3.33 \(\int x^3 (a+b \text{sech}^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=124 \[ -\frac{b x^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{6 c^2}-\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^4}+\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 x^2}{12 c^2}-\frac{b^2 \log (x)}{3 c^4} \]

[Out]

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

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Rubi [A]  time = 0.119395, antiderivative size = 124, 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 = {6285, 5451, 4185, 4184, 3475} \[ -\frac{b x^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{6 c^2}-\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^4}+\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 x^2}{12 c^2}-\frac{b^2 \log (x)}{3 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSech[c*x])^2,x]

[Out]

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

Rule 6285

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

Rule 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{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 \text{sech}^{-1}(c x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \text{sech}^4(x) \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c^4}\\ &=\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b \operatorname{Subst}\left (\int (a+b x) \text{sech}^4(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{2 c^4}\\ &=-\frac{b^2 x^2}{12 c^2}-\frac{b x^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{6 c^2}+\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b \operatorname{Subst}\left (\int (a+b x) \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^4}\\ &=-\frac{b^2 x^2}{12 c^2}-\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^4}-\frac{b x^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{6 c^2}+\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2+\frac{b^2 \operatorname{Subst}\left (\int \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^4}\\ &=-\frac{b^2 x^2}{12 c^2}-\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^4}-\frac{b x^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{6 c^2}+\frac{1}{4} x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 \log (x)}{3 c^4}\\ \end{align*}

Mathematica [A]  time = 0.330303, size = 212, normalized size = 1.71 \[ -\frac{-3 a^2 c^4 x^4+2 a b c^3 x^3 \sqrt{\frac{1-c x}{c x+1}}+2 a b c^2 x^2 \sqrt{\frac{1-c x}{c x+1}}+2 b \text{sech}^{-1}(c x) \left (b \sqrt{\frac{1-c x}{c x+1}} \left (c^3 x^3+c^2 x^2+2 c x+2\right )-3 a c^4 x^4\right )+4 a b c x \sqrt{\frac{1-c x}{c x+1}}+4 a b \sqrt{\frac{1-c x}{c x+1}}+b^2 c^2 x^2-3 b^2 c^4 x^4 \text{sech}^{-1}(c x)^2+4 b^2 \log (x)}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.29, size = 264, normalized size = 2.1 \begin{align*}{\frac{{a}^{2}{x}^{4}}{4}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right )}{3\,{c}^{4}}}+{\frac{{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}{x}^{4}}{4}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right ){x}^{3}}{6\,c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right )x}{3\,{c}^{3}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}-{\frac{{b}^{2}{x}^{2}}{12\,{c}^{2}}}+{\frac{{b}^{2}}{3\,{c}^{4}}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+{\frac{ab{x}^{4}{\rm arcsech} \left (cx\right )}{2}}-{\frac{ab{x}^{3}}{6\,c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}-{\frac{xab}{3\,{c}^{3}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsech(c*x))^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{1}{6} \,{\left (3 \, x^{4} \operatorname{arsech}\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 + b^{2} \int x^{3} \log \left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 1/6*(3*x^4*arcsech(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)*a*b
+ b^2*integrate(x^3*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2, x)

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Fricas [B]  time = 2.16148, size = 525, normalized size = 4.23 \begin{align*} \frac{3 \, b^{2} c^{4} x^{4} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 3 \, a^{2} c^{4} x^{4} - 6 \, a b c^{4} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - b^{2} c^{2} x^{2} - 4 \, b^{2} \log \left (x\right ) + 2 \,{\left (3 \, a b c^{4} x^{4} - 3 \, a b c^{4} -{\left (b^{2} c^{3} x^{3} + 2 \, b^{2} c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \,{\left (a b c^{3} x^{3} + 2 \, a b c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{asech}{\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*asech(c*x))**2,x)

[Out]

Integral(x**3*(a + b*asech(c*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsech}\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*arcsech(c*x))^2,x, algorithm="giac")

[Out]

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