3.291 \(\int \frac{\sec ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac{2 \tan (e+f x)}{3 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}+\frac{\tan (e+f x)}{3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

Tan[e + f*x]/(3*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (2*Tan[e + f*x])/(3*(a + b)^2*f*Sqrt[a + b + b*T
an[e + f*x]^2])

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Rubi [A]  time = 0.0895528, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4146, 192, 191} \[ \frac{2 \tan (e+f x)}{3 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}+\frac{\tan (e+f x)}{3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

Tan[e + f*x]/(3*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) + (2*Tan[e + f*x])/(3*(a + b)^2*f*Sqrt[a + b + b*T
an[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 192

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, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [B]  time = 6.13418, size = 215, normalized size = 3.03 \[ \frac{(3 a+b) \sec ^5(e+f x) \left (\frac{2 \sqrt{2} \sin (e+f x)}{(a+b)^2 \sqrt{-a \sin ^2(e+f x)+a+b}}+\frac{\sqrt{2} \sin (e+f x)}{(a+b) \left (-a \sin ^2(e+f x)+a+b\right )^{3/2}}\right ) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{48 a f \left (a+b \sec ^2(e+f x)\right )^{5/2}}-\frac{\tan (e+f x) \sec ^4(e+f x) (a \cos (2 e+2 f x)+a+2 b)^{5/2}}{8 \sqrt{2} a f \left (-a \sin ^2(e+f x)+a+b\right )^{3/2} \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((3*a + b)*(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5*((Sqrt[2]*Sin[e + f*x])/((a + b)*(a + b - a*Sin
[e + f*x]^2)^(3/2)) + (2*Sqrt[2]*Sin[e + f*x])/((a + b)^2*Sqrt[a + b - a*Sin[e + f*x]^2])))/(48*a*f*(a + b*Sec
[e + f*x]^2)^(5/2)) - ((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4*Tan[e + f*x])/(8*Sqrt[2]*a*f*(a + b
*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.266, size = 85, normalized size = 1.2 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 3\,a \left ( \cos \left ( fx+e \right ) \right ) ^{2}+b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,b \right ) \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{3\,f \left ( a+b \right ) ^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{-{\frac{5}{2}}}} \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)^(5/2),x)

[Out]

1/3/f/(a+b)^2*sin(f*x+e)*(3*a*cos(f*x+e)^2+b*cos(f*x+e)^2+2*b)*(b+a*cos(f*x+e)^2)/cos(f*x+e)^5/((b+a*cos(f*x+e
)^2)/cos(f*x+e)^2)^(5/2)

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Maxima [A]  time = 1.20573, size = 82, normalized size = 1.15 \begin{align*} \frac{\frac{2 \, \tan \left (f x + e\right )}{\sqrt{b \tan \left (f x + e\right )^{2} + a + b}{\left (a + b\right )}^{2}} + \frac{\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac{3}{2}}{\left (a + b\right )}}}{3 \, 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)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) + tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2
)*(a + b)))/f

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Fricas [B]  time = 0.912419, size = 313, normalized size = 4.41 \begin{align*} \frac{{\left ({\left (3 \, a + b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{3 \,{\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\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)^(5/2),x, algorithm="fricas")

[Out]

1/3*((3*a + b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)/((a
^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^2*b^2 + 2*a*b^3
+ b^4)*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \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)**(5/2),x)

[Out]

Integral(sec(e + f*x)**2/(a + b*sec(e + f*x)**2)**(5/2), x)

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Giac [B]  time = 2.44098, size = 562, normalized size = 7.92 \begin{align*} -\frac{2 \,{\left ({\left (\frac{3 \,{\left (a^{6} b^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) + 2 \, a^{5} b^{5} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) + a^{4} b^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}} - \frac{2 \,{\left (3 \, a^{6} b^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) + 2 \, a^{5} b^{5} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) - a^{4} b^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )\right )}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \frac{3 \,{\left (a^{6} b^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) + 2 \, a^{5} b^{5} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) + a^{4} b^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )\right )}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{3 \,{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a + b\right )}^{\frac{3}{2}} 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)^(5/2),x, algorithm="giac")

[Out]

-2/3*((3*(a^6*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + 2*a^5*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^4*b^6*sgn(ta
n(1/2*f*x + 1/2*e)^2 - 1))*tan(1/2*f*x + 1/2*e)^2/(a^7*b^4 + 3*a^6*b^5 + 3*a^5*b^6 + a^4*b^7) - 2*(3*a^6*b^4*s
gn(tan(1/2*f*x + 1/2*e)^2 - 1) + 2*a^5*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - a^4*b^6*sgn(tan(1/2*f*x + 1/2*e)^
2 - 1))/(a^7*b^4 + 3*a^6*b^5 + 3*a^5*b^6 + a^4*b^7))*tan(1/2*f*x + 1/2*e)^2 + 3*(a^6*b^4*sgn(tan(1/2*f*x + 1/2
*e)^2 - 1) + 2*a^5*b^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) + a^4*b^6*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^7*b^4 + 3
*a^6*b^5 + 3*a^5*b^6 + a^4*b^7))*tan(1/2*f*x + 1/2*e)/((a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 -
2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)^(3/2)*f)