3.90 \(\int (d+e x^2) (a+b \text{sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=112 \[ d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (6 c^2 d+e\right ) \sin ^{-1}(c x)}{6 c^3}-\frac{b e x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{6 c^2} \]

[Out]

-(b*e*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(6*c^2) + d*x*(a + b*ArcSech[c*x]) + (e*x^3*(a +
b*ArcSech[c*x]))/3 + (b*(6*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(6*c^3)

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Rubi [A]  time = 0.0508121, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6291, 12, 388, 216} \[ d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (6 c^2 d+e\right ) \sin ^{-1}(c x)}{6 c^3}-\frac{b e x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{6 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)*(a + b*ArcSech[c*x]),x]

[Out]

-(b*e*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(6*c^2) + d*x*(a + b*ArcSech[c*x]) + (e*x^3*(a +
b*ArcSech[c*x]))/3 + (b*(6*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(6*c^3)

Rule 6291

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)], Int[SimplifyIntegrand[u/(x
*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \left (d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{3 d+e x^2}{3 \sqrt{1-c^2 x^2}} \, dx\\ &=d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{3 d+e x^2}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b e x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{6 c^2}+d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{\left (b \left (6 c^2 d+e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{6 c^2}\\ &=-\frac{b e x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{6 c^2}+d x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b \left (6 c^2 d+e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{6 c^3}\\ \end{align*}

Mathematica [C]  time = 0.356175, size = 169, normalized size = 1.51 \[ a d x+\frac{1}{3} a e x^3-\frac{b d \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c (c x-1)}+b e \sqrt{\frac{1-c x}{c x+1}} \left (-\frac{x}{6 c^2}-\frac{x^2}{6 c}\right )+\frac{i b e \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{6 c^3}+b d x \text{sech}^{-1}(c x)+\frac{1}{3} b e x^3 \text{sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*e*x^3)/3 + b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(-x/(6*c^2) - x^2/(6*c)) + b*d*x*ArcSech[c*x] + (b*e*x^3*A
rcSech[c*x])/3 - (b*d*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*(-1 + c*x)) + ((I/6)*b*e*Log
[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)])/c^3

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Maple [A]  time = 0.175, size = 135, normalized size = 1.2 \begin{align*}{\frac{1}{c} \left ({\frac{a}{{c}^{2}} \left ({\frac{e{c}^{3}{x}^{3}}{3}}+x{c}^{3}d \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\rm arcsech} \left (cx\right )e{c}^{3}{x}^{3}}{3}}+{\rm arcsech} \left (cx\right ){c}^{3}dx+{\frac{cx}{6}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( 6\,\arcsin \left ( cx \right ){c}^{2}d-ecx\sqrt{-{c}^{2}{x}^{2}+1}+e\arcsin \left ( cx \right ) \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a/c^2*(1/3*e*c^3*x^3+x*c^3*d)+b/c^2*(1/3*arcsech(c*x)*e*c^3*x^3+arcsech(c*x)*c^3*d*x+1/6*(-(c*x-1)/c/x)^(
1/2)*c*x*((c*x+1)/c/x)^(1/2)*(6*arcsin(c*x)*c^2*d-e*c*x*(-c^2*x^2+1)^(1/2)+e*arcsin(c*x))/(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.49131, size = 144, normalized size = 1.29 \begin{align*} \frac{1}{3} \, a e x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \operatorname{arsech}\left (c x\right ) - \frac{\frac{\sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac{\arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b e + a d x + \frac{{\left (c x \operatorname{arsech}\left (c x\right ) - \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )\right )} b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e*x^3 + 1/6*(2*x^3*arcsech(c*x) - (sqrt(1/(c^2*x^2) - 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + arctan(sqrt(1/(
c^2*x^2) - 1))/c^2)/c)*b*e + a*d*x + (c*x*arcsech(c*x) - arctan(sqrt(1/(c^2*x^2) - 1)))*b*d/c

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Fricas [B]  time = 2.5259, size = 460, normalized size = 4.11 \begin{align*} \frac{2 \, a c^{3} e x^{3} - b c^{2} e x^{2} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 6 \, a c^{3} d x - 2 \,{\left (6 \, b c^{2} d + b e\right )} \arctan \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 2 \,{\left (3 \, b c^{3} d + b c^{3} e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 2 \,{\left (b c^{3} e x^{3} + 3 \, b c^{3} d x - 3 \, b c^{3} d - b c^{3} e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e*x^3 - b*c^2*e*x^2*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 6*a*c^3*d*x - 2*(6*b*c^2*d + b*e)*arctan((c*
x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(c*x)) - 2*(3*b*c^3*d + b*c^3*e)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))
- 1)/x) + 2*(b*c^3*e*x^3 + 3*b*c^3*d*x - 3*b*c^3*d - b*c^3*e)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c
*x)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asech(c*x)),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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