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]

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

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Rubi [A]  time = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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 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 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 )^{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*}

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Mathematica [B]  time = 6.12, 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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fricas [B]  time = 0.70, size = 134, normalized size = 1.89 \[ \frac {{\left ({\left (3 \, a + b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}} \]

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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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)^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f*2*tan((f*x+exp(1))/2)*(tan((f*x+exp(1))/2)^2*(-tan((f*x+exp(1))/2)^2*(1327104*a^
4*b^6*sign(tan((f*x+exp(1))/2)^2-1)+2654208*a^5*b^5*sign(tan((f*x+exp(1))/2)^2-1)+1327104*a^6*b^4*sign(tan((f*
x+exp(1))/2)^2-1))/(2654208*a^4*b^7+7962624*a^5*b^6+7962624*a^6*b^5+2654208*a^7*b^4)-(884736*a^4*b^6*sign(tan(
(f*x+exp(1))/2)^2-1)-1769472*a^5*b^5*sign(tan((f*x+exp(1))/2)^2-1)-2654208*a^6*b^4*sign(tan((f*x+exp(1))/2)^2-
1))/(2654208*a^4*b^7+7962624*a^5*b^6+7962624*a^6*b^5+2654208*a^7*b^4))-(1327104*a^4*b^6*sign(tan((f*x+exp(1))/
2)^2-1)+2654208*a^5*b^5*sign(tan((f*x+exp(1))/2)^2-1)+1327104*a^6*b^4*sign(tan((f*x+exp(1))/2)^2-1))/(2654208*
a^4*b^7+7962624*a^5*b^6+7962624*a^6*b^5+2654208*a^7*b^4))/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)/(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)

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maple [A]  time = 1.61, size = 85, normalized size = 1.20 \[ \frac {\sin \left (f x +e \right ) \left (3 a \left (\cos ^{2}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) b +2 b \right ) \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{3 f \cos \left (f x +e \right )^{5} \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \left (a +b \right )^{2}} \]

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

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maxima [A]  time = 0.37, size = 61, normalized size = 0.86 \[ \frac {\frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}}}{3 \, f} \]

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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mupad [B]  time = 13.90, size = 172, normalized size = 2.42 \[ -\frac {\left ({\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (a\,3{}\mathrm {i}+b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,6{}\mathrm {i}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,10{}\mathrm {i}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}\right )}{3\,f\,{\left (a+b\right )}^2\,{\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}^2} \]

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)^(5/2)),x)

[Out]

-((exp(e*4i + f*x*4i) - 1)*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(a*3i + b*1i + a*ex
p(e*2i + f*x*2i)*6i + a*exp(e*4i + f*x*4i)*3i + b*exp(e*2i + f*x*2i)*10i + b*exp(e*4i + f*x*4i)*1i))/(3*f*(a +
b)^2*(a + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 4*b*exp(e*2i + f*x*2i))^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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