3.4 \(\int x^3 (a+b \sec ^{-1}(c x)) \, dx\)

Optimal. Leaf size=64 \[ \frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{12 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{6 c^3} \]

[Out]

-(b*Sqrt[1 - 1/(c^2*x^2)]*x)/(6*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) + (x^4*(a + b*ArcSec[c*x]))/4

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Rubi [A]  time = 0.0265671, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5220, 271, 191} \[ \frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{12 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{6 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSec[c*x]),x]

[Out]

-(b*Sqrt[1 - 1/(c^2*x^2)]*x)/(6*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) + (x^4*(a + b*ArcSec[c*x]))/4

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^3 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac{b \int \frac{x^2}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{4 c}\\ &=-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{6 c^3}\\ &=-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.100045, size = 62, normalized size = 0.97 \[ \frac{a x^4}{4}+b \sqrt{\frac{c^2 x^2-1}{c^2 x^2}} \left (-\frac{x}{6 c^3}-\frac{x^3}{12 c}\right )+\frac{1}{4} b x^4 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSec[c*x]),x]

[Out]

(a*x^4)/4 + b*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)]*(-x/(6*c^3) - x^3/(12*c)) + (b*x^4*ArcSec[c*x])/4

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Maple [A]  time = 0.162, size = 74, normalized size = 1.2 \begin{align*}{\frac{1}{{c}^{4}} \left ({\frac{{c}^{4}{x}^{4}a}{4}}+b \left ({\frac{{c}^{4}{x}^{4}{\rm arcsec} \left (cx\right )}{4}}-{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ({c}^{2}{x}^{2}+2 \right ) }{12\,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^3*(a+b*arcsec(c*x)),x)

[Out]

1/c^4*(1/4*c^4*x^4*a+b*(1/4*c^4*x^4*arcsec(c*x)-1/12*(c^2*x^2-1)*(c^2*x^2+2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))

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Maxima [A]  time = 0.976702, size = 81, normalized size = 1.27 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arcsec}\left (c x\right ) - \frac{c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 3 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 + 1/12*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqrt(-1/(c^2*x^2) + 1))/c^3)*b

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Fricas [A]  time = 2.66535, size = 119, normalized size = 1.86 \begin{align*} \frac{3 \, b c^{4} x^{4} \operatorname{arcsec}\left (c x\right ) + 3 \, a c^{4} x^{4} -{\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt{c^{2} x^{2} - 1}}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*c^4*x^4*arcsec(c*x) + 3*a*c^4*x^4 - (b*c^2*x^2 + 2*b)*sqrt(c^2*x^2 - 1))/c^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3*(a+b*asec(c*x)),x)

[Out]

Integral(x**3*(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^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)*x^3, x)