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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 12

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

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]

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 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])

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*}

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Mathematica [C]  time = 0.37, size = 169, normalized size = 1.51 \[ a d x+\frac {1}{3} a e x^3+\frac {i b e \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{6 c^3}-\frac {b d \sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c (c x-1)}+b e \sqrt {\frac {1-c x}{c x+1}} \left (-\frac {x}{6 c^2}-\frac {x^2}{6 c}\right )+b d x \text {sech}^{-1}(c x)+\frac {1}{3} b e x^3 \text {sech}^{-1}(c x) \]

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)]*(-1/6*x/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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fricas [B]  time = 0.98, size = 209, normalized size = 1.87 \[ \frac {2 \, a c^{3} e x^{3} - b c^{2} e x^{2} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 6 \, a c^{3} d x - 2 \, {\left (6 \, b c^{2} d + b e\right )} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 2 \, {\left (3 \, b c^{3} d + b c^{3} e\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 2 \, {\left (b c^{3} e x^{3} + 3 \, b c^{3} d x - 3 \, b c^{3} d - b c^{3} e\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 135, normalized size = 1.21 \[ \frac {\frac {a \left (\frac {1}{3} e \,c^{3} x^{3}+x \,c^{3} d \right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arcsech}\left (c x \right ) e \,c^{3} x^{3}}{3}+\mathrm {arcsech}\left (c x \right ) c^{3} d x +\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (6 \arcsin \left (c x \right ) c^{2} d -e c x \sqrt {-c^{2} x^{2}+1}+e \arcsin \left (c x \right )\right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}}{c} \]

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 = 0.41, size = 107, normalized size = 0.96 \[ \frac {1}{3} \, a e x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {arsech}\left (c x\right ) - \frac {\frac {\sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b e + a d x + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b d}{c} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)*(a + b*acosh(1/(c*x))),x)

[Out]

int((d + e*x^2)*(a + b*acosh(1/(c*x))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

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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