3.311 \(\int \frac{\sec ^4(x)}{a+b \sin ^2(x)} \, dx\)

Optimal. Leaf size=59 \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{\tan ^3(x)}{3 (a+b)}+\frac{(a+2 b) \tan (x)}{(a+b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0813427, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3191, 390, 205} \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{\tan ^3(x)}{3 (a+b)}+\frac{(a+2 b) \tan (x)}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + b*Sin[x]^2),x]

[Out]

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

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

Mathematica [A]  time = 0.2104, size = 59, normalized size = 1. \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{\tan (x) \left ((a+b) \sec ^2(x)+2 a+5 b\right )}{3 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4/(a + b*Sin[x]^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.099, size = 75, normalized size = 1.3 \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{3}a}{3\, \left ( a+b \right ) ^{2}}}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{3}b}{3\, \left ( a+b \right ) ^{2}}}+{\frac{\tan \left ( x \right ) a}{ \left ( a+b \right ) ^{2}}}+2\,{\frac{\tan \left ( x \right ) b}{ \left ( a+b \right ) ^{2}}}+{\frac{{b}^{2}}{ \left ( a+b \right ) ^{2}}\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)^4/(a+b*sin(x)^2),x)

[Out]

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

________________________________________________________________________________________

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)^4/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.1583, size = 834, normalized size = 14.14 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} - a b} b^{2} \cos \left (x\right )^{3} \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 (a^{3} + 2 \, a^{2} b + a b^{2} +{\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{12 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{3}}, -\frac{3 \, \sqrt{a^{2} + a b} b^{2} \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 ) \cos \left (x\right )^{3} - 2 \,{\left (a^{3} + 2 \, a^{2} b + a b^{2} +{\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

[-1/12*(3*sqrt(-a^2 - a*b)*b^2*cos(x)^3*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*(a^3 + 2*a^2*b + a*b^2 + (2*a^3 + 7*a^2*b + 5*a*b^2)*cos(x)^2)*sin(
x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(x)^3), -1/6*(3*sqrt(a^2 + a*b)*b^2*arctan(1/2*((2*a + b)*cos(x)^2
- a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))*cos(x)^3 - 2*(a^3 + 2*a^2*b + a*b^2 + (2*a^3 + 7*a^2*b + 5*a*b^2)*c
os(x)^2)*sin(x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(x)^3)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.1147, size = 181, normalized size = 3.07 \begin{align*} \frac{{\left (\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt{a^{2} + a b}}\right )\right )} b^{2}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a^{2} + a b}} + \frac{a^{2} \tan \left (x\right )^{3} + 2 \, a b \tan \left (x\right )^{3} + b^{2} \tan \left (x\right )^{3} + 3 \, a^{2} \tan \left (x\right ) + 9 \, a b \tan \left (x\right ) + 6 \, b^{2} \tan \left (x\right )}{3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+b*sin(x)^2),x, algorithm="giac")

[Out]

(pi*floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))*b^2/((a^2 + 2*a*b + b^2)
*sqrt(a^2 + a*b)) + 1/3*(a^2*tan(x)^3 + 2*a*b*tan(x)^3 + b^2*tan(x)^3 + 3*a^2*tan(x) + 9*a*b*tan(x) + 6*b^2*ta
n(x))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)