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

Optimal. Leaf size=109 \[ -\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {csch}^{-1}(c x)\right )}{x}-\frac {b c \sqrt {-c^2 x^2-1} \left (2 c^2 d-9 e\right )}{9 \sqrt {-c^2 x^2}}+\frac {b c d \sqrt {-c^2 x^2-1}}{9 x^2 \sqrt {-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 6302, 12, 453, 264} \[ -\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {csch}^{-1}(c x)\right )}{x}-\frac {b c \sqrt {-c^2 x^2-1} \left (2 c^2 d-9 e\right )}{9 \sqrt {-c^2 x^2}}+\frac {b c d \sqrt {-c^2 x^2-1}}{9 x^2 \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 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 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 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]))

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 68, normalized size = 0.62 \[ \frac {-3 a \left (d+3 e x^2\right )+b c x \sqrt {\frac {1}{c^2 x^2}+1} \left (-2 c^2 d x^2+d+9 e x^2\right )-3 b \text {csch}^{-1}(c x) \left (d+3 e x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 105, normalized size = 0.96 \[ -\frac {9 \, a e x^{2} + 3 \, a d + 3 \, {\left (3 \, b e x^{2} + b d\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c d x - {\left (2 \, b c^{3} d - 9 \, b c e\right )} x^{3}\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{9 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 122, normalized size = 1.12 \[ c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\mathrm {arccsch}\left (c x \right ) e}{c x}-\frac {\mathrm {arccsch}\left (c x \right ) d}{3 c \,x^{3}}-\frac {\left (c^{2} x^{2}+1\right ) \left (2 c^{4} d \,x^{2}-9 c^{2} x^{2} e -c^{2} d \right )}{9 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 91, normalized size = 0.83 \[ {\left (c \sqrt {\frac {1}{c^{2} x^{2}} + 1} - \frac {\operatorname {arcsch}\left (c x\right )}{x}\right )} b e + \frac {1}{9} \, b d {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {3 \, \operatorname {arcsch}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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