3.105 \(\int \frac{(d+e x^2)^2 (a+b \text{sech}^{-1}(c x))}{x^8} \, dx\)

Optimal. Leaf size=281 \[ -\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}+\frac{2 b c^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x^3}+\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{49 x^7}+\frac{2 b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (15 c^2 d+49 e\right )}{1225 x^5} \]

[Out]

(b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(49*x^7) + (2*b*d*(15*c^2*d + 49*e)*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(1225*x^5) + (b*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(11025*x^3) + (2*b*c^2*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[(
1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(11025*x) - (d^2*(a + b*ArcSech[c*x]))/(7*x^7) - (2*d*e*(a + b
*ArcSech[c*x]))/(5*x^5) - (e^2*(a + b*ArcSech[c*x]))/(3*x^3)

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Rubi [A]  time = 0.200737, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {270, 6301, 12, 1265, 453, 271, 264} \[ -\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}+\frac{2 b c^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x^3}+\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{49 x^7}+\frac{2 b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (15 c^2 d+49 e\right )}{1225 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcSech[c*x]))/x^8,x]

[Out]

(b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(49*x^7) + (2*b*d*(15*c^2*d + 49*e)*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(1225*x^5) + (b*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(11025*x^3) + (2*b*c^2*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[(
1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(11025*x) - (d^2*(a + b*ArcSech[c*x]))/(7*x^7) - (2*d*e*(a + b
*ArcSech[c*x]))/(5*x^5) - (e^2*(a + b*ArcSech[c*x]))/(3*x^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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{x^8} \, dx &=-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \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{-15 d^2-42 d e x^2-35 e^2 x^4}{105 x^8 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}+\frac{1}{105} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{-15 d^2-42 d e x^2-35 e^2 x^4}{x^8 \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}}{49 x^7}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{1}{735} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{6 d \left (15 c^2 d+49 e\right )+245 e^2 x^2}{x^6 \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}}{49 x^7}+\frac{2 b d \left (15 c^2 d+49 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{1225 x^5}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{\left (b \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{x^4 \sqrt{1-c^2 x^2}} \, dx}{3675}\\ &=\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{49 x^7}+\frac{2 b d \left (15 c^2 d+49 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{1225 x^5}+\frac{b \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{11025 x^3}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{\left (2 b c^2 \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{x^2 \sqrt{1-c^2 x^2}} \, dx}{11025}\\ &=\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{49 x^7}+\frac{2 b d \left (15 c^2 d+49 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{1225 x^5}+\frac{b \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{11025 x^3}+\frac{2 b c^2 \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{11025 x}-\frac{d^2 \left (a+b \text{sech}^{-1}(c x)\right )}{7 x^7}-\frac{2 d e \left (a+b \text{sech}^{-1}(c x)\right )}{5 x^5}-\frac{e^2 \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.332848, size = 160, normalized size = 0.57 \[ \frac{-105 a \left (15 d^2+42 d e x^2+35 e^2 x^4\right )+b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (45 d^2 \left (16 c^6 x^6+8 c^4 x^4+6 c^2 x^2+5\right )+294 d e x^2 \left (8 c^4 x^4+4 c^2 x^2+3\right )+1225 e^2 x^4 \left (2 c^2 x^2+1\right )\right )-105 b \text{sech}^{-1}(c x) \left (15 d^2+42 d e x^2+35 e^2 x^4\right )}{11025 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcSech[c*x]))/x^8,x]

[Out]

(-105*a*(15*d^2 + 42*d*e*x^2 + 35*e^2*x^4) + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(1225*e^2*x^4*(1 + 2*c^2*x^
2) + 294*d*e*x^2*(3 + 4*c^2*x^2 + 8*c^4*x^4) + 45*d^2*(5 + 6*c^2*x^2 + 8*c^4*x^4 + 16*c^6*x^6)) - 105*b*(15*d^
2 + 42*d*e*x^2 + 35*e^2*x^4)*ArcSech[c*x])/(11025*x^7)

