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

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

[Out]

-d^2*(a+b*arcsech(c*x))/x+2*d*e*x*(a+b*arcsech(c*x))+1/3*e^2*x^3*(a+b*arcsech(c*x))+1/6*b*e*(12*c^2*d+e)*arcsi
n(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^3+b*d^2*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x-1/6*b*e^
2*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 = 0.13, antiderivative size = 177, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]))

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*}

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Mathematica [C]  time = 0.32, size = 158, normalized size = 0.89 \[ \frac {2 a c^3 \left (-3 d^2+6 d e x^2+e^2 x^4\right )+2 b c^3 \text {sech}^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )-b c \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (e^2 x^2-6 c^2 d^2\right )+i b e x \left (12 c^2 d+e\right ) \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{6 c^3 x} \]

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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fricas [B]  time = 1.84, size = 287, normalized size = 1.62 \[ \frac {2 \, a c^{3} e^{2} x^{4} + 12 \, a c^{3} d e x^{2} - 6 \, a c^{3} d^{2} - 2 \, {\left (12 \, b c^{2} d e + b e^{2}\right )} x \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) + 2 \, {\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 2 \, {\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2} + {\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (6 \, b c^{4} d^{2} x - b c^{2} e^{2} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{6 \, c^{3} x} \]

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

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)

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maple [A]  time = 0.08, size = 197, normalized size = 1.11 \[ c \left (\frac {a \left (\frac {c^{3} x^{3} e^{2}}{3}+2 c^{3} d e x -\frac {d^{2} c^{3}}{x}\right )}{c^{4}}+\frac {b \left (\frac {e^{2} \mathrm {arcsech}\left (c x \right ) c^{3} x^{3}}{3}+2 \,\mathrm {arcsech}\left (c x \right ) c^{3} d e x -\frac {\mathrm {arcsech}\left (c x \right ) d^{2} c^{3}}{x}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (6 \sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2}+12 \arcsin \left (c x \right ) c^{3} x d e -c^{2} x^{2} e^{2} \sqrt {-c^{2} x^{2}+1}+\arcsin \left (c x \right ) c x \,e^{2}\right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{4}}\right ) \]

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*d*e*x-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*d*e*x-
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 = 0.41, size = 152, normalized size = 0.86 \[ \frac {1}{3} \, a e^{2} x^{3} + {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {arsech}\left (c x\right ) - \frac {\frac {\sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b e^{2} + 2 \, a d e x + \frac {2 \, {\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b d e}{c} - \frac {a d^{2}}{x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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