3.81 \(\int \frac{(d+e x^2) (a+b \text{csch}^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=158 \[ -\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}+\frac{2 b c^3 \sqrt{-c^2 x^2-1} \left (12 c^2 d-25 e\right )}{225 \sqrt{-c^2 x^2}}-\frac{b c \sqrt{-c^2 x^2-1} \left (12 c^2 d-25 e\right )}{225 x^2 \sqrt{-c^2 x^2}}+\frac{b c d \sqrt{-c^2 x^2-1}}{25 x^4 \sqrt{-c^2 x^2}} \]

[Out]

(2*b*c^3*(12*c^2*d - 25*e)*Sqrt[-1 - c^2*x^2])/(225*Sqrt[-(c^2*x^2)]) + (b*c*d*Sqrt[-1 - c^2*x^2])/(25*x^4*Sqr
t[-(c^2*x^2)]) - (b*c*(12*c^2*d - 25*e)*Sqrt[-1 - c^2*x^2])/(225*x^2*Sqrt[-(c^2*x^2)]) - (d*(a + b*ArcCsch[c*x
]))/(5*x^5) - (e*(a + b*ArcCsch[c*x]))/(3*x^3)

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Rubi [A]  time = 0.102568, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {14, 6302, 12, 453, 271, 264} \[ -\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}+\frac{2 b c^3 \sqrt{-c^2 x^2-1} \left (12 c^2 d-25 e\right )}{225 \sqrt{-c^2 x^2}}-\frac{b c \sqrt{-c^2 x^2-1} \left (12 c^2 d-25 e\right )}{225 x^2 \sqrt{-c^2 x^2}}+\frac{b c d \sqrt{-c^2 x^2-1}}{25 x^4 \sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcCsch[c*x]))/x^6,x]

[Out]

(2*b*c^3*(12*c^2*d - 25*e)*Sqrt[-1 - c^2*x^2])/(225*Sqrt[-(c^2*x^2)]) + (b*c*d*Sqrt[-1 - c^2*x^2])/(25*x^4*Sqr
t[-(c^2*x^2)]) - (b*c*(12*c^2*d - 25*e)*Sqrt[-1 - c^2*x^2])/(225*x^2*Sqrt[-(c^2*x^2)]) - (d*(a + b*ArcCsch[c*x
]))/(5*x^5) - (e*(a + b*ArcCsch[c*x]))/(3*x^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 ) \left (a+b \text{csch}^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{(b c x) \int \frac{-3 d-5 e x^2}{15 x^6 \sqrt{-1-c^2 x^2}} \, dx}{\sqrt{-c^2 x^2}}\\ &=-\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{(b c x) \int \frac{-3 d-5 e x^2}{x^6 \sqrt{-1-c^2 x^2}} \, dx}{15 \sqrt{-c^2 x^2}}\\ &=\frac{b c d \sqrt{-1-c^2 x^2}}{25 x^4 \sqrt{-c^2 x^2}}-\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}-\frac{\left (b c \left (12 c^2 d-25 e\right ) x\right ) \int \frac{1}{x^4 \sqrt{-1-c^2 x^2}} \, dx}{75 \sqrt{-c^2 x^2}}\\ &=\frac{b c d \sqrt{-1-c^2 x^2}}{25 x^4 \sqrt{-c^2 x^2}}-\frac{b c \left (12 c^2 d-25 e\right ) \sqrt{-1-c^2 x^2}}{225 x^2 \sqrt{-c^2 x^2}}-\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}+\frac{\left (2 b c^3 \left (12 c^2 d-25 e\right ) x\right ) \int \frac{1}{x^2 \sqrt{-1-c^2 x^2}} \, dx}{225 \sqrt{-c^2 x^2}}\\ &=\frac{2 b c^3 \left (12 c^2 d-25 e\right ) \sqrt{-1-c^2 x^2}}{225 \sqrt{-c^2 x^2}}+\frac{b c d \sqrt{-1-c^2 x^2}}{25 x^4 \sqrt{-c^2 x^2}}-\frac{b c \left (12 c^2 d-25 e\right ) \sqrt{-1-c^2 x^2}}{225 x^2 \sqrt{-c^2 x^2}}-\frac{d \left (a+b \text{csch}^{-1}(c x)\right )}{5 x^5}-\frac{e \left (a+b \text{csch}^{-1}(c x)\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.121658, size = 93, normalized size = 0.59 \[ \frac{-15 a \left (3 d+5 e x^2\right )+b c x \sqrt{\frac{1}{c^2 x^2}+1} \left (3 d \left (8 c^4 x^4-4 c^2 x^2+3\right )+25 e x^2 \left (1-2 c^2 x^2\right )\right )-15 b \text{csch}^{-1}(c x) \left (3 d+5 e x^2\right )}{225 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcCsch[c*x]))/x^6,x]

