3.15 \(\int x^3 (a+b \sec ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=107 \[ -\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{6 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac{b^2 x^2}{12 c^2}+\frac{b^2 \log (x)}{3 c^4} \]

[Out]

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

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Rubi [A]  time = 0.106724, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5222, 4409, 4185, 4184, 3475} \[ -\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{6 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^2+\frac{b^2 x^2}{12 c^2}+\frac{b^2 \log (x)}{3 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSec[c*x])^2,x]

[Out]

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

Rule 5222

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.210462, size = 124, normalized size = 1.16 \[ \frac{c x \left (3 a^2 c^3 x^3-2 a b \sqrt{1-\frac{1}{c^2 x^2}} \left (c^2 x^2+2\right )+b^2 c x\right )-2 b c x \sec ^{-1}(c x) \left (b \sqrt{1-\frac{1}{c^2 x^2}} \left (c^2 x^2+2\right )-3 a c^3 x^3\right )+3 b^2 c^4 x^4 \sec ^{-1}(c x)^2+4 b^2 \log (x)}{12 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSec[c*x])^2,x]

[Out]

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

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Maple [B]  time = 0.248, size = 208, normalized size = 1.9 \begin{align*}{\frac{{a}^{2}{x}^{4}}{4}}+{\frac{{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}{x}^{4}}{4}}-{\frac{{b}^{2}{\rm arcsec} \left (cx\right ){x}^{3}}{6\,c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{b}^{2}{x}^{2}}{12\,{c}^{2}}}-{\frac{{b}^{2}{\rm arcsec} \left (cx\right )x}{3\,{c}^{3}}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{{b}^{2}}{3\,{c}^{4}}\ln \left ({\frac{1}{cx}} \right ) }+{\frac{ab{x}^{4}{\rm arcsec} \left (cx\right )}{2}}-{\frac{ab{x}^{3}}{6\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{xab}{6\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{ab}{3\,{c}^{5}x}{\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^3*(a+b*arcsec(c*x))^2,x)

[Out]

1/4*a^2*x^4+1/4*b^2*arcsec(c*x)^2*x^4-1/6/c*b^2*arcsec(c*x)*x^3*((c^2*x^2-1)/c^2/x^2)^(1/2)+1/12*b^2*x^2/c^2-1
/3/c^3*b^2*arcsec(c*x)*x*((c^2*x^2-1)/c^2/x^2)^(1/2)-1/3/c^4*b^2*ln(1/c/x)+1/2*a*b*x^4*arcsec(c*x)-1/6/c*a*b/(
(c^2*x^2-1)/c^2/x^2)^(1/2)*x^3-1/6/c^3*a*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x+1/3/c^5*a*b/((c^2*x^2-1)/c^2/x^2)^(1/
2)/x

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Maxima [A]  time = 2.09367, size = 220, normalized size = 2.06 \begin{align*} \frac{1}{4} \, b^{2} x^{4} \operatorname{arcsec}\left (c x\right )^{2} + \frac{1}{4} \, a^{2} x^{4} + \frac{1}{6} \,{\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 )} a b + \frac{{\left ({\left (c^{2} x^{2} + 2 \, \log \left (x^{2}\right )\right )} \sqrt{c x + 1} \sqrt{c x - 1} - 2 \,{\left (c^{4} x^{4} + c^{2} x^{2} - 2\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )} b^{2}}{12 \, \sqrt{c x + 1} \sqrt{c x - 1} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsec(c*x))^2,x, algorithm="maxima")

[Out]

1/4*b^2*x^4*arcsec(c*x)^2 + 1/4*a^2*x^4 + 1/6*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqr
t(-1/(c^2*x^2) + 1))/c^3)*a*b + 1/12*((c^2*x^2 + 2*log(x^2))*sqrt(c*x + 1)*sqrt(c*x - 1) - 2*(c^4*x^4 + c^2*x^
2 - 2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*b^2/(sqrt(c*x + 1)*sqrt(c*x - 1)*c^4)

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Fricas [A]  time = 2.84531, size = 339, normalized size = 3.17 \begin{align*} \frac{3 \, b^{2} c^{4} x^{4} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} c^{4} x^{4} + 12 \, a b c^{4} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + b^{2} c^{2} x^{2} + 4 \, b^{2} \log \left (x\right ) + 6 \,{\left (a b c^{4} x^{4} - a b c^{4}\right )} \operatorname{arcsec}\left (c x\right ) - 2 \,{\left (a b c^{2} x^{2} + 2 \, a b +{\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \operatorname{arcsec}\left (c x\right )\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))^2,x, algorithm="fricas")

[Out]

1/12*(3*b^2*c^4*x^4*arcsec(c*x)^2 + 3*a^2*c^4*x^4 + 12*a*b*c^4*arctan(-c*x + sqrt(c^2*x^2 - 1)) + b^2*c^2*x^2
+ 4*b^2*log(x) + 6*(a*b*c^4*x^4 - a*b*c^4)*arcsec(c*x) - 2*(a*b*c^2*x^2 + 2*a*b + (b^2*c^2*x^2 + 2*b^2)*arcsec
(c*x))*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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asec(c*x))**2,x)

[Out]

Integral(x**3*(a + b*asec(c*x))**2, 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 )}^{2} 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))^2,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)^2*x^3, x)