3.199 \(\int \frac{\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=100 \[ \frac{a^2 \tan (e+f x)}{2 b^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac{a (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 b^{5/2} f (a+b)^{3/2}}+\frac{\tan (e+f x)}{b^2 f} \]

[Out]

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

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Rubi [A]  time = 0.135116, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4146, 390, 385, 205} \[ \frac{a^2 \tan (e+f x)}{2 b^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac{a (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 b^{5/2} f (a+b)^{3/2}}+\frac{\tan (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(a*(3*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*b^(5/2)*(a + b)^(3/2)*f) + Tan[e + f*x]/(b^2*f)
+ (a^2*Tan[e + f*x])/(2*b^2*(a + b)*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 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 ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{b^2}-\frac{a (a+2 b)+2 a b x^2}{b^2 \left (a+b+b x^2\right )^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{b^2 f}-\frac{\operatorname{Subst}\left (\int \frac{a (a+2 b)+2 a b x^2}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{b^2 f}\\ &=\frac{\tan (e+f x)}{b^2 f}+\frac{a^2 \tan (e+f x)}{2 b^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{(a (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 b^2 (a+b) f}\\ &=-\frac{a (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 b^{5/2} (a+b)^{3/2} f}+\frac{\tan (e+f x)}{b^2 f}+\frac{a^2 \tan (e+f x)}{2 b^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.58448, size = 248, normalized size = 2.48 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{a (a \sin (2 f x)-(a+2 b) \sin (2 e))}{(a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+2 \sec (e) \sin (f x) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)+\frac{a (3 a+4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{(a+b)^{3/2} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{8 b^2 f \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.069, size = 128, normalized size = 1.3 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{{b}^{2}f}}+{\frac{{a}^{2}\tan \left ( fx+e \right ) }{2\,{b}^{2} \left ( a+b \right ) f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,{a}^{2}}{2\,{b}^{2} \left ( a+b \right ) 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}}}}-2\,{\frac{a}{fb \left ( a+b \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)

[Out]

tan(f*x+e)/b^2/f+1/2*a^2*tan(f*x+e)/b^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)-3/2/f*a^2/b^2/(a+b)/((a+b)*b)^(1/2)*arcta
n(tan(f*x+e)*b/((a+b)*b)^(1/2))-2/f*a/b/(a+b)/((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)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.600593, size = 1172, normalized size = 11.72 \begin{align*} \left [-\frac{{\left ({\left (3 \, a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\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 (2 \, a^{2} b^{2} + 4 \, a b^{3} + 2 \, b^{4} +{\left (3 \, a^{3} b + 5 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \,{\left ({\left (a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{3} +{\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )\right )}}, \frac{{\left ({\left (3 \, a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\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 (2 \, a^{2} b^{2} + 4 \, a b^{3} + 2 \, b^{4} +{\left (3 \, a^{3} b + 5 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{4 \,{\left ({\left (a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{3} +{\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/8*(((3*a^3 + 4*a^2*b)*cos(f*x + e)^3 + (3*a^2*b + 4*a*b^2)*cos(f*x + e))*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))*s
qrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 4*(2*a^2*b^2 + 4*a*b^
3 + 2*b^4 + (3*a^3*b + 5*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^3*b^3 + 2*a^2*b^4 + a*b^5)*f*cos
(f*x + e)^3 + (a^2*b^4 + 2*a*b^5 + b^6)*f*cos(f*x + e)), 1/4*(((3*a^3 + 4*a^2*b)*cos(f*x + e)^3 + (3*a^2*b + 4
*a*b^2)*cos(f*x + e))*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*(2*a^2*b^2 + 4*a*b^3 + 2*b^4 + (3*a^3*b + 5*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sin(f*x + e)
)/((a^3*b^3 + 2*a^2*b^4 + a*b^5)*f*cos(f*x + e)^3 + (a^2*b^4 + 2*a*b^5 + b^6)*f*cos(f*x + e))]

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

[Out]

Timed out

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Giac [A]  time = 1.19698, size = 169, normalized size = 1.69 \begin{align*} \frac{\frac{a^{2} \tan \left (f x + e\right )}{{\left (a b^{2} + b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} - \frac{{\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 (3 \, a^{2} + 4 \, a b\right )}}{{\left (a b^{2} + b^{3}\right )} \sqrt{a b + b^{2}}} + \frac{2 \, \tan \left (f x + e\right )}{b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/2*(a^2*tan(f*x + e)/((a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)) - (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arct
an(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3*a^2 + 4*a*b)/((a*b^2 + b^3)*sqrt(a*b + b^2)) + 2*tan(f*x + e)/b^2)/f