[Out]

(-15*a*(3*d + 5*e*x^2) + b*c*Sqrt[1 + 1/(c^2*x^2)]*x*(25*e*x^2*(1 - 2*c^2*x^2) + 3*d*(3 - 4*c^2*x^2 + 8*c^4*x^
4)) - 15*b*(3*d + 5*e*x^2)*ArcCsch[c*x])/(225*x^5)

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Maple [A]  time = 0.197, size = 140, normalized size = 0.9 \begin{align*}{c}^{5} \left ({\frac{a}{{c}^{2}} \left ( -{\frac{d}{5\,{c}^{3}{x}^{5}}}-{\frac{e}{3\,{c}^{3}{x}^{3}}} \right ) }+{\frac{b}{{c}^{2}} \left ( -{\frac{{\rm arccsch} \left (cx\right )d}{5\,{c}^{3}{x}^{5}}}-{\frac{{\rm arccsch} \left (cx\right )e}{3\,{c}^{3}{x}^{3}}}+{\frac{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( 24\,{x}^{4}{c}^{6}d-50\,{c}^{4}e{x}^{4}-12\,{c}^{4}d{x}^{2}+25\,{c}^{2}{x}^{2}e+9\,{c}^{2}d \right ) }{225\,{c}^{6}{x}^{6}}{\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)*(a+b*arccsch(c*x))/x^6,x)

[Out]

c^5*(a/c^2*(-1/5/c^3*d/x^5-1/3*e/c^3/x^3)+b/c^2*(-1/5*arccsch(c*x)/c^3*d/x^5-1/3*arccsch(c*x)*e/c^3/x^3+1/225*
(c^2*x^2+1)*(24*c^6*d*x^4-50*c^4*e*x^4-12*c^4*d*x^2+25*c^2*e*x^2+9*c^2*d)/((c^2*x^2+1)/c^2/x^2)^(1/2)/c^6/x^6)
)

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Maxima [A]  time = 1.00057, size = 178, normalized size = 1.13 \begin{align*} \frac{1}{75} \, b d{\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{arcsch}\left (c x\right )}{x^{5}}\right )} + \frac{1}{9} \, b e{\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{a e}{3 \, x^{3}} - \frac{a d}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccsch(c*x))/x^6,x, algorithm="maxima")

[Out]

1/75*b*d*((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*arccsch(c*x)/x^5) + 1/9*b*e*((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) - 1/3*a*e/x^3 - 1/5*a*d/x^5

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Fricas [A]  time = 2.60352, size = 294, normalized size = 1.86 \begin{align*} -\frac{75 \, a e x^{2} + 45 \, a d + 15 \,{\left (5 \, b e x^{2} + 3 \, b d\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (2 \,{\left (12 \, b c^{5} d - 25 \, b c^{3} e\right )} x^{5} + 9 \, b c d x -{\left (12 \, b c^{3} d - 25 \, b c e\right )} x^{3}\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{225 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccsch(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/225*(75*a*e*x^2 + 45*a*d + 15*(5*b*e*x^2 + 3*b*d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (2*(
12*b*c^5*d - 25*b*c^3*e)*x^5 + 9*b*c*d*x - (12*b*c^3*d - 25*b*c*e)*x^3)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/x^5

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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 )}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*acsch(c*x))/x**6,x)

[Out]

Integral((a + b*acsch(c*x))*(d + e*x**2)/x**6, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccsch(c*x))/x^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arccsch(c*x) + a)/x^6, x)