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

Optimal. Leaf size=177 \[ -\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{x}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (12 c^2 d+e\right ) \sin ^{-1}(c x)}{6 c^3}-\frac{b e^2 x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{6 c^2} \]

[Out]

(b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/x - (b*e^2*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*S
qrt[1 - c^2*x^2])/(6*c^2) - (d^2*(a + b*ArcSech[c*x]))/x + 2*d*e*x*(a + b*ArcSech[c*x]) + (e^2*x^3*(a + b*ArcS
ech[c*x]))/3 + (b*e*(12*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 = 0.13156, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {270, 6301, 12, 1265, 388, 216} \[ -\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{x}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (12 c^2 d+e\right ) \sin ^{-1}(c x)}{6 c^3}-\frac{b e^2 x \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{6 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/x - (b*e^2*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*S
qrt[1 - c^2*x^2])/(6*c^2) - (d^2*(a + b*ArcSech[c*x]))/x + 2*d*e*x*(a + b*ArcSech[c*x]) + (e^2*x^3*(a + b*ArcS
ech[c*x]))/3 + (b*e*(12*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(6*c^3)

Rule 270

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

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 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

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 \frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 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^2+6 d e x^2+e^2 x^4}{3 x^2 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 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^2+6 d e x^2+e^2 x^4}{x^2 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{x}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 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{-6 d e-e^2 x^2}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{x}-\frac{b e^2 x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{6 c^2}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \text{sech}^{-1}(c x)\right )+-\frac{\left (b \left (-12 c^2 d e-e^2\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 d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{x}-\frac{b e^2 x \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{6 c^2}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x}+2 d e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b e \left (12 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 = 0.24164, size = 158, normalized size = 0.89 \[ \frac{2 a c^3 \left (-3 d^2+6 d e x^2+e^2 x^4\right )-b c \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (e^2 x^2-6 c^2 d^2\right )+2 b c^3 \text{sech}^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )+i b e x \left (12 c^2 d+e\right ) \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{6 c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.187, size = 197, normalized size = 1.1 \begin{align*} c \left ({\frac{a}{{c}^{4}} \left ({\frac{{c}^{3}{x}^{3}{e}^{2}}{3}}+2\,{c}^{3}xde-{\frac{{d}^{2}{c}^{3}}{x}} \right ) }+{\frac{b}{{c}^{4}} \left ({\frac{{e}^{2}{\rm arcsech} \left (cx\right ){c}^{3}{x}^{3}}{3}}+2\,{\rm arcsech} \left (cx\right ){c}^{3}xde-{\frac{{\rm arcsech} \left (cx\right ){d}^{2}{c}^{3}}{x}}+{\frac{1}{6}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( 6\,\sqrt{-{c}^{2}{x}^{2}+1}{c}^{4}{d}^{2}+12\,\arcsin \left ( cx \right ){c}^{3}xde-{c}^{2}{x}^{2}{e}^{2}\sqrt{-{c}^{2}{x}^{2}+1}+\arcsin \left ( cx \right ) cx{e}^{2} \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)^2*(a+b*arcsech(c*x))/x^2,x)

[Out]

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

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Maxima [A]  time = 1.52381, size = 205, normalized size = 1.16 \begin{align*} \frac{1}{3} \, a e^{2} x^{3} +{\left (c \sqrt{\frac{1}{c^{2} x^{2}} - 1} - \frac{\operatorname{arsech}\left (c x\right )}{x}\right )} b d^{2} + \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^{2} + 2 \, a d e x + \frac{2 \,{\left (c x \operatorname{arsech}\left (c x\right ) - \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )\right )} b d e}{c} - \frac{a d^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.45177, size = 614, normalized size = 3.47 \begin{align*} \frac{2 \, a c^{3} e^{2} x^{4} + 12 \, a c^{3} d e x^{2} - 6 \, a c^{3} d^{2} - 2 \,{\left (12 \, b c^{2} d e + b e^{2}\right )} x \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^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x \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^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2} +{\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x\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} d^{2} x - b c^{2} e^{2} x^{3}\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{6 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e^2*x^4 + 12*a*c^3*d*e*x^2 - 6*a*c^3*d^2 - 2*(12*b*c^2*d*e + b*e^2)*x*arctan((c*x*sqrt(-(c^2*x^2
- 1)/(c^2*x^2)) - 1)/(c*x)) + 2*(3*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e^2)*x*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^
2)) - 1)/x) + 2*(b*c^3*e^2*x^4 + 6*b*c^3*d*e*x^2 - 3*b*c^3*d^2 + (3*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e^2)*x)*lo
g((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + (6*b*c^4*d^2*x - b*c^2*e^2*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x
^2)))/(c^3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{2}{\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((e*x^2+d)^2*(a+b*arcsech(c*x))/x^2,x, algorithm="giac")

[Out]

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