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

Optimal. Leaf size=73 \[ \frac{\sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^2}+\frac{b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}+\frac{\tanh ^{-1}(\sin (x))}{(a+b)^2} \]

[Out]

(Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^2) + ArcTanh[Sin[x]]/(a + b)^2 + (b*Si
n[x])/(2*a*(a + b)*(a + b*Sin[x]^2))

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Rubi [A]  time = 0.0903967, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3190, 414, 522, 206, 205} \[ \frac{\sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^2}+\frac{b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}+\frac{\tanh ^{-1}(\sin (x))}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + b*Sin[x]^2)^2,x]

[Out]

(Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^2) + ArcTanh[Sin[x]]/(a + b)^2 + (b*Si
n[x])/(2*a*(a + b)*(a + b*Sin[x]^2))

Rule 3190

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\sin (x)\right )\\ &=\frac{b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{b-2 (a+b)+b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{2 a (a+b)}\\ &=\frac{b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (x)\right )}{(a+b)^2}+\frac{(b (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sin (x)\right )}{2 a (a+b)^2}\\ &=\frac{\sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^2}+\frac{\tanh ^{-1}(\sin (x))}{(a+b)^2}+\frac{b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.561215, size = 130, normalized size = 1.78 \[ \frac{\frac{\sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{a^{3/2}}-\frac{\sqrt{b} (3 a+b) \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{a^{3/2}}+4 \left (\frac{b (a+b) \sin (x)}{a (2 a-b \cos (2 x)+b)}-\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{4 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, size = 122, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( -1+\sin \left ( x \right ) \right ) }{2\, \left ( a+b \right ) ^{2}}}+{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{2\, \left ( a+b \right ) ^{2}}}+{\frac{b\sin \left ( x \right ) }{2\, \left ( a+b \right ) ^{2} \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{2}\sin \left ( x \right ) }{2\, \left ( a+b \right ) ^{2}a \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{3\,b}{2\, \left ( a+b \right ) ^{2}}\arctan \left ({b\sin \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{2\, \left ( a+b \right ) ^{2}a}\arctan \left ({b\sin \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a+b)^2*ln(-1+sin(x))+1/2/(a+b)^2*ln(1+sin(x))+1/2*b/(a+b)^2*sin(x)/(a+b*sin(x)^2)+1/2*b^2/(a+b)^2/a*sin(
x)/(a+b*sin(x)^2)+3/2*b/(a+b)^2/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))+1/2*b^2/(a+b)^2/a/(a*b)^(1/2)*arctan(
sin(x)*b/(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)/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.55647, size = 826, normalized size = 11.32 \begin{align*} \left [-\frac{{\left ({\left (3 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 3 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt{-\frac{b}{a}} \log \left (-\frac{b \cos \left (x\right )^{2} - 2 \, a \sqrt{-\frac{b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + 2 \,{\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (\sin \left (x\right ) + 1\right ) - 2 \,{\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \,{\left (a b + b^{2}\right )} \sin \left (x\right )}{4 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} -{\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )}}, -\frac{{\left ({\left (3 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 3 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\sqrt{\frac{b}{a}} \sin \left (x\right )\right ) +{\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (\sin \left (x\right ) + 1\right ) -{\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (-\sin \left (x\right ) + 1\right ) -{\left (a b + b^{2}\right )} \sin \left (x\right )}{2 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} -{\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.10629, size = 147, normalized size = 2.01 \begin{align*} \frac{{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{a b}} + \frac{\log \left (\sin \left (x\right ) + 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{\log \left (-\sin \left (x\right ) + 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{b \sin \left (x\right )}{2 \,{\left (b \sin \left (x\right )^{2} + a\right )}{\left (a^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a*b + b^2)*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(a*b)) + 1/2*log(sin(x) + 1)/(a^2 +
2*a*b + b^2) - 1/2*log(-sin(x) + 1)/(a^2 + 2*a*b + b^2) + 1/2*b*sin(x)/((b*sin(x)^2 + a)*(a^2 + a*b))