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

Optimal. Leaf size=77 \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{b^{5/2} f \sqrt{a+b}}-\frac{(a-b) \tan (e+f x)}{b^2 f}+\frac{\tan ^3(e+f x)}{3 b f} \]

[Out]

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

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Rubi [A]  time = 0.0880261, antiderivative size = 77, 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, 390, 205} \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{b^{5/2} f \sqrt{a+b}}-\frac{(a-b) \tan (e+f x)}{b^2 f}+\frac{\tan ^3(e+f x)}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.38925, size = 224, normalized size = 2.91 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{a+b} \sqrt{b (\sin (e)+i \cos (e))^4} \sec (e+f x) \left (\sec (e) \sin (f x) \left (-3 a+b \sec ^2(e+f x)+2 b\right )+b \tan (e) \sec (e+f x)\right )-3 a^2 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{6 b^2 f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(-3*a^2*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*S
in[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[a
+ b]*Sec[e + f*x]*Sqrt[b*(I*Cos[e] + Sin[e])^4]*(Sec[e]*(-3*a + 2*b + b*Sec[e + f*x]^2)*Sin[f*x] + b*Sec[e + f
*x]*Tan[e])))/(6*b^2*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

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Maple [A]  time = 0.059, size = 79, normalized size = 1. \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3\,fb}}-{\frac{\tan \left ( fx+e \right ) a}{f{b}^{2}}}+{\frac{\tan \left ( fx+e \right ) }{fb}}+{\frac{{a}^{2}}{f{b}^{2}}\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)^6/(a+b*sec(f*x+e)^2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.565428, size = 824, normalized size = 10.7 \begin{align*} \left [-\frac{3 \, \sqrt{-a b - b^{2}} a^{2} \cos \left (f x + e\right )^{3} \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 (a b^{2} + b^{3} -{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{12 \,{\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}, -\frac{3 \, \sqrt{a b + b^{2}} a^{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 ) \cos \left (f x + e\right )^{3} - 2 \,{\left (a b^{2} + b^{3} -{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{6 \,{\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}\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),x, algorithm="fricas")

[Out]

[-1/12*(3*sqrt(-a*b - b^2)*a^2*cos(f*x + e)^3*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*co
s(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-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*(a*b^2 + b^3 - (3*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sin(f*x +
e))/((a*b^3 + b^4)*f*cos(f*x + e)^3), -1/6*(3*sqrt(a*b + b^2)*a^2*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(s
qrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^3 - 2*(a*b^2 + b^3 - (3*a^2*b + a*b^2 - 2*b^3)*cos(f*x
+ e)^2)*sin(f*x + e))/((a*b^3 + b^4)*f*cos(f*x + e)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{6}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \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),x)

[Out]

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

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Giac [A]  time = 1.28178, size = 136, normalized size = 1.77 \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 )} a^{2}}{\sqrt{a b + b^{2}} b^{2}} + \frac{b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) + 3 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, 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),x, algorithm="giac")

[Out]

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