3.92 \(\int \frac {(d+e x^2) (a+b \text {sech}^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=126 \[ -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (2 c^2 d+9 e\right )}{9 x}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{9 x^3} \]

[Out]

-1/3*d*(a+b*arcsech(c*x))/x^3-e*(a+b*arcsech(c*x))/x+1/9*b*d*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2
)/x^3+1/9*b*(2*c^2*d+9*e)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.08, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 6301, 12, 453, 264} \[ -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (2 c^2 d+9 e\right )}{9 x}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^4,x]

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x^3) + (b*(2*c^2*d + 9*e)*Sqrt[(1 + c*x)^(-1)]*S
qrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x) - (d*(a + b*ArcSech[c*x]))/(3*x^3) - (e*(a + b*ArcSech[c*x]))/x

Rule 12

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

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 264

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -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 ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d-3 e x^2}{3 x^4 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+\frac {1}{3} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d-3 e x^2}{x^4 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x^3}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+\frac {1}{9} \left (b \left (-2 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x^2 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x^3}+\frac {b \left (2 c^2 d+9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 76, normalized size = 0.60 \[ \frac {-3 a \left (d+3 e x^2\right )+b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (2 c^2 d x^2+d+9 e x^2\right )-3 b \text {sech}^{-1}(c x) \left (d+3 e x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^4,x]

[Out]

(-3*a*(d + 3*e*x^2) + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + 2*c^2*d*x^2 + 9*e*x^2) - 3*b*(d + 3*e*x^2)*Ar
cSech[c*x])/(9*x^3)

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fricas [A]  time = 0.61, size = 106, normalized size = 0.84 \[ -\frac {9 \, a e x^{2} + 3 \, a d + 3 \, {\left (3 \, b e x^{2} + b d\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c d x + {\left (2 \, b c^{3} d + 9 \, b c e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{9 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(9*a*e*x^2 + 3*a*d + 3*(3*b*e*x^2 + b*d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c*d*x +
(2*b*c^3*d + 9*b*c*e)*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsech(c*x) + a)/x^4, x)

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maple [A]  time = 0.07, size = 123, normalized size = 0.98 \[ c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\mathrm {arcsech}\left (c x \right ) e}{c x}-\frac {\mathrm {arcsech}\left (c x \right ) d}{3 c \,x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{4} d \,x^{2}+9 c^{2} x^{2} e +c^{2} d \right )}{9 c^{2} x^{2}}\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x)

[Out]

c^3*(a/c^2*(-e/c/x-1/3/c*d/x^3)+b/c^2*(-arcsech(c*x)*e/c/x-1/3*arcsech(c*x)/c*d/x^3+1/9*(-(c*x-1)/c/x)^(1/2)/c
^2/x^2*((c*x+1)/c/x)^(1/2)*(2*c^4*d*x^2+9*c^2*e*x^2+c^2*d)))

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maxima [A]  time = 0.31, size = 91, normalized size = 0.72 \[ {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b e + \frac {1}{9} \, b d {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x, algorithm="maxima")

[Out]

(c*sqrt(1/(c^2*x^2) - 1) - arcsech(c*x)/x)*b*e + 1/9*b*d*((c^4*(1/(c^2*x^2) - 1)^(3/2) + 3*c^4*sqrt(1/(c^2*x^2
) - 1))/c - 3*arcsech(c*x)/x^3) - a*e/x - 1/3*a*d/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)*(a + b*acosh(1/(c*x))))/x^4,x)

[Out]

int(((d + e*x^2)*(a + b*acosh(1/(c*x))))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asech(c*x))/x**4,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)/x**4, x)

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