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

Optimal. Leaf size=96 \[ \frac{b^2 (6 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{7/2}}+\frac{b^3 \tan (x)}{2 a (a+b)^3 \left ((a+b) \tan ^2(x)+a\right )}+\frac{\tan ^3(x)}{3 (a+b)^2}+\frac{(a+3 b) \tan (x)}{(a+b)^3} \]

[Out]

(b^2*(6*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(7/2)) + ((a + 3*b)*Tan[x])/(a + b)^3
+ Tan[x]^3/(3*(a + b)^2) + (b^3*Tan[x])/(2*a*(a + b)^3*(a + (a + b)*Tan[x]^2))

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Rubi [A]  time = 0.147673, antiderivative size = 96, 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} \[ \frac{b^2 (6 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{7/2}}+\frac{b^3 \tan (x)}{2 a (a+b)^3 \left ((a+b) \tan ^2(x)+a\right )}+\frac{\tan ^3(x)}{3 (a+b)^2}+\frac{(a+3 b) \tan (x)}{(a+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + b*Sin[x]^2)^2,x]

[Out]

(b^2*(6*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(7/2)) + ((a + 3*b)*Tan[x])/(a + b)^3
+ Tan[x]^3/(3*(a + b)^2) + (b^3*Tan[x])/(2*a*(a + b)^3*(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 ^4(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{a+3 b}{(a+b)^3}+\frac{x^2}{(a+b)^2}+\frac{b^2 (3 a+b)+3 b^2 (a+b) x^2}{(a+b)^3 \left (a+(a+b) x^2\right )^2}\right ) \, dx,x,\tan (x)\right )\\ &=\frac{(a+3 b) \tan (x)}{(a+b)^3}+\frac{\tan ^3(x)}{3 (a+b)^2}+\frac{\operatorname{Subst}\left (\int \frac{b^2 (3 a+b)+3 b^2 (a+b) x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )}{(a+b)^3}\\ &=\frac{(a+3 b) \tan (x)}{(a+b)^3}+\frac{\tan ^3(x)}{3 (a+b)^2}+\frac{b^3 \tan (x)}{2 a (a+b)^3 \left (a+(a+b) \tan ^2(x)\right )}+\frac{\left (b^2 (6 a+b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{2 a (a+b)^3}\\ &=\frac{b^2 (6 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^{7/2}}+\frac{(a+3 b) \tan (x)}{(a+b)^3}+\frac{\tan ^3(x)}{3 (a+b)^2}+\frac{b^3 \tan (x)}{2 a (a+b)^3 \left (a+(a+b) \tan ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.970912, size = 97, normalized size = 1.01 \[ \frac{1}{6} \left (\frac{3 b^2 (6 a+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{a^{3/2} (a+b)^{7/2}}+\frac{\frac{3 b^3 \sin (2 x)}{a (2 a-b \cos (2 x)+b)}+2 (a+b) \tan (x) \sec ^2(x)+4 a \tan (x)+16 b \tan (x)}{(a+b)^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2*(6*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(a^(3/2)*(a + b)^(7/2)) + ((3*b^3*Sin[2*x])/(a*(2*a +
b - b*Cos[2*x])) + 4*a*Tan[x] + 16*b*Tan[x] + 2*(a + b)*Sec[x]^2*Tan[x])/(a + b)^3)/6

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Maple [B]  time = 0.102, size = 193, normalized size = 2. \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{3}a}{ \left ( 3\,{a}^{2}+6\,ab+3\,{b}^{2} \right ) \left ( a+b \right ) }}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{3}b}{ \left ( 3\,{a}^{2}+6\,ab+3\,{b}^{2} \right ) \left ( a+b \right ) }}+{\frac{\tan \left ( x \right ) a}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a+b \right ) }}+3\,{\frac{\tan \left ( x \right ) b}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a+b \right ) }}+{\frac{{b}^{3}\tan \left ( x \right ) }{2\, \left ( a+b \right ) ^{3}a \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}+3\,{\frac{{b}^{2}}{ \left ( a+b \right ) ^{3}\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}^{3}}{2\, \left ( a+b \right ) ^{3}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)^4/(a+b*sin(x)^2)^2,x)

[Out]

1/3/(a^2+2*a*b+b^2)/(a+b)*tan(x)^3*a+1/3/(a^2+2*a*b+b^2)/(a+b)*tan(x)^3*b+1/(a^2+2*a*b+b^2)/(a+b)*tan(x)*a+3/(
a^2+2*a*b+b^2)/(a+b)*tan(x)*b+1/2*b^3/(a+b)^3/a*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)+3*b^2/(a+b)^3/(a*(a+b))^(1/2)
*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))+1/2*b^3/(a+b)^3/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)^4/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.45104, size = 1481, normalized size = 15.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*((6*a*b^3 + b^4)*cos(x)^5 - (6*a^2*b^2 + 7*a*b^3 + b^4)*cos(x)^3)*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^5 + 6*a^4
*b + 6*a^3*b^2 + 2*a^2*b^3 - (4*a^4*b + 20*a^3*b^2 + 13*a^2*b^3 - 3*a*b^4)*cos(x)^4 + 2*(2*a^5 + 11*a^4*b + 16
*a^3*b^2 + 7*a^2*b^3)*cos(x)^2)*sin(x))/((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*cos(x)^5 - (a^7
+ 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*cos(x)^3), -1/12*(3*((6*a*b^3 + b^4)*cos(x)^5 - (6
*a^2*b^2 + 7*a*b^3 + b^4)*cos(x)^3)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)/(sqrt(a^2 + a*b)*c
os(x)*sin(x))) + 2*(2*a^5 + 6*a^4*b + 6*a^3*b^2 + 2*a^2*b^3 - (4*a^4*b + 20*a^3*b^2 + 13*a^2*b^3 - 3*a*b^4)*co
s(x)^4 + 2*(2*a^5 + 11*a^4*b + 16*a^3*b^2 + 7*a^2*b^3)*cos(x)^2)*sin(x))/((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a
^3*b^4 + a^2*b^5)*cos(x)^5 - (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*cos(x)^3)]

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

[Out]

Timed out

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Giac [B]  time = 1.14682, size = 365, normalized size = 3.8 \begin{align*} \frac{b^{3} \tan \left (x\right )}{2 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )}{\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )}} + \frac{{\left (6 \, a b^{2} + b^{3}\right )}{\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 )}}{2 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt{a^{2} + a b}} + \frac{a^{4} \tan \left (x\right )^{3} + 4 \, a^{3} b \tan \left (x\right )^{3} + 6 \, a^{2} b^{2} \tan \left (x\right )^{3} + 4 \, a b^{3} \tan \left (x\right )^{3} + b^{4} \tan \left (x\right )^{3} + 3 \, a^{4} \tan \left (x\right ) + 18 \, a^{3} b \tan \left (x\right ) + 36 \, a^{2} b^{2} \tan \left (x\right ) + 30 \, a b^{3} \tan \left (x\right ) + 9 \, b^{4} \tan \left (x\right )}{3 \,{\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*tan(x)/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*(a*tan(x)^2 + b*tan(x)^2 + a)) + 1/2*(6*a*b^2 + b^3)*(pi*f
loor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))/((a^4 + 3*a^3*b + 3*a^2*b^2 +
a*b^3)*sqrt(a^2 + a*b)) + 1/3*(a^4*tan(x)^3 + 4*a^3*b*tan(x)^3 + 6*a^2*b^2*tan(x)^3 + 4*a*b^3*tan(x)^3 + b^4*
tan(x)^3 + 3*a^4*tan(x) + 18*a^3*b*tan(x) + 36*a^2*b^2*tan(x) + 30*a*b^3*tan(x) + 9*b^4*tan(x))/(a^6 + 6*a^5*b
+ 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)