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

Optimal. Leaf size=32 \[ \frac {\tan (e+f x)}{f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4146, 191} \[ \frac {\tan (e+f x)}{f (a+b) \sqrt {a+b \tan ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 57, normalized size = 1.78 \[ \frac {\tan (e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)}{2 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 65, normalized size = 2.03 \[ \frac {\sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{{\left (a^{2} + a b\right )} f \cos \left (f x + e\right )^{2} + {\left (a b + b^{2}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)-2/f*512*a^2*b^2*tan((f*x+exp(1))/2)*sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp
(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b)*sign(tan((f*x+exp(1))/2)^2-1)/(512*a^2*b^3+512*a^3*b^2)/(a*tan((f*x+e
xp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b)

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maple [A]  time = 1.45, size = 59, normalized size = 1.84 \[ \frac {\left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right ) \sin \left (f x +e \right )}{f \cos \left (f x +e \right )^{3} \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {3}{2}} \left (a +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 30, normalized size = 0.94 \[ \frac {\tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.29, size = 199, normalized size = 6.22 \[ \frac {\sqrt {\frac {a+2\,b+a\,\cos \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (5\,a\,\sin \left (2\,e+2\,f\,x\right )+4\,a\,\sin \left (4\,e+4\,f\,x\right )+a\,\sin \left (6\,e+6\,f\,x\right )+8\,b\,\sin \left (2\,e+2\,f\,x\right )+4\,b\,\sin \left (4\,e+4\,f\,x\right )\right )}{f\,\left (a+b\right )\,\left (24\,a\,b+10\,a^2+16\,b^2+15\,a^2\,\cos \left (2\,e+2\,f\,x\right )+6\,a^2\,\cos \left (4\,e+4\,f\,x\right )+a^2\,\cos \left (6\,e+6\,f\,x\right )+16\,b^2\,\cos \left (2\,e+2\,f\,x\right )+32\,a\,b\,\cos \left (2\,e+2\,f\,x\right )+8\,a\,b\,\cos \left (4\,e+4\,f\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^2*(a + b/cos(e + f*x)^2)^(3/2)),x)

[Out]

(((a + 2*b + a*cos(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2)*(5*a*sin(2*e + 2*f*x) + 4*a*sin(4*e + 4*f*x) +
a*sin(6*e + 6*f*x) + 8*b*sin(2*e + 2*f*x) + 4*b*sin(4*e + 4*f*x)))/(f*(a + b)*(24*a*b + 10*a^2 + 16*b^2 + 15*a
^2*cos(2*e + 2*f*x) + 6*a^2*cos(4*e + 4*f*x) + a^2*cos(6*e + 6*f*x) + 16*b^2*cos(2*e + 2*f*x) + 32*a*b*cos(2*e
+ 2*f*x) + 8*a*b*cos(4*e + 4*f*x)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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