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

Optimal. Leaf size=31 \[ \frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{3 b} \]

[Out]

-Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.032495, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2606, 14} \[ \frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^3,x]

[Out]

-Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac{\sec ^3(a+b x)}{3 b}+\frac{\sec ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0479105, size = 31, normalized size = 1. \[ \frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*b)

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Maple [B]  time = 0.019, size = 78, normalized size = 2.5 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{15\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{15\,\cos \left ( bx+a \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( bx+a \right ) \right ) ^{2} \right ) \cos \left ( bx+a \right ) }{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6*sin(b*x+a)^3,x)

[Out]

1/b*(1/5*sin(b*x+a)^4/cos(b*x+a)^5+1/15*sin(b*x+a)^4/cos(b*x+a)^3-1/15*sin(b*x+a)^4/cos(b*x+a)-1/15*(2+sin(b*x
+a)^2)*cos(b*x+a))

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Maxima [A]  time = 0.976277, size = 34, normalized size = 1.1 \begin{align*} -\frac{5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)

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Fricas [A]  time = 1.60811, size = 66, normalized size = 2.13 \begin{align*} -\frac{5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)

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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)**6*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1863, size = 34, normalized size = 1.1 \begin{align*} -\frac{5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)