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

Optimal. Leaf size=56 \[ -\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^2+\frac{b^2 \log (x)}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.0682639, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5222, 4409, 4184, 3475} \[ -\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^2+\frac{b^2 \log (x)}{c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSec[c*x])^2,x]

[Out]

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

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

Mathematica [A]  time = 0.149093, size = 90, normalized size = 1.61 \[ \frac{a c x \left (a c x-2 b \sqrt{1-\frac{1}{c^2 x^2}}\right )+2 b c x \sec ^{-1}(c x) \left (a c x-b \sqrt{1-\frac{1}{c^2 x^2}}\right )+b^2 c^2 x^2 \sec ^{-1}(c x)^2+2 b^2 \log (c x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcSec[c*x])^2,x]

[Out]

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

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Maple [B]  time = 0.249, size = 134, normalized size = 2.4 \begin{align*}{\frac{{a}^{2}{x}^{2}}{2}}+{\frac{{x}^{2}{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{2}}-{\frac{{b}^{2}{\rm arcsec} \left (cx\right )x}{c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{{b}^{2}}{{c}^{2}}\ln \left ({\frac{1}{cx}} \right ) }+ab{x}^{2}{\rm arcsec} \left (cx\right )-{\frac{xab}{c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{ab}{{c}^{3}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*(a+b*arcsec(c*x))^2,x)

[Out]

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

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Maxima [A]  time = 1.0201, size = 117, normalized size = 2.09 \begin{align*} \frac{1}{2} \, b^{2} x^{2} \operatorname{arcsec}\left (c x\right )^{2} + \frac{1}{2} \, a^{2} x^{2} +{\left (x^{2} \operatorname{arcsec}\left (c x\right ) - \frac{x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c}\right )} a b -{\left (\frac{x \sqrt{-\frac{1}{c^{2} x^{2}} + 1} \operatorname{arcsec}\left (c x\right )}{c} - \frac{\log \left (x\right )}{c^{2}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2*arcsec(c*x)^2 + 1/2*a^2*x^2 + (x^2*arcsec(c*x) - x*sqrt(-1/(c^2*x^2) + 1)/c)*a*b - (x*sqrt(-1/(c^2
*x^2) + 1)*arcsec(c*x)/c - log(x)/c^2)*b^2

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Fricas [B]  time = 2.75695, size = 266, normalized size = 4.75 \begin{align*} \frac{b^{2} c^{2} x^{2} \operatorname{arcsec}\left (c x\right )^{2} + a^{2} c^{2} x^{2} + 4 \, a b c^{2} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 2 \, b^{2} \log \left (x\right ) + 2 \,{\left (a b c^{2} x^{2} - a b c^{2}\right )} \operatorname{arcsec}\left (c x\right ) - 2 \, \sqrt{c^{2} x^{2} - 1}{\left (b^{2} \operatorname{arcsec}\left (c x\right ) + a b\right )}}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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*(a+b*asec(c*x))**2,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^2*x, x)