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

Optimal. Leaf size=174 \[ \frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )-\frac{b x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (20 c^2 d+9 e\right )}{120 c^4}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (20 c^2 d+9 e\right ) \sin ^{-1}(c x)}{120 c^5}-\frac{b e x^3 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{20 c^2} \]

[Out]

-(b*(20*c^2*d + 9*e)*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(120*c^4) - (b*e*x^3*Sqrt[(1 + c*
x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(20*c^2) + (d*x^3*(a + b*ArcSech[c*x]))/3 + (e*x^5*(a + b*ArcSech[c*
x]))/5 + (b*(20*c^2*d + 9*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(120*c^5)

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Rubi [A]  time = 0.102986, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {14, 6301, 12, 459, 321, 216} \[ \frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )-\frac{b x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (20 c^2 d+9 e\right )}{120 c^4}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (20 c^2 d+9 e\right ) \sin ^{-1}(c x)}{120 c^5}-\frac{b e x^3 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{20 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(20*c^2*d + 9*e)*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(120*c^4) - (b*e*x^3*Sqrt[(1 + c*
x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(20*c^2) + (d*x^3*(a + b*ArcSech[c*x]))/3 + (e*x^5*(a + b*ArcSech[c*
x]))/5 + (b*(20*c^2*d + 9*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(120*c^5)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6301

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(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, f, m, p}, x] &&
((IGtQ[p, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ
[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 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 459

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

Rule 321

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

Rubi steps

\begin{align*} \int x^2 \left (d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^2 \left (5 d+3 e x^2\right )}{15 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{15} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^2 \left (5 d+3 e x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b e x^3 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{20 c^2}+\frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{60} \left (b \left (20 d+\frac{9 e}{c^2}\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b \left (20 c^2 d+9 e\right ) x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{120 c^4}-\frac{b e x^3 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{20 c^2}+\frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{\left (b \left (20 d+\frac{9 e}{c^2}\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{120 c^2}\\ &=-\frac{b \left (20 c^2 d+9 e\right ) x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{120 c^4}-\frac{b e x^3 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{20 c^2}+\frac{1}{3} d x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b \left (20 c^2 d+9 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{120 c^5}\\ \end{align*}

Mathematica [C]  time = 0.208183, size = 144, normalized size = 0.83 \[ \frac{8 a c^5 x^3 \left (5 d+3 e x^2\right )-b c x \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (c^2 \left (20 d+6 e x^2\right )+9 e\right )+8 b c^5 x^3 \text{sech}^{-1}(c x) \left (5 d+3 e x^2\right )+i b \left (20 c^2 d+9 e\right ) \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{120 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*c^5*x^3*(5*d + 3*e*x^2) - b*c*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(9*e + c^2*(20*d + 6*e*x^2)) + 8*b*c^
5*x^3*(5*d + 3*e*x^2)*ArcSech[c*x] + I*b*(20*c^2*d + 9*e)*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*
x)])/(120*c^5)

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Maple [A]  time = 0.185, size = 182, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{a}{{c}^{2}} \left ({\frac{{c}^{5}{x}^{5}e}{5}}+{\frac{{c}^{5}{x}^{3}d}{3}} \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\rm arcsech} \left (cx\right ){c}^{5}{x}^{5}e}{5}}+{\frac{{\rm arcsech} \left (cx\right ){c}^{5}{x}^{3}d}{3}}-{\frac{cx}{120}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( 6\,e{c}^{3}{x}^{3}\sqrt{-{c}^{2}{x}^{2}+1}+20\,\sqrt{-{c}^{2}{x}^{2}+1}{c}^{3}xd-20\,\arcsin \left ( cx \right ){c}^{2}d+9\,ecx\sqrt{-{c}^{2}{x}^{2}+1}-9\,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(x^2*(e*x^2+d)*(a+b*arcsech(c*x)),x)

[Out]

1/c^3*(a/c^2*(1/5*c^5*x^5*e+1/3*c^5*x^3*d)+b/c^2*(1/5*arcsech(c*x)*c^5*x^5*e+1/3*arcsech(c*x)*c^5*x^3*d-1/120*
(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(6*e*c^3*x^3*(-c^2*x^2+1)^(1/2)+20*(-c^2*x^2+1)^(1/2)*c^3*x*d-20*
arcsin(c*x)*c^2*d+9*e*c*x*(-c^2*x^2+1)^(1/2)-9*e*arcsin(c*x))/(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.50551, size = 246, normalized size = 1.41 \begin{align*} \frac{1}{5} \, a e x^{5} + \frac{1}{3} \, a d 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 d + \frac{1}{40} \,{\left (8 \, x^{5} \operatorname{arsech}\left (c x\right ) - \frac{\frac{3 \,{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} + 5 \, \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{4}} + \frac{3 \, \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )}{c^{4}}}{c}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e*x^5 + 1/3*a*d*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) + a
rctan(sqrt(1/(c^2*x^2) - 1))/c^2)/c)*b*d + 1/40*(8*x^5*arcsech(c*x) - ((3*(1/(c^2*x^2) - 1)^(3/2) + 5*sqrt(1/(
c^2*x^2) - 1))/(c^4*(1/(c^2*x^2) - 1)^2 + 2*c^4*(1/(c^2*x^2) - 1) + c^4) + 3*arctan(sqrt(1/(c^2*x^2) - 1))/c^4
)/c)*b*e

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Fricas [B]  time = 2.73193, size = 531, normalized size = 3.05 \begin{align*} \frac{24 \, a c^{5} e x^{5} + 40 \, a c^{5} d x^{3} - 2 \,{\left (20 \, b c^{2} d + 9 \, b e\right )} \arctan \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 8 \,{\left (5 \, b c^{5} d + 3 \, b c^{5} e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 8 \,{\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3} - 5 \, b c^{5} d - 3 \, b c^{5} e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (6 \, b c^{4} e x^{4} +{\left (20 \, b c^{4} d + 9 \, b c^{2} e\right )} x^{2}\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{120 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*a*c^5*e*x^5 + 40*a*c^5*d*x^3 - 2*(20*b*c^2*d + 9*b*e)*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1
)/(c*x)) - 8*(5*b*c^5*d + 3*b*c^5*e)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) + 8*(3*b*c^5*e*x^5 + 5*b*
c^5*d*x^3 - 5*b*c^5*d - 3*b*c^5*e)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (6*b*c^4*e*x^4 + (20*
b*c^4*d + 9*b*c^2*e)*x^2)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/c^5

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

[Out]

Integral(x**2*(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 )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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