Integrand size = 12, antiderivative size = 58 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=-\frac {1}{3} b c^3 \sqrt {1+\frac {1}{c^2 x^2}}+\frac {1}{9} b c^3 \left (1+\frac {1}{c^2 x^2}\right )^{3/2}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3} \] Output:
-1/3*b*c^3*(1+1/c^2/x^2)^(1/2)+1/9*b*c^3*(1+1/c^2/x^2)^(3/2)-1/3*(a+b*arcc sch(c*x))/x^3
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=-\frac {a}{3 x^3}+b \left (-\frac {2 c^3}{9}+\frac {c}{9 x^2}\right ) \sqrt {\frac {1+c^2 x^2}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{3 x^3} \] Input:
Integrate[(a + b*ArcCsch[c*x])/x^4,x]
Output:
-1/3*a/x^3 + b*((-2*c^3)/9 + c/(9*x^2))*Sqrt[(1 + c^2*x^2)/(c^2*x^2)] - (b *ArcCsch[c*x])/(3*x^3)
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6838, 798, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx\) |
\(\Big \downarrow \) 6838 |
\(\displaystyle -\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5}dx}{3 c}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^2}d\frac {1}{x^2}}{6 c}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {b \int \left (c^2 \sqrt {1+\frac {1}{c^2 x^2}}-\frac {c^2}{\sqrt {1+\frac {1}{c^2 x^2}}}\right )d\frac {1}{x^2}}{6 c}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {2}{3} c^4 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}-2 c^4 \sqrt {\frac {1}{c^2 x^2}+1}\right )}{6 c}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}\) |
Input:
Int[(a + b*ArcCsch[c*x])/x^4,x]
Output:
(b*(-2*c^4*Sqrt[1 + 1/(c^2*x^2)] + (2*c^4*(1 + 1/(c^2*x^2))^(3/2))/3))/(6* c) - (a + b*ArcCsch[c*x])/(3*x^3)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Si mp[(d*x)^(m + 1)*((a + b*ArcCsch[c*x])/(d*(m + 1))), x] + Simp[b*(d/(c*(m + 1))) Int[(d*x)^(m - 1)/Sqrt[1 + 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.22
method | result | size |
parts | \(-\frac {a}{3 x^{3}}+b \,c^{3} \left (-\frac {\operatorname {arccsch}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\left (c^{2} x^{2}+1\right ) \left (2 c^{2} x^{2}-1\right )}{9 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\) | \(71\) |
derivativedivides | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\operatorname {arccsch}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\left (c^{2} x^{2}+1\right ) \left (2 c^{2} x^{2}-1\right )}{9 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right )\) | \(75\) |
default | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\operatorname {arccsch}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\left (c^{2} x^{2}+1\right ) \left (2 c^{2} x^{2}-1\right )}{9 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right )\) | \(75\) |
Input:
int((a+b*arccsch(c*x))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*a/x^3+b*c^3*(-1/3/c^3/x^3*arccsch(c*x)-1/9*(c^2*x^2+1)*(2*c^2*x^2-1)/ ((c^2*x^2+1)/c^2/x^2)^(1/2)/c^4/x^4)
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.33 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=-\frac {3 \, b \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (2 \, b c^{3} x^{3} - b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 3 \, a}{9 \, x^{3}} \] Input:
integrate((a+b*arccsch(c*x))/x^4,x, algorithm="fricas")
Output:
-1/9*(3*b*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (2*b*c^3*x^ 3 - b*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 3*a)/x^3
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=\int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{4}}\, dx \] Input:
integrate((a+b*acsch(c*x))/x**4,x)
Output:
Integral((a + b*acsch(c*x))/x**4, x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=\frac {1}{9} \, b {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {3 \, \operatorname {arcsch}\left (c x\right )}{x^{3}}\right )} - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*arccsch(c*x))/x^4,x, algorithm="maxima")
Output:
1/9*b*((c^4*(1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(1/(c^2*x^2) + 1))/c - 3*a rccsch(c*x)/x^3) - 1/3*a/x^3
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=\int { \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{4}} \,d x } \] Input:
integrate((a+b*arccsch(c*x))/x^4,x, algorithm="giac")
Output:
integrate((b*arccsch(c*x) + a)/x^4, x)
Timed out. \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x^4} \,d x \] Input:
int((a + b*asinh(1/(c*x)))/x^4,x)
Output:
int((a + b*asinh(1/(c*x)))/x^4, x)
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx=\frac {3 \left (\int \frac {\mathit {acsch} \left (c x \right )}{x^{4}}d x \right ) b \,x^{3}-a}{3 x^{3}} \] Input:
int((a+b*acsch(c*x))/x^4,x)
Output:
(3*int(acsch(c*x)/x**4,x)*b*x**3 - a)/(3*x**3)