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

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

[Out]

-1/7*d^2*(a+b*arcsech(c*x))/x^7-2/5*d*e*(a+b*arcsech(c*x))/x^5-1/3*e^2*(a+b*arcsech(c*x))/x^3+1/49*b*d^2*(1/(c
*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x^7+2/1225*b*d*(15*c^2*d+49*e)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(
-c^2*x^2+1)^(1/2)/x^5+1/11025*b*(360*c^4*d^2+1176*c^2*d*e+1225*e^2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+
1)^(1/2)/x^3+2/11025*b*c^2*(360*c^4*d^2+1176*c^2*d*e+1225*e^2)*(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 = 0.20, antiderivative size = 281, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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^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*}

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Mathematica [A]  time = 0.40, size = 160, normalized size = 0.57 \[ \frac {-105 a \left (15 d^2+42 d e x^2+35 e^2 x^4\right )+b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (1225 e^2 x^4 \left (2 c^2 x^2+1\right )+294 d e x^2 \left (8 c^4 x^4+4 c^2 x^2+3\right )+45 d^2 \left (16 c^6 x^6+8 c^4 x^4+6 c^2 x^2+5\right )\right )-105 b \text {sech}^{-1}(c x) \left (15 d^2+42 d e x^2+35 e^2 x^4\right )}{11025 x^7} \]

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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fricas [A]  time = 0.55, size = 199, normalized size = 0.71 \[ -\frac {3675 \, a e^{2} x^{4} + 4410 \, a d e x^{2} + 1575 \, a d^{2} + 105 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (2 \, {\left (360 \, b c^{7} d^{2} + 1176 \, b c^{5} d e + 1225 \, b c^{3} e^{2}\right )} x^{7} + {\left (360 \, b c^{5} d^{2} + 1176 \, b c^{3} d e + 1225 \, b c e^{2}\right )} x^{5} + 225 \, b c d^{2} x + 18 \, {\left (15 \, b c^{3} d^{2} + 49 \, b c d e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{11025 \, x^{7}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 225, normalized size = 0.80 \[ c^{7} \left (\frac {a \left (-\frac {e^{2}}{3 c^{3} x^{3}}-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {2 d e}{5 c^{3} x^{5}}\right )}{c^{4}}+\frac {b \left (-\frac {\mathrm {arcsech}\left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {\mathrm {arcsech}\left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {2 \,\mathrm {arcsech}\left (c x \right ) d e}{5 c^{3} x^{5}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (720 c^{10} d^{2} x^{6}+2352 c^{8} d e \,x^{6}+360 c^{8} d^{2} x^{4}+2450 c^{6} e^{2} x^{6}+1176 c^{6} d e \,x^{4}+270 c^{6} d^{2} x^{2}+1225 c^{4} e^{2} x^{4}+882 c^{4} d e \,x^{2}+225 d^{2} c^{4}\right )}{11025 c^{6} x^{6}}\right )}{c^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^7*(a/c^4*(-1/3*e^2/c^3/x^3-1/7*d^2/c^3/x^7-2/5/c^3*d*e/x^5)+b/c^4*(-1/3*arcsech(c*x)*e^2/c^3/x^3-1/7*arcsech
(c*x)*d^2/c^3/x^7-2/5*arcsech(c*x)/c^3*d*e/x^5+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 = 0.32, size = 232, normalized size = 0.83 \[ \frac {1}{245} \, b d^{2} {\left (\frac {5 \, c^{8} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {7}{2}} + 21 \, c^{8} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {5}{2}} + 35 \, c^{8} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 35 \, c^{8} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {35 \, \operatorname {arsech}\left (c x\right )}{x^{7}}\right )} + \frac {2}{75} \, b d e {\left (\frac {3 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {5}{2}} + 10 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {15 \, \operatorname {arsech}\left (c x\right )}{x^{5}}\right )} + \frac {1}{9} \, b e^{2} {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {a d^{2}}{7 \, x^{7}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)^2*(a + b*acosh(1/(c*x))))/x^8,x)

[Out]

int(((d + e*x^2)^2*(a + b*acosh(1/(c*x))))/x^8, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{8}}\, dx \]

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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