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

Optimal. Leaf size=176 \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b d^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{9 x^3}+\frac {2 b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (c^2 d+9 e\right )}{9 x}+\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{c} \]

[Out]

-1/3*d^2*(a+b*arcsech(c*x))/x^3-2*d*e*(a+b*arcsech(c*x))/x+e^2*x*(a+b*arcsech(c*x))+b*e^2*arcsin(c*x)*(1/(c*x+
1))^(1/2)*(c*x+1)^(1/2)/c+1/9*b*d^2*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x^3+2/9*b*d*(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 = 0.14, antiderivative size = 176, 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, 451, 216} \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b d^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{9 x^3}+\frac {2 b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (c^2 d+9 e\right )}{9 x}+\frac {b e^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 451

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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^4} \, dx &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d^2-6 d e x^2+3 e^2 x^4}{3 x^4 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \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 {-d^2-6 d e x^2+3 e^2 x^4}{x^4 \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}}{9 x^3}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )-\frac {1}{9} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {2 d \left (c^2 d+9 e\right )-9 e^2 x^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}}{9 x^3}+\frac {2 b d \left (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^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\left (b e^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\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}}{9 x^3}+\frac {2 b d \left (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^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b e^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{c}\\ \end {align*}

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Mathematica [C]  time = 0.31, size = 149, normalized size = 0.85 \[ \frac {-3 a c \left (d^2+6 d e x^2-3 e^2 x^4\right )+b c d \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (2 c^2 d x^2+d+18 e x^2\right )-3 b c \text {sech}^{-1}(c x) \left (d^2+6 d e x^2-3 e^2 x^4\right )+9 i b e^2 x^3 \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{9 c x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + 2*c^2*d*x^2 + 18*e*x^2) - 3*a*c*(d^2 + 6*d*e*x^2 - 3*e^2*x^4)
- 3*b*c*(d^2 + 6*d*e*x^2 - 3*e^2*x^4)*ArcSech[c*x] + (9*I)*b*e^2*x^3*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*
x)]*(1 + c*x)])/(9*c*x^3)

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fricas [B]  time = 0.70, size = 267, normalized size = 1.52 \[ \frac {9 \, a c e^{2} x^{4} - 18 \, b e^{2} x^{3} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 18 \, a c d e x^{2} + 3 \, {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 3 \, a c d^{2} + 3 \, {\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2} + {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (b c^{2} d^{2} x + 2 \, {\left (b c^{4} d^{2} + 9 \, b c^{2} d e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{9 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(9*a*c*e^2*x^4 - 18*b*e^2*x^3*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(c*x)) - 18*a*c*d*e*x^2 + 3*
(b*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) - 3*a*c*d^2 + 3*(3*b*c*e
^2*x^4 - 6*b*c*d*e*x^2 - b*c*d^2 + (b*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^
2)) + 1)/(c*x)) + (b*c^2*d^2*x + 2*(b*c^4*d^2 + 9*b*c^2*d*e)*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/(c*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 205, normalized size = 1.16 \[ c^{3} \left (\frac {a \left (c x \,e^{2}-\frac {2 c d e}{x}-\frac {d^{2} c}{3 x^{3}}\right )}{c^{4}}+\frac {b \left (\mathrm {arcsech}\left (c x \right ) c x \,e^{2}-\frac {2 \,\mathrm {arcsech}\left (c x \right ) c d e}{x}-\frac {\mathrm {arcsech}\left (c x \right ) d^{2} c}{3 x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 \sqrt {-c^{2} x^{2}+1}\, c^{6} x^{2} d^{2}+18 c^{4} d e \,x^{2} \sqrt {-c^{2} x^{2}+1}+\sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2}+9 \arcsin \left (c x \right ) c^{3} x^{3} e^{2}\right )}{9 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}\right )}{c^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(a/c^4*(c*x*e^2-2*c*d*e/x-1/3*d^2*c/x^3)+b/c^4*(arcsech(c*x)*c*x*e^2-2*arcsech(c*x)*c*d*e/x-1/3*arcsech(c*
x)*d^2*c/x^3+1/9*(-(c*x-1)/c/x)^(1/2)/c^2/x^2*((c*x+1)/c/x)^(1/2)*(2*(-c^2*x^2+1)^(1/2)*c^6*x^2*d^2+18*c^4*d*e
*x^2*(-c^2*x^2+1)^(1/2)+(-c^2*x^2+1)^(1/2)*c^4*d^2+9*arcsin(c*x)*c^3*x^3*e^2)/(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.31, size = 134, normalized size = 0.76 \[ 2 \, {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac {1}{9} \, b d^{2} {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b e^{2}}{c} - \frac {2 \, a d e}{x} - \frac {a d^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((d + e*x^2)^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 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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