3.93 \(\int \frac {(d+e x^2) (a+b \text {sech}^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=183 \[ -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\frac {2 b c^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (12 c^2 d+25 e\right )}{225 x}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (12 c^2 d+25 e\right )}{225 x^3}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{25 x^5} \]

[Out]

-1/5*d*(a+b*arcsech(c*x))/x^5-1/3*e*(a+b*arcsech(c*x))/x^3+1/25*b*d*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+
1)^(1/2)/x^5+1/225*b*(12*c^2*d+25*e)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x^3+2/225*b*c^2*(12*c^
2*d+25*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 = 0.10, antiderivative size = 183, normalized size of antiderivative = 1.00, 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, 453, 271, 264} \[ -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\frac {2 b c^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (12 c^2 d+25 e\right )}{225 x}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (12 c^2 d+25 e\right )}{225 x^3}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{25 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^6,x]

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(25*x^5) + (b*(12*c^2*d + 25*e)*Sqrt[(1 + c*x)^(-1)
]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(225*x^3) + (2*b*c^2*(12*c^2*d + 25*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*S
qrt[1 - c^2*x^2])/(225*x) - (d*(a + b*ArcSech[c*x]))/(5*x^5) - (e*(a + b*ArcSech[c*x]))/(3*x^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 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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(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^6} \, dx &=-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-3 d-5 e x^2}{15 x^6 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\frac {1}{15} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-3 d-5 e x^2}{x^6 \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}}{25 x^5}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\frac {1}{75} \left (b \left (-12 c^2 d-25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{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}}{25 x^5}+\frac {b \left (12 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{225 x^3}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}+\frac {1}{225} \left (2 b c^2 \left (-12 c^2 d-25 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}}{25 x^5}+\frac {b \left (12 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{225 x^3}+\frac {2 b c^2 \left (12 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{225 x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 101, normalized size = 0.55 \[ \frac {-15 a \left (3 d+5 e x^2\right )+b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (25 e x^2 \left (2 c^2 x^2+1\right )+3 d \left (8 c^4 x^4+4 c^2 x^2+3\right )\right )-15 b \text {sech}^{-1}(c x) \left (3 d+5 e x^2\right )}{225 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^6,x]

[Out]

(-15*a*(3*d + 5*e*x^2) + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(25*e*x^2*(1 + 2*c^2*x^2) + 3*d*(3 + 4*c^2*x^2
+ 8*c^4*x^4)) - 15*b*(3*d + 5*e*x^2)*ArcSech[c*x])/(225*x^5)

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fricas [A]  time = 0.63, size = 128, normalized size = 0.70 \[ -\frac {75 \, a e x^{2} + 45 \, a d + 15 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (2 \, {\left (12 \, b c^{5} d + 25 \, b c^{3} e\right )} x^{5} + 9 \, b c d x + {\left (12 \, b c^{3} d + 25 \, b c e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/225*(75*a*e*x^2 + 45*a*d + 15*(5*b*e*x^2 + 3*b*d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (2*
(12*b*c^5*d + 25*b*c^3*e)*x^5 + 9*b*c*d*x + (12*b*c^3*d + 25*b*c*e)*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsech(c*x) + a)/x^6, x)

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maple [A]  time = 0.07, size = 142, normalized size = 0.78 \[ c^{5} \left (\frac {a \left (-\frac {e}{3 c^{3} x^{3}}-\frac {d}{5 c^{3} x^{5}}\right )}{c^{2}}+\frac {b \left (-\frac {\mathrm {arcsech}\left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\mathrm {arcsech}\left (c x \right ) d}{5 c^{3} x^{5}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (24 c^{6} d \,x^{4}+50 c^{4} e \,x^{4}+12 c^{4} d \,x^{2}+25 c^{2} x^{2} e +9 c^{2} d \right )}{225 c^{4} x^{4}}\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arcsech(c*x))/x^6,x)

[Out]

c^5*(a/c^2*(-1/3*e/c^3/x^3-1/5/c^3*d/x^5)+b/c^2*(-1/3*arcsech(c*x)*e/c^3/x^3-1/5*arcsech(c*x)/c^3*d/x^5+1/225*
(-(c*x-1)/c/x)^(1/2)/c^4/x^4*((c*x+1)/c/x)^(1/2)*(24*c^6*d*x^4+50*c^4*e*x^4+12*c^4*d*x^2+25*c^2*e*x^2+9*c^2*d)
))

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maxima [A]  time = 0.32, size = 132, normalized size = 0.72 \[ \frac {1}{75} \, b d {\left (\frac {3 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {5}{2}} + 10 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {15 \, \operatorname {arsech}\left (c x\right )}{x^{5}}\right )} + \frac {1}{9} \, b e {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{3 \, x^{3}} - \frac {a d}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^6,x, algorithm="maxima")

[Out]

1/75*b*d*((3*c^6*(1/(c^2*x^2) - 1)^(5/2) + 10*c^6*(1/(c^2*x^2) - 1)^(3/2) + 15*c^6*sqrt(1/(c^2*x^2) - 1))/c -
15*arcsech(c*x)/x^5) + 1/9*b*e*((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) - 1/3*a*e/x^3 - 1/5*a*d/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)*(a + b*acosh(1/(c*x))))/x^6,x)

[Out]

int(((d + e*x^2)*(a + b*acosh(1/(c*x))))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asech(c*x))/x**6,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)/x**6, x)

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