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Maple [A]  time = 0.189, size = 225, normalized size = 0.8 \begin{align*}{c}^{7} \left ({\frac{a}{{c}^{4}} \left ( -{\frac{{d}^{2}}{7\,{c}^{3}{x}^{7}}}-{\frac{2\,de}{5\,{c}^{3}{x}^{5}}}-{\frac{{e}^{2}}{3\,{c}^{3}{x}^{3}}} \right ) }+{\frac{b}{{c}^{4}} \left ( -{\frac{{\rm arcsech} \left (cx\right ){d}^{2}}{7\,{c}^{3}{x}^{7}}}-{\frac{2\,{\rm arcsech} \left (cx\right )de}{5\,{c}^{3}{x}^{5}}}-{\frac{{\rm arcsech} \left (cx\right ){e}^{2}}{3\,{c}^{3}{x}^{3}}}+{\frac{720\,{c}^{10}{d}^{2}{x}^{6}+2352\,{c}^{8}de{x}^{6}+360\,{c}^{8}{d}^{2}{x}^{4}+2450\,{c}^{6}{e}^{2}{x}^{6}+1176\,{c}^{6}de{x}^{4}+270\,{c}^{6}{d}^{2}{x}^{2}+1225\,{c}^{4}{e}^{2}{x}^{4}+882\,{c}^{4}de{x}^{2}+225\,{d}^{2}{c}^{4}}{11025\,{c}^{6}{x}^{6}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}} \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^8,x)

[Out]

c^7*(a/c^4*(-1/7*d^2/c^3/x^7-2/5/c^3*d*e/x^5-1/3*e^2/c^3/x^3)+b/c^4*(-1/7*arcsech(c*x)*d^2/c^3/x^7-2/5*arcsech
(c*x)/c^3*d*e/x^5-1/3*arcsech(c*x)*e^2/c^3/x^3+1/11025*(-(c*x-1)/c/x)^(1/2)/c^6/x^6*((c*x+1)/c/x)^(1/2)*(720*c
^10*d^2*x^6+2352*c^8*d*e*x^6+360*c^8*d^2*x^4+2450*c^6*e^2*x^6+1176*c^6*d*e*x^4+270*c^6*d^2*x^2+1225*c^4*e^2*x^
4+882*c^4*d*e*x^2+225*c^4*d^2)))

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Maxima [A]  time = 1.01764, size = 313, normalized size = 1.11 \begin{align*} \frac{1}{245} \, b d^{2}{\left (\frac{5 \, c^{8}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{7}{2}} + 21 \, c^{8}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{5}{2}} + 35 \, c^{8}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} + 35 \, c^{8} \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c} - \frac{35 \, \operatorname{arsech}\left (c x\right )}{x^{7}}\right )} + \frac{2}{75} \, b d e{\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^{2}{\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^{2}}{3 \, x^{3}} - \frac{2 \, a d e}{5 \, x^{5}} - \frac{a d^{2}}{7 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

1/245*b*d^2*((5*c^8*(1/(c^2*x^2) - 1)^(7/2) + 21*c^8*(1/(c^2*x^2) - 1)^(5/2) + 35*c^8*(1/(c^2*x^2) - 1)^(3/2)
+ 35*c^8*sqrt(1/(c^2*x^2) - 1))/c - 35*arcsech(c*x)/x^7) + 2/75*b*d*e*((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^2*((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^2/x^3 - 2/5*a*d*e/x^5 - 1/7*a*d
^2/x^7

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Fricas [A]  time = 1.87113, size = 483, normalized size = 1.72 \begin{align*} -\frac{3675 \, a e^{2} x^{4} + 4410 \, a d e x^{2} + 1575 \, a d^{2} + 105 \,{\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (2 \,{\left (360 \, b c^{7} d^{2} + 1176 \, b c^{5} d e + 1225 \, b c^{3} e^{2}\right )} x^{7} +{\left (360 \, b c^{5} d^{2} + 1176 \, b c^{3} d e + 1225 \, b c e^{2}\right )} x^{5} + 225 \, b c d^{2} x + 18 \,{\left (15 \, b c^{3} d^{2} + 49 \, b c d e\right )} x^{3}\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{11025 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

-1/11025*(3675*a*e^2*x^4 + 4410*a*d*e*x^2 + 1575*a*d^2 + 105*(35*b*e^2*x^4 + 42*b*d*e*x^2 + 15*b*d^2)*log((c*x
*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (2*(360*b*c^7*d^2 + 1176*b*c^5*d*e + 1225*b*c^3*e^2)*x^7 + (360*
b*c^5*d^2 + 1176*b*c^3*d*e + 1225*b*c*e^2)*x^5 + 225*b*c*d^2*x + 18*(15*b*c^3*d^2 + 49*b*c*d*e)*x^3)*sqrt(-(c^
2*x^2 - 1)/(c^2*x^2)))/x^7

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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^{8}}\, 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**8,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)**2/x**8, 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^{8}}\,{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^8,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arcsech(c*x) + a)/x^8, x)