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(b*x+a)/b+2/3*tan(b*x+a)^3/b+1/5*tan(b*x+a)^5/b

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.82, size = 41, normalized size = 1.00 \[ \frac {{\left (8 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 3\right )} \sin \left (b x + a\right )}{15 \, b \cos \left (b x + a\right )^{5}} \]

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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giac [A]  time = 0.46, size = 34, normalized size = 0.83 \[ \frac {3 \, \tan \left (b x + a\right )^{5} + 10 \, \tan \left (b x + a\right )^{3} + 15 \, \tan \left (b x + a\right )}{15 \, b} \]

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

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maple [A]  time = 0.52, size = 34, normalized size = 0.83 \[ -\frac {\left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (b x +a \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (b x +a \right )\right )}{15}\right ) \tan \left (b x +a \right )}{b} \]

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 = 0.42, size = 34, normalized size = 0.83 \[ \frac {3 \, \tan \left (b x + a\right )^{5} + 10 \, \tan \left (b x + a\right )^{3} + 15 \, \tan \left (b x + a\right )}{15 \, b} \]

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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mupad [B]  time = 0.09, size = 31, normalized size = 0.76 \[ \frac {\frac {{\mathrm {tan}\left (a+b\,x\right )}^5}{5}+\frac {2\,{\mathrm {tan}\left (a+b\,x\right )}^3}{3}+\mathrm {tan}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec ^{6}{\left (a + b x \right )}\, dx \]

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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