3.6 \(\int \sec ^6(a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac{\tan ^5(a+b x)}{5 b}+\frac{2 \tan ^3(a+b x)}{3 b}+\frac{\tan (a+b x)}{b} \]

[Out]

Tan[a + b*x]/b + (2*Tan[a + b*x]^3)/(3*b) + Tan[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.0148022, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3767} \[ \frac{\tan ^5(a+b x)}{5 b}+\frac{2 \tan ^3(a+b x)}{3 b}+\frac{\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^6,x]

[Out]

Tan[a + b*x]/b + (2*Tan[a + b*x]^3)/(3*b) + Tan[a + b*x]^5/(5*b)

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \sec ^6(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (a+b x)\right )}{b}\\ &=\frac{\tan (a+b x)}{b}+\frac{2 \tan ^3(a+b x)}{3 b}+\frac{\tan ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.100569, size = 35, normalized size = 0.85 \[ \frac{\frac{1}{5} \tan ^5(a+b x)+\frac{2}{3} \tan ^3(a+b x)+\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6,x]

[Out]

(Tan[a + b*x] + (2*Tan[a + b*x]^3)/3 + Tan[a + b*x]^5/5)/b

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Maple [A]  time = 0.043, size = 34, normalized size = 0.8 \begin{align*} -{\frac{\tan \left ( bx+a \right ) }{b} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( bx+a \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( bx+a \right ) \right ) ^{2}}{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6,x)

[Out]

-1/b*(-8/15-1/5*sec(b*x+a)^4-4/15*sec(b*x+a)^2)*tan(b*x+a)

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Maxima [A]  time = 1.20084, size = 46, normalized size = 1.12 \begin{align*} \frac{3 \, \tan \left (b x + a\right )^{5} + 10 \, \tan \left (b x + a\right )^{3} + 15 \, \tan \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6,x, algorithm="maxima")

[Out]

1/15*(3*tan(b*x + a)^5 + 10*tan(b*x + a)^3 + 15*tan(b*x + a))/b

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Fricas [A]  time = 1.36232, size = 108, normalized size = 2.63 \begin{align*} \frac{{\left (8 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 3\right )} \sin \left (b x + a\right )}{15 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(8*cos(b*x + a)^4 + 4*cos(b*x + a)^2 + 3)*sin(b*x + a)/(b*cos(b*x + a)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**6,x)

[Out]

Integral(sec(a + b*x)**6, x)

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Giac [A]  time = 1.13672, size = 46, normalized size = 1.12 \begin{align*} \frac{3 \, \tan \left (b x + a\right )^{5} + 10 \, \tan \left (b x + a\right )^{3} + 15 \, \tan \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*tan(b*x + a)^5 + 10*tan(b*x + a)^3 + 15*tan(b*x + a))/b