3.8 \(\int \sec ^8(a+b x) \, dx\)

Optimal. Leaf size=53 \[ \frac{\tan ^7(a+b x)}{7 b}+\frac{3 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{b}+\frac{\tan (a+b x)}{b} \]

[Out]

Tan[a + b*x]/b + Tan[a + b*x]^3/b + (3*Tan[a + b*x]^5)/(5*b) + Tan[a + b*x]^7/(7*b)

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Rubi [A]  time = 0.0167995, antiderivative size = 53, 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} \[ \frac{\tan ^7(a+b x)}{7 b}+\frac{3 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{b}+\frac{\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^8,x]

[Out]

Tan[a + b*x]/b + Tan[a + b*x]^3/b + (3*Tan[a + b*x]^5)/(5*b) + Tan[a + b*x]^7/(7*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 ^8(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (a+b x)\right )}{b}\\ &=\frac{\tan (a+b x)}{b}+\frac{\tan ^3(a+b x)}{b}+\frac{3 \tan ^5(a+b x)}{5 b}+\frac{\tan ^7(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.218945, size = 43, normalized size = 0.81 \[ \frac{\frac{1}{7} \tan ^7(a+b x)+\frac{3}{5} \tan ^5(a+b x)+\tan ^3(a+b x)+\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^8,x]

[Out]

(Tan[a + b*x] + Tan[a + b*x]^3 + (3*Tan[a + b*x]^5)/5 + Tan[a + b*x]^7/7)/b

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Maple [A]  time = 0.043, size = 44, normalized size = 0.8 \begin{align*} -{\frac{\tan \left ( bx+a \right ) }{b} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( bx+a \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( bx+a \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( bx+a \right ) \right ) ^{2}}{35}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^8,x)

[Out]

-1/b*(-16/35-1/7*sec(b*x+a)^6-6/35*sec(b*x+a)^4-8/35*sec(b*x+a)^2)*tan(b*x+a)

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Maxima [A]  time = 1.07473, size = 59, normalized size = 1.11 \begin{align*} \frac{5 \, \tan \left (b x + a\right )^{7} + 21 \, \tan \left (b x + a\right )^{5} + 35 \, \tan \left (b x + a\right )^{3} + 35 \, \tan \left (b x + a\right )}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8,x, algorithm="maxima")

[Out]

1/35*(5*tan(b*x + a)^7 + 21*tan(b*x + a)^5 + 35*tan(b*x + a)^3 + 35*tan(b*x + a))/b

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Fricas [A]  time = 1.39288, size = 135, normalized size = 2.55 \begin{align*} \frac{{\left (16 \, \cos \left (b x + a\right )^{6} + 8 \, \cos \left (b x + a\right )^{4} + 6 \, \cos \left (b x + a\right )^{2} + 5\right )} \sin \left (b x + a\right )}{35 \, b \cos \left (b x + a\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8,x, algorithm="fricas")

[Out]

1/35*(16*cos(b*x + a)^6 + 8*cos(b*x + a)^4 + 6*cos(b*x + a)^2 + 5)*sin(b*x + a)/(b*cos(b*x + a)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{8}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**8,x)

[Out]

Integral(sec(a + b*x)**8, x)

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Giac [A]  time = 1.27431, size = 59, normalized size = 1.11 \begin{align*} \frac{5 \, \tan \left (b x + a\right )^{7} + 21 \, \tan \left (b x + a\right )^{5} + 35 \, \tan \left (b x + a\right )^{3} + 35 \, \tan \left (b x + a\right )}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8,x, algorithm="giac")

[Out]

1/35*(5*tan(b*x + a)^7 + 21*tan(b*x + a)^5 + 35*tan(b*x + a)^3 + 35*tan(b*x + a))/b