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

Optimal. Leaf size=27 \[ \frac{\sec ^3(a+b x)}{3 b}-\frac{\sec (a+b x)}{b} \]

[Out]

-(Sec[a + b*x]/b) + Sec[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0200087, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2606} \[ \frac{\sec ^3(a+b x)}{3 b}-\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^3,x]

[Out]

-(Sec[a + b*x]/b) + Sec[a + b*x]^3/(3*b)

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

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

Mathematica [A]  time = 0.0233389, size = 27, normalized size = 1. \[ \frac{\sec ^3(a+b x)}{3 b}-\frac{\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x]^3,x]

[Out]

-(Sec[a + b*x]/b) + Sec[a + b*x]^3/(3*b)

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Maple [B]  time = 0.02, size = 60, normalized size = 2.2 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{3\,\cos \left ( bx+a \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) \cos \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4*sin(b*x+a)^3,x)

[Out]

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

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Maxima [A]  time = 1.01126, size = 34, normalized size = 1.26 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)

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Fricas [A]  time = 1.55351, size = 65, normalized size = 2.41 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.16811, size = 34, normalized size = 1.26 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)