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]

(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

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Rubi [A]  time = 0.138995, antiderivative size = 176, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [C]  time = 0.268651, 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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Maple [A]  time = 0.185, size = 205, normalized size = 1.2 \begin{align*}{c}^{3} \left ({\frac{a}{{c}^{4}} \left ( cx{e}^{2}-2\,{\frac{cde}{x}}-{\frac{c{d}^{2}}{3\,{x}^{3}}} \right ) }+{\frac{b}{{c}^{4}} \left ({\rm arcsech} \left (cx\right )cx{e}^{2}-2\,{\frac{{\rm arcsech} \left (cx\right )cde}{x}}-{\frac{{\rm arcsech} \left (cx\right ){d}^{2}c}{3\,{x}^{3}}}+{\frac{1}{9\,{c}^{2}{x}^{2}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( 2\,\sqrt{-{c}^{2}{x}^{2}+1}{c}^{6}{x}^{2}{d}^{2}+18\,\sqrt{-{c}^{2}{x}^{2}+1}{c}^{4}{x}^{2}de+\sqrt{-{c}^{2}{x}^{2}+1}{c}^{4}{d}^{2}+9\,\arcsin \left ( cx \right ){c}^{3}{x}^{3}{e}^{2} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \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^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^2*x
^2+1)^(1/2)*c^4*x^2*d*e+(-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 = 0.978516, size = 181, normalized size = 1.03 \begin{align*} 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}} \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^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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Fricas [B]  time = 2.23457, size = 590, normalized size = 3.35 \begin{align*} \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}} \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^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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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^{4}}\, 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**4,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)**2/x**4, 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^{4}}\,{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^4,x, algorithm="giac")

[Out]

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