3.5 \(\int \sec ^5(a+b x) \, dx\)

Optimal. Leaf size=55 \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{4 b}+\frac{3 \tan (a+b x) \sec (a+b x)}{8 b} \]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(8*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(4*b)

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Rubi [A]  time = 0.0250803, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3770} \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{4 b}+\frac{3 \tan (a+b x) \sec (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^5,x]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(8*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(4*b)

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec ^5(a+b x) \, dx &=\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}+\frac{3}{4} \int \sec ^3(a+b x) \, dx\\ &=\frac{3 \sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}+\frac{3}{8} \int \sec (a+b x) \, dx\\ &=\frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{3 \sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0714815, size = 42, normalized size = 0.76 \[ \frac{3 \tanh ^{-1}(\sin (a+b x))+\tan (a+b x) \sec (a+b x) \left (2 \sec ^2(a+b x)+3\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^5,x]

[Out]

(3*ArcTanh[Sin[a + b*x]] + Sec[a + b*x]*(3 + 2*Sec[a + b*x]^2)*Tan[a + b*x])/(8*b)

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Maple [A]  time = 0.044, size = 57, normalized size = 1. \begin{align*}{\frac{ \left ( \sec \left ( bx+a \right ) \right ) ^{3}\tan \left ( bx+a \right ) }{4\,b}}+{\frac{3\,\sec \left ( bx+a \right ) \tan \left ( bx+a \right ) }{8\,b}}+{\frac{3\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5,x)

[Out]

1/4*sec(b*x+a)^3*tan(b*x+a)/b+3/8*sec(b*x+a)*tan(b*x+a)/b+3/8/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 1.03009, size = 96, normalized size = 1.75 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (b x + a\right )^{3} - 5 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/16*(2*(3*sin(b*x + a)^3 - 5*sin(b*x + a))/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) - 3*log(sin(b*x + a) + 1)
+ 3*log(sin(b*x + a) - 1))/b

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Fricas [A]  time = 1.42324, size = 200, normalized size = 3.64 \begin{align*} \frac{3 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5,x, algorithm="fricas")

[Out]

1/16*(3*cos(b*x + a)^4*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^4*log(-sin(b*x + a) + 1) + 2*(3*cos(b*x + a)^2 +
2)*sin(b*x + a))/(b*cos(b*x + a)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{5}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**5,x)

[Out]

Integral(sec(a + b*x)**5, x)

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Giac [A]  time = 1.31957, size = 85, normalized size = 1.55 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (b x + a\right )^{3} - 5 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5,x, algorithm="giac")

[Out]

-1/16*(2*(3*sin(b*x + a)^3 - 5*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 - 3*log(abs(sin(b*x + a) + 1)) + 3*log(abs
(sin(b*x + a) - 1)))/b