Integrand size = 12, antiderivative size = 74 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{16 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{32 x}+\frac {3}{32} b c^4 \text {csch}^{-1}(c x)-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4} \] Output:
1/16*b*c*(1+1/c^2/x^2)^(1/2)/x^3-3/32*b*c^3*(1+1/c^2/x^2)^(1/2)/x+3/32*b*c ^4*arccsch(c*x)-1/4*(a+b*arccsch(c*x))/x^4
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=-\frac {a}{4 x^4}+b \left (\frac {c}{16 x^3}-\frac {3 c^3}{32 x}\right ) \sqrt {\frac {1+c^2 x^2}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{4 x^4}+\frac {3}{32} b c^4 \text {arcsinh}\left (\frac {1}{c x}\right ) \] Input:
Integrate[(a + b*ArcCsch[c*x])/x^5,x]
Output:
-1/4*a/x^4 + b*(c/(16*x^3) - (3*c^3)/(32*x))*Sqrt[(1 + c^2*x^2)/(c^2*x^2)] - (b*ArcCsch[c*x])/(4*x^4) + (3*b*c^4*ArcSinh[1/(c*x)])/32
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6838, 858, 262, 262, 222}
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.
\(\displaystyle \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx\) |
\(\Big \downarrow \) 6838 |
\(\displaystyle -\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^6}dx}{4 c}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^4}d\frac {1}{x}}{4 c}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {b \left (\frac {c^2 \sqrt {\frac {1}{c^2 x^2}+1}}{4 x^3}-\frac {3}{4} c^2 \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^2}d\frac {1}{x}\right )}{4 c}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {b \left (\frac {c^2 \sqrt {\frac {1}{c^2 x^2}+1}}{4 x^3}-\frac {3}{4} c^2 \left (\frac {c^2 \sqrt {\frac {1}{c^2 x^2}+1}}{2 x}-\frac {1}{2} c^2 \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}}}d\frac {1}{x}\right )\right )}{4 c}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {b \left (\frac {c^2 \sqrt {\frac {1}{c^2 x^2}+1}}{4 x^3}-\frac {3}{4} c^2 \left (\frac {c^2 \sqrt {\frac {1}{c^2 x^2}+1}}{2 x}-\frac {1}{2} c^3 \text {arcsinh}\left (\frac {1}{c x}\right )\right )\right )}{4 c}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\) |
Input:
Int[(a + b*ArcCsch[c*x])/x^5,x]
Output:
-1/4*(a + b*ArcCsch[c*x])/x^4 + (b*((c^2*Sqrt[1 + 1/(c^2*x^2)])/(4*x^3) - (3*c^2*((c^2*Sqrt[1 + 1/(c^2*x^2)])/(2*x) - (c^3*ArcSinh[1/(c*x)])/2))/4)) /(4*c)
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
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 = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57
method | result | size |
parts | \(-\frac {a}{4 x^{4}}+b \,c^{4} \left (-\frac {\operatorname {arccsch}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {c^{2} x^{2}+1}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{4} x^{4}-3 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {c^{2} x^{2}+1}\right )}{32 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\) | \(116\) |
derivativedivides | \(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\operatorname {arccsch}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {c^{2} x^{2}+1}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{4} x^{4}-3 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {c^{2} x^{2}+1}\right )}{32 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\right )\) | \(120\) |
default | \(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\operatorname {arccsch}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {c^{2} x^{2}+1}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{4} x^{4}-3 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {c^{2} x^{2}+1}\right )}{32 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\right )\) | \(120\) |
Input:
int((a+b*arccsch(c*x))/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*a/x^4+b*c^4*(-1/4/c^4/x^4*arccsch(c*x)+1/32*(c^2*x^2+1)^(1/2)*(3*arct anh(1/(c^2*x^2+1)^(1/2))*c^4*x^4-3*(c^2*x^2+1)^(1/2)*c^2*x^2+2*(c^2*x^2+1) ^(1/2))/((c^2*x^2+1)/c^2/x^2)^(1/2)/c^5/x^5)
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\frac {{\left (3 \, b c^{4} x^{4} - 8 \, b\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (3 \, b c^{3} x^{3} - 2 \, b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - 8 \, a}{32 \, x^{4}} \] Input:
integrate((a+b*arccsch(c*x))/x^5,x, algorithm="fricas")
Output:
1/32*((3*b*c^4*x^4 - 8*b)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x )) - (3*b*c^3*x^3 - 2*b*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 8*a)/x^4
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{5}}\, dx \] Input:
integrate((a+b*acsch(c*x))/x**5,x)
Output:
Integral((a + b*acsch(c*x))/x**5, x)
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (62) = 124\).
Time = 0.03 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.99 \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\frac {1}{64} \, b {\left (\frac {3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right ) - 3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (3 \, c^{8} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 5 \, c^{6} x \sqrt {\frac {1}{c^{2} x^{2}} + 1}\right )}}{c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{2} - 2 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} + 1}}{c} - \frac {16 \, \operatorname {arcsch}\left (c x\right )}{x^{4}}\right )} - \frac {a}{4 \, x^{4}} \] Input:
integrate((a+b*arccsch(c*x))/x^5,x, algorithm="maxima")
Output:
1/64*b*((3*c^5*log(c*x*sqrt(1/(c^2*x^2) + 1) + 1) - 3*c^5*log(c*x*sqrt(1/( c^2*x^2) + 1) - 1) - 2*(3*c^8*x^3*(1/(c^2*x^2) + 1)^(3/2) - 5*c^6*x*sqrt(1 /(c^2*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) + 1)^2 - 2*c^2*x^2*(1/(c^2*x^2) + 1 ) + 1))/c - 16*arccsch(c*x)/x^4) - 1/4*a/x^4
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\int { \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{5}} \,d x } \] Input:
integrate((a+b*arccsch(c*x))/x^5,x, algorithm="giac")
Output:
integrate((b*arccsch(c*x) + a)/x^5, x)
Timed out. \[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x^5} \,d x \] Input:
int((a + b*asinh(1/(c*x)))/x^5,x)
Output:
int((a + b*asinh(1/(c*x)))/x^5, x)
\[ \int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx=\frac {4 \left (\int \frac {\mathit {acsch} \left (c x \right )}{x^{5}}d x \right ) b \,x^{4}-a}{4 x^{4}} \] Input:
int((a+b*acsch(c*x))/x^5,x)
Output:
(4*int(acsch(c*x)/x**5,x)*b*x**4 - a)/(4*x**4)