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

Optimal. Leaf size=164 \[ -\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )+\frac{b c d^2 \sqrt{-c^2 x^2-1}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{2 b c d \sqrt{-c^2 x^2-1} \left (c^2 d-9 e\right )}{9 \sqrt{-c^2 x^2}}-\frac{b e^2 x \tan ^{-1}\left (\frac{c x}{\sqrt{-c^2 x^2-1}}\right )}{\sqrt{-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.142695, antiderivative size = 164, 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, 6302, 12, 1265, 451, 217, 203} \[ -\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )+\frac{b c d^2 \sqrt{-c^2 x^2-1}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{2 b c d \sqrt{-c^2 x^2-1} \left (c^2 d-9 e\right )}{9 \sqrt{-c^2 x^2}}-\frac{b e^2 x \tan ^{-1}\left (\frac{c x}{\sqrt{-c^2 x^2-1}}\right )}{\sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6302

Int[((a_.) + ArcCsch[(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*ArcCsch[c*x], u, x] - Dist[(b*c*x)/Sqrt[-(c^2*x^2)], Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{(b c x) \int \frac{-d^2-6 d e x^2+3 e^2 x^4}{3 x^4 \sqrt{-1-c^2 x^2}} \, dx}{\sqrt{-c^2 x^2}}\\ &=-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{(b c x) \int \frac{-d^2-6 d e x^2+3 e^2 x^4}{x^4 \sqrt{-1-c^2 x^2}} \, dx}{3 \sqrt{-c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1-c^2 x^2}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{(b c x) \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}{9 \sqrt{-c^2 x^2}}\\ &=-\frac{2 b c d \left (c^2 d-9 e\right ) \sqrt{-1-c^2 x^2}}{9 \sqrt{-c^2 x^2}}+\frac{b c d^2 \sqrt{-1-c^2 x^2}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{\left (b c e^2 x\right ) \int \frac{1}{\sqrt{-1-c^2 x^2}} \, dx}{\sqrt{-c^2 x^2}}\\ &=-\frac{2 b c d \left (c^2 d-9 e\right ) \sqrt{-1-c^2 x^2}}{9 \sqrt{-c^2 x^2}}+\frac{b c d^2 \sqrt{-1-c^2 x^2}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{\left (b c e^2 x\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x^2} \, dx,x,\frac{x}{\sqrt{-1-c^2 x^2}}\right )}{\sqrt{-c^2 x^2}}\\ &=-\frac{2 b c d \left (c^2 d-9 e\right ) \sqrt{-1-c^2 x^2}}{9 \sqrt{-c^2 x^2}}+\frac{b c d^2 \sqrt{-1-c^2 x^2}}{9 x^2 \sqrt{-c^2 x^2}}-\frac{d^2 \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \text{csch}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text{csch}^{-1}(c x)\right )-\frac{b e^2 x \tan ^{-1}\left (\frac{c x}{\sqrt{-1-c^2 x^2}}\right )}{\sqrt{-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.262792, size = 123, normalized size = 0.75 \[ \frac{b c d x \sqrt{\frac{1}{c^2 x^2}+1} \left (-2 c^2 d x^2+d+18 e x^2\right )-3 a \left (d^2+6 d e x^2-3 e^2 x^4\right )}{9 x^3}+\frac{b e^2 \log \left (x \left (\sqrt{\frac{1}{c^2 x^2}+1}+1\right )\right )}{c}-\frac{b \text{csch}^{-1}(c x) \left (d^2+6 d e x^2-3 e^2 x^4\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.191, size = 191, 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 arccsch} \left (cx\right )cx{e}^{2}-2\,{\frac{{\rm arccsch} \left (cx\right )cde}{x}}-{\frac{{\rm arccsch} \left (cx\right ){d}^{2}c}{3\,{x}^{3}}}-{\frac{1}{9\,{c}^{4}{x}^{4}}\sqrt{{c}^{2}{x}^{2}+1} \left ( 2\,\sqrt{{c}^{2}{x}^{2}+1}{c}^{6}{x}^{2}{d}^{2}-18\,{c}^{4}de\sqrt{{c}^{2}{x}^{2}+1}{x}^{2}-{d}^{2}{c}^{4}\sqrt{{c}^{2}{x}^{2}+1}-9\,{e}^{2}{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.02882, size = 205, normalized size = 1.25 \begin{align*} 2 \,{\left (c \sqrt{\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsch}\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{arcsch}\left (c x\right )}{x^{3}}\right )} + \frac{{\left (2 \, c x \operatorname{arcsch}\left (c x\right ) + \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + 1\right ) - \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} - 1\right )\right )} b e^{2}}{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*arccsch(c*x))/x^4,x, algorithm="maxima")

[Out]

2*(c*sqrt(1/(c^2*x^2) + 1) - arccsch(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*arccsch(c*x)/x^3) + 1/2*(2*c*x*arccsch(c*x) + log(sqrt(1/(c^2*x^2) + 1) + 1) - l
og(sqrt(1/(c^2*x^2) + 1) - 1))*b*e^2/c - 2*a*d*e/x - 1/3*a*d^2/x^3

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Fricas [B]  time = 3.01253, size = 743, normalized size = 4.53 \begin{align*} \frac{9 \, a c e^{2} x^{4} - 9 \, b e^{2} x^{3} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 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 (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) + 3 \,{\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) - 3 \, a c d^{2} - 2 \,{\left (b c^{4} d^{2} - 9 \, b c^{2} d e\right )} x^{3} + 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*arccsch(c*x))/x^4,x, algorithm="fricas")

[Out]

1/9*(9*a*c*e^2*x^4 - 9*b*e^2*x^3*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 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)) - c*x + 1) + 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)) - c*x - 1) - 3*a*c*d^2 - 2*(b*c^4*d^2 - 9*b*c^2*d*e)*x^3 + 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{acsch}{\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*acsch(c*x))/x**4,x)

[Out]

Integral((a + b*acsch(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{arcsch}\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*arccsch(c*x))/x^4,x, algorithm="giac")

[Out]

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