3.2 \(\int x^5 (a+b \sec ^{-1}(c x)) \, dx\)

Optimal. Leaf size=89 \[ \frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^5 \sqrt{1-\frac{1}{c^2 x^2}}}{30 c}-\frac{2 b x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{45 c^3}-\frac{4 b x \sqrt{1-\frac{1}{c^2 x^2}}}{45 c^5} \]

[Out]

(-4*b*Sqrt[1 - 1/(c^2*x^2)]*x)/(45*c^5) - (2*b*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(45*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*
x^5)/(30*c) + (x^6*(a + b*ArcSec[c*x]))/6

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Rubi [A]  time = 0.0413837, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5220, 271, 191} \[ \frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^5 \sqrt{1-\frac{1}{c^2 x^2}}}{30 c}-\frac{2 b x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{45 c^3}-\frac{4 b x \sqrt{1-\frac{1}{c^2 x^2}}}{45 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcSec[c*x]),x]

[Out]

(-4*b*Sqrt[1 - 1/(c^2*x^2)]*x)/(45*c^5) - (2*b*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(45*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*
x^5)/(30*c) + (x^6*(a + b*ArcSec[c*x]))/6

Rule 5220

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSec[c*x]
))/(d*(m + 1)), x] - Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int x^5 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{b \int \frac{x^4}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{6 c}\\ &=-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^5}{30 c}+\frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(2 b) \int \frac{x^2}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{15 c^3}\\ &=-\frac{2 b \sqrt{1-\frac{1}{c^2 x^2}} x^3}{45 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^5}{30 c}+\frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )-\frac{(4 b) \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{45 c^5}\\ &=-\frac{4 b \sqrt{1-\frac{1}{c^2 x^2}} x}{45 c^5}-\frac{2 b \sqrt{1-\frac{1}{c^2 x^2}} x^3}{45 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^5}{30 c}+\frac{1}{6} x^6 \left (a+b \sec ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0811036, size = 72, normalized size = 0.81 \[ \frac{a x^6}{6}+b \sqrt{\frac{c^2 x^2-1}{c^2 x^2}} \left (-\frac{2 x^3}{45 c^3}-\frac{4 x}{45 c^5}-\frac{x^5}{30 c}\right )+\frac{1}{6} b x^6 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcSec[c*x]),x]

[Out]

(a*x^6)/6 + b*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)]*((-4*x)/(45*c^5) - (2*x^3)/(45*c^3) - x^5/(30*c)) + (b*x^6*ArcSec
[c*x])/6

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Maple [A]  time = 0.16, size = 83, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{6}} \left ({\frac{{c}^{6}{x}^{6}a}{6}}+b \left ({\frac{{c}^{6}{x}^{6}{\rm arcsec} \left (cx\right )}{6}}-{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ( 3\,{c}^{4}{x}^{4}+4\,{c}^{2}{x}^{2}+8 \right ) }{90\,cx}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsec(c*x)),x)

[Out]

1/c^6*(1/6*c^6*x^6*a+b*(1/6*c^6*x^6*arcsec(c*x)-1/90*(c^2*x^2-1)*(3*c^4*x^4+4*c^2*x^2+8)/((c^2*x^2-1)/c^2/x^2)
^(1/2)/c/x))

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Maxima [A]  time = 0.987436, size = 109, normalized size = 1.22 \begin{align*} \frac{1}{6} \, a x^{6} + \frac{1}{90} \,{\left (15 \, x^{6} \operatorname{arcsec}\left (c x\right ) - \frac{3 \, c^{4} x^{5}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{5}{2}} + 10 \, c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 15 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{5}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*x^6 + 1/90*(15*x^6*arcsec(c*x) - (3*c^4*x^5*(-1/(c^2*x^2) + 1)^(5/2) + 10*c^2*x^3*(-1/(c^2*x^2) + 1)^(3/
2) + 15*x*sqrt(-1/(c^2*x^2) + 1))/c^5)*b

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Fricas [A]  time = 2.77975, size = 143, normalized size = 1.61 \begin{align*} \frac{15 \, b c^{6} x^{6} \operatorname{arcsec}\left (c x\right ) + 15 \, a c^{6} x^{6} -{\left (3 \, b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} \sqrt{c^{2} x^{2} - 1}}{90 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(15*b*c^6*x^6*arcsec(c*x) + 15*a*c^6*x^6 - (3*b*c^4*x^4 + 4*b*c^2*x^2 + 8*b)*sqrt(c^2*x^2 - 1))/c^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asec(c*x)),x)

[Out]

Integral(x**5*(a + b*asec(c*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)*x^5, x)