3.3 \(\int x^4 (a+b \sec ^{-1}(c x)) \, dx\)

Optimal. Leaf size=89 \[ \frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^4 \sqrt{1-\frac{1}{c^2 x^2}}}{20 c}-\frac{3 b x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{40 c^3}-\frac{3 b \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{40 c^5} \]

[Out]

(-3*b*Sqrt[1 - 1/(c^2*x^2)]*x^2)/(40*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^4)/(20*c) + (x^5*(a + b*ArcSec[c*x]))/5
- (3*b*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(40*c^5)

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Rubi [A]  time = 0.0463687, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5220, 266, 51, 63, 208} \[ \frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^4 \sqrt{1-\frac{1}{c^2 x^2}}}{20 c}-\frac{3 b x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{40 c^3}-\frac{3 b \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{40 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcSec[c*x]),x]

[Out]

(-3*b*Sqrt[1 - 1/(c^2*x^2)]*x^2)/(40*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^4)/(20*c) + (x^5*(a + b*ArcSec[c*x]))/5
- (3*b*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(40*c^5)

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 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int x^4 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac{b \int \frac{x^3}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{5 c}\\ &=\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac{b \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1-\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{10 c}\\ &=-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{40 c^3}\\ &=-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x^2}{40 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{80 c^5}\\ &=-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x^2}{40 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{c^2-c^2 x^2} \, dx,x,\sqrt{1-\frac{1}{c^2 x^2}}\right )}{40 c^3}\\ &=-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x^2}{40 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac{3 b \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{40 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0688045, size = 97, normalized size = 1.09 \[ \frac{a x^5}{5}+b \sqrt{\frac{c^2 x^2-1}{c^2 x^2}} \left (-\frac{3 x^2}{40 c^3}-\frac{x^4}{20 c}\right )-\frac{3 b \log \left (x \left (\sqrt{\frac{c^2 x^2-1}{c^2 x^2}}+1\right )\right )}{40 c^5}+\frac{1}{5} b x^5 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcSec[c*x]),x]

[Out]

(a*x^5)/5 + b*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)]*((-3*x^2)/(40*c^3) - x^4/(20*c)) + (b*x^5*ArcSec[c*x])/5 - (3*b*L
og[x*(1 + Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])])/(40*c^5)

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Maple [A]  time = 0.19, size = 150, normalized size = 1.7 \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{{x}^{5}b{\rm arcsec} \left (cx\right )}{5}}-{\frac{b{x}^{4}}{20\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{x}^{2}}{40\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,b}{40\,{c}^{5}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{3\,b}{40\,{c}^{6}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsec(c*x)),x)

[Out]

1/5*a*x^5+1/5*x^5*b*arcsec(c*x)-1/20/c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x^4-1/40/c^3*b/((c^2*x^2-1)/c^2/x^2)^(1/2
)*x^2+3/40/c^5*b/((c^2*x^2-1)/c^2/x^2)^(1/2)-3/40/c^6*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*ln(c*x
+(c^2*x^2-1)^(1/2))

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Maxima [A]  time = 1.00718, size = 177, normalized size = 1.99 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{80} \,{\left (16 \, x^{5} \operatorname{arcsec}\left (c x\right ) + \frac{\frac{2 \,{\left (3 \,{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{-\frac{1}{c^{2} x^{2}} + 1}\right )}}{c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{4}} - \frac{3 \, \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{4}} + \frac{3 \, \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{4}}}{c}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/80*(16*x^5*arcsec(c*x) + (2*(3*(-1/(c^2*x^2) + 1)^(3/2) - 5*sqrt(-1/(c^2*x^2) + 1))/(c^4*(1/(c^2
*x^2) - 1)^2 + 2*c^4*(1/(c^2*x^2) - 1) + c^4) - 3*log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^4 + 3*log(sqrt(-1/(c^2*x^2
) + 1) - 1)/c^4)/c)*b

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Fricas [A]  time = 2.79015, size = 248, normalized size = 2.79 \begin{align*} \frac{8 \, a c^{5} x^{5} + 16 \, b c^{5} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 8 \,{\left (b c^{5} x^{5} - b c^{5}\right )} \operatorname{arcsec}\left (c x\right ) + 3 \, b \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt{c^{2} x^{2} - 1}}{40 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="fricas")

[Out]

1/40*(8*a*c^5*x^5 + 16*b*c^5*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 8*(b*c^5*x^5 - b*c^5)*arcsec(c*x) + 3*b*log(-c
*x + sqrt(c^2*x^2 - 1)) - (2*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 - 1))/c^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \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**4*(a+b*asec(c*x)),x)

[Out]

Integral(x**4*(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^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)*x^4, x)