3.3 \(\int \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=34 \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{\tan (a+b x) \sec (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0143485, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3770} \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{\tan (a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3,x]

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*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 ^3(a+b x) \, dx &=\frac{\sec (a+b x) \tan (a+b x)}{2 b}+\frac{1}{2} \int \sec (a+b x) \, dx\\ &=\frac{\tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{\sec (a+b x) \tan (a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0098694, size = 34, normalized size = 1. \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b}+\frac{\tan (a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.133, size = 38, normalized size = 1.1 \begin{align*}{\frac{\sec \left ( bx+a \right ) \tan \left ( bx+a \right ) }{2\,b}}+{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3,x)

[Out]

1/2*sec(b*x+a)*tan(b*x+a)/b+1/2/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 1.00103, size = 62, normalized size = 1.82 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (\sin \left (b x + a\right ) - 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.41226, size = 162, normalized size = 4.76 \begin{align*} \frac{\cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3,x)

[Out]

Integral(sec(a + b*x)**3, x)

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Giac [A]  time = 1.35288, size = 65, normalized size = 1.91 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3,x, algorithm="giac")

[Out]

-1/4*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) - log(abs(sin(b*x + a) + 1)) + log(abs(sin(b*x + a) - 1)))/b