

[Out]

1/2*b^2*tan(x)/((a^3 + 2*a^2*b + a*b^2)*(a*tan(x)^2 + b*tan(x)^2 + a)) + 1/2*(4*a*b + b^2)*arctan((a*tan(x) +
b*tan(x))/sqrt(a^2 + a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(a^2 + a*b)) + tan(x)/(a^2 + 2*a*b + b^2)