3.214 \(\int \frac{\sec ^2(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=106 \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 \sqrt{b} f (a+b)^{5/2}}+\frac{3 \tan (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\tan (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

(3*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*Sqrt[b]*(a + b)^(5/2)*f) + Tan[e + f*x]/(4*(a + b)*f*(a + b
+ b*Tan[e + f*x]^2)^2) + (3*Tan[e + f*x])/(8*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.0818153, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4146, 199, 205} \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 \sqrt{b} f (a+b)^{5/2}}+\frac{3 \tan (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\tan (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(3*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*Sqrt[b]*(a + b)^(5/2)*f) + Tan[e + f*x]/(4*(a + b)*f*(a + b
+ b*Tan[e + f*x]^2)^2) + (3*Tan[e + f*x])/(8*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a+b) f}\\ &=\frac{\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{3 \tan (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^2 f}\\ &=\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 \sqrt{b} (a+b)^{5/2} f}+\frac{\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{3 \tan (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.74883, size = 265, normalized size = 2.5 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{\sec (2 e) \left (a (5 a+2 b) \sin (2 f x)-\left (5 a^2+16 a b+8 b^2\right ) \sin (2 e)\right ) (a \cos (2 (e+f x))+a+2 b)}{a^2}+\frac{4 b (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a^2}-\frac{3 (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{64 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*((-3*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[
f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])^2*(Co
s[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (4*b*(a + b)*Sec[2*e]*((a + 2*b)*Sin[2*e]
- a*Sin[2*f*x]))/a^2 + ((a + 2*b + a*Cos[2*(e + f*x)])*Sec[2*e]*(-((5*a^2 + 16*a*b + 8*b^2)*Sin[2*e]) + a*(5*a
+ 2*b)*Sin[2*f*x]))/a^2))/(64*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.067, size = 97, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{ \left ( 4\,a+4\,b \right ) f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,\tan \left ( fx+e \right ) }{8\, \left ( a+b \right ) ^{2}f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{3}{8\, \left ( a+b \right ) ^{2}f}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x)

[Out]

1/4*tan(f*x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2+3/8*tan(f*x+e)/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)+3/8/f/(a+b)^2/((a+
b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*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(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.645417, size = 1319, normalized size = 12.44 \begin{align*} \left [-\frac{3 \,{\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-a b - b^{2}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \,{\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \,{\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}, -\frac{3 \,{\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{a b + b^{2}} \arctan \left (\frac{{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt{a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \,{\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \,{\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*
x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*s
in(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 4*((5*a^2*b + 7*a*b^2 + 2*b^3)*cos(f*x
+ e)^3 + 3*(a*b^2 + b^3)*cos(f*x + e))*sin(f*x + e))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f*cos(f*x + e
)^4 + 2*(a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*f*cos(f*x + e)^2 + (a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*f),
-1/16*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)
^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) - 2*((5*a^2*b + 7*a*b^2 + 2*b^3)*cos(f*x + e)^3 + 3*(a*b^
2 + b^3)*cos(f*x + e))*sin(f*x + e))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^4 + 2*(a^4*b^2
+ 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*f*cos(f*x + e)^2 + (a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*f)]

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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(f*x+e)**2/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33375, size = 176, normalized size = 1.66 \begin{align*} \frac{\frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b + b^{2}}} + \frac{3 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) + 5 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}{\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/8*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^2 + 2*a*b + b^2)*sqr
t(a*b + b^2)) + (3*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) + 5*b*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)^2*(a^2
+ 2*a*b + b^2)))